# Solved Examples on Linear Functions

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For JAMB, CMAT, and NZQA Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions.
Use at least two methods whenever applicable.
Show all work.

(1.) Determine the slope of the line through the points:
$(-1, -10)$ and $(1, 8)$

Point $1$ is $(-1, -10)$
$x_1 = -1$
$y_1 = -10$

Point $2$ is $(1, 8)$
$x_2 = 1$
$y_2 = 8$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{8 - (-10)}{1 - (-1)} \\[5ex] m = \dfrac{8 + 10}{1 + 1} \\[5ex] m = \dfrac{18}{2} \\[5ex] m = 9$
(2.) Determine the slope of the line through the points:
$(-12, -18)$ and $(-19, -20)$

Point $1$ is $(-12, -18)$
$x_1 = -12$
$y_1 = -18$

Point $2$ is $(-19, -20)$
$x_2 = -19$
$y_2 = -20$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-20 - (-18)}{-19 - (-12)} \\[5ex] m = \dfrac{-20 + 18}{-19 + 12} \\[5ex] m = \dfrac{-2}{-7} \\[5ex] m = \dfrac{2}{7}$
(3.) Determine the slope of the line through the points:
$(3, -6)$ and $(-7, -6)$

Point $1$ is $(3, -6)$
$x_1 = 3$
$y_1 = -6$

Point $2$ is $(-7, -6)$
$x_2 = -7$
$y_2 = -6$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-6 - (-6)}{-7 - 3} \\[5ex] m = \dfrac{-6 + 6}{-10} \\[5ex] m = \dfrac{0}{-10} \\[5ex] m = 0$
(4.) Determine the slope of the line through the points:
$(3, -6)$ and $(3, -9)$

Point $1$ is $(3, -6)$
$x_1 = 3$
$y_1 = -6$

Point $2$ is $(3, -9)$
$x_2 = 3$
$y_2 = -9$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-9 - (-6)}{3 - 3} \\[5ex] m = \dfrac{-9 + 6}{0} \\[5ex] m = \dfrac{-3}{0} \\[5ex] m\:\:\:is\:\:\: undefined$
(5.) Determine the slope of the line through the points:
$\left(\dfrac{7}{8}, -\dfrac{1}{2}\right)$ and $\left(\dfrac{1}{4}, \dfrac{1}{4}\right)$

Point $1$ is $\left(\dfrac{7}{8}, -\dfrac{1}{2}\right)$

$x_1 = \dfrac{7}{8}$

$y_1 = -\dfrac{1}{2}$

Point $2$ is $\left(\dfrac{1}{4}, \dfrac{1}{4}\right)$

$x_2 = \dfrac{1}{4}$

$y_2 = \dfrac{1}{4}$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[7ex] m = \dfrac{\dfrac{1}{4} - \left(-\dfrac{1}{2}\right)}{\dfrac{1}{4} - \dfrac{7}{8}} \\[7ex] m = \dfrac{\dfrac{1}{4} + \dfrac{1}{2}}{\dfrac{2}{8} - \dfrac{7}{8}} \\[7ex] m = \dfrac{\dfrac{1}{4} + \dfrac{2}{4}}{\dfrac{2 - 7}{8}} \\[7ex] m = \dfrac{\dfrac{1 + 2}{4}}{\dfrac{-5}{8}} \\[7ex] m = \dfrac{\dfrac{3}{4}}{-\dfrac{5}{8}} \\[7ex] m = \dfrac{3}{4} \div -\dfrac{5}{8} \\[7ex] m = \dfrac{3}{4} * -\dfrac{8}{5} \\[7ex] m = \dfrac{3 * -2}{5} \\[7ex] m = -\dfrac{6}{5}$
(6.) Determine the slope of the line through the points:
$(2, 6)$ and $(10, 1)$

Point $1$ is $(2, 6)$
$x_1 = 2$
$y_1 = 6$

Point $2$ is $(10, 1)$
$x_2 = 10$
$y_2 = 1$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{1 - 6}{10 - 2} \\[5ex] m = \dfrac{-5}{8} \\[5ex] m = -\dfrac{5}{8}$
(7.) Determine the slope and intercepts of the line:
$7x + 8y = 28$

First: Determine the slope and y-intercept.
Write it in the slope-intercept form

$7x + 8y = 28 \\[2ex] 8y = 28 - 7x \\[2ex] 8y = -7x + 28 \\[2ex] y = -\dfrac{7}{8}x + \dfrac{28}{8} \\[5ex] y = -\dfrac{7}{8}x + \dfrac{7}{2} \\[5ex]$ Compare to $y = mx + b$

Slope, $m = -\dfrac{7}{8}$

$b = \dfrac{7}{2}$

y-intercept = $\left(0, \dfrac{7}{2}\right)$

Second: Determine the x-intercept.
Set $y = 0$ and solve for $x$

$7x + 8y = 28 \\[2ex] 7x + 8(0) = 28 \\[2ex] 7x + 0 = 28 \\[2ex] 7x = 28 \\[2ex] x = \dfrac{28}{7} \\[5ex] x = 4 \\[3ex]$ x-intercept = $(4, 0)$
(8.) Determine the zero(s) of the function:
$f(x) = -\dfrac{1}{5}x - \dfrac{7}{10}$

The zeros of the function means what values of $x$ would make the function, $y$ to be zero.
Recall: $y = f(x)$
What value(s) of the input would make the output to be zero?
This implies that we should set $f(x)$ to be $0$

$f(x) = -\dfrac{1}{5}x - \dfrac{7}{10} \\[5ex] 0 = -\dfrac{1}{5}x - \dfrac{7}{10} \\[5ex] -\dfrac{1}{5}x - \dfrac{7}{10} = 0 \\[5ex] -\dfrac{1}{5}x = \dfrac{7}{10} \\[5ex]$ To isolate $x$, multiply both sides by $-5$

$(-5) * -\dfrac{1}{5}x = (-5) * \dfrac{7}{10} \\[5ex] x = -\dfrac{7}{2} \\[5ex]$ This is a linear function.
It has only one zero.
The zero of the function is: $x = -\dfrac{7}{2}$
(9.) Determine the slope and intercepts of the line:
$3x - 2y = 19$

First: Determine the slope and y-intercept.
Write it in the slope-intercept form

$3x - 2y = 19 \\[2ex] 3x - 19 = 2y \\[2ex] 2y = 3x - 19 \\[2ex] y = \dfrac{3}{2}x - \dfrac{19}{2}$ Compare to $y = mx + b$

Slope, $m = \dfrac{3}{2}$

$b = -\dfrac{19}{2}$

y-intercept = $\left(0, -\dfrac{19}{2}\right)$

Second: Determine the x-intercept.
Set $y = 0$ and solve for $x$

$3x - 2y = 19 \\[2ex] 3x - 2(0) = 19 \\[2ex] 3x - 0 = 19 \\[2ex] 3x = 19 \\[2ex] x = \dfrac{19}{3} \\[5ex]$ x-intercept = $\left(\dfrac{19}{3}, 0\right)$
(10.) Determine the slope of the line through the points:
$(3a, 9a^2)$ and $(3a + 3h, (3a + 3h)^2)$

Point $1$ is $(3a, 9a^2)$
$x_1 = 3a$
$y_1 = 9a^2$

Point $2$ is $(3a + 3h, (3a + 3h)^2)$
$x_2 = 3a + 3h$
$y_2 = (3a + 3h)^2$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[7ex] m = \dfrac{(3a + 3h)^2 - 9a^2}{(3a + 3h) - 3a} \\[7ex] m = \dfrac{(3a + 3h)(3a + 3h) - 9a^2}{3a + 3h - 3a} \\[7ex] m = \dfrac{9a^2 + 9ah + 9ah + 9h^2 - 9a^2}{3h} \\[7ex] m = \dfrac{18ah + 9h^2}{3h} \\[7ex] m = \dfrac{9h(2a + h)}{3h} \\[5ex] m = 3(2a + h)$
(11.) Determine the slope of the line through the points:
$f\left(\dfrac{1}{8}\right) = -1$ and $f\left(-\dfrac{1}{4}\right) = -\dfrac{1}{4}$

$f(x) = y$
This means that:
Point $1$ is $\left(\dfrac{1}{8}\right), -1$

$x_1 = \dfrac{1}{8}$

$y_1 = -1$

Point $2$ is $\left(-\dfrac{1}{4}, -\dfrac{1}{4}\right)$

$x_2 = -\dfrac{1}{4}$

$y_2 = -\dfrac{1}{4}$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[7ex] m = \dfrac{-\dfrac{1}{4} - (-1)}{-\dfrac{1}{4} - \dfrac{1}{8}} \\[7ex] m = \dfrac{-\dfrac{1}{4} + 1}{-\dfrac{2}{8} - \dfrac{1}{8}} \\[7ex] m = \dfrac{-\dfrac{1}{4} + \dfrac{4}{4}}{\dfrac{-2 - 1}{8}} \\[7ex] m = \dfrac{\dfrac{-1 + 4}{4}}{\dfrac{-3}{8}} \\[7ex] m = \dfrac{\dfrac{3}{4}}{-\dfrac{3}{8}} \\[7ex] m = \dfrac{3}{4} \div -\dfrac{3}{8} \\[7ex] m = \dfrac{3}{4} * \dfrac{8}{-3} \\[7ex] m = \dfrac{8}{-4} \\[7ex] m = -2$
(12.) Write the slope-intercept form of an equation whose:
$slope = -\dfrac{15}{11}$ and $y-intercept = (0, -8)$

Slope-Intercept Form: $y = mx + b$

where: $m = slope$ and $b = y-intercept$

$y = -\dfrac{15}{11}x + (-8) \\[5ex] y = -\dfrac{15}{11}x - 8$
(13.) ACT In the standard $(x, y)$ coordinate plane,
what is the slope of the line $4x + 7y = 9$?

$A.\:\: -\dfrac{4}{7} \\[5ex] B.\:\: \dfrac{4}{9} \\[5ex] C.\:\: -4 \\[3ex] D.\:\: 4 \\[3ex] E.\:\: 9 \\[3ex]$

Let us write it in the slope-intercept form: $y = mx + b$

$4x + 7y = 9 \\[3ex] 7y = 9 - 4x \\[3ex] 7y = -4x + 9 \\[3ex] y = \dfrac{-4x + 9}{7} \\[5ex] y = -\dfrac{4x}{7} + \dfrac{9}{7} \\[5ex]$ The slope is $-\dfrac{4}{7}$
(14.) ACT When graphed in the (x, y) coordinate plane, what is the slope of the line $\dfrac{y}{2} = x$?

$A.\:\: -2 \\[3ex] B.\:\: -\dfrac{1}{2} \\[5ex] C.\:\: \dfrac{1}{2} \\[5ex] D.\:\: 1 \\[3ex] E.\:\: 2 \\[3ex]$

$\dfrac{y}{2} = x \\[5ex] y = 2 * x \\[3ex] y = 2x \\[3ex] Compare:\:\: y = mx + b \\[3ex] m = 2$
(15.) ACT In the $(x, y)$ coordinate plane, what is the $y-intercept$ of the line $6x - 2y = 6$?

$F.\:\: -3 \\[3ex] G.\:\: -2 \\[3ex] H.\:\: 0 \\[3ex] J.\:\: 3 \\[3ex] K.\:\: 6 \\[3ex]$

We can solve this question in two ways.
Use whichever method is convenient and fast for you.

$\underline{First\:\:Method:\:\:Definition} \\[3ex] 6x - 2y = 6 \\[3ex] Set\:\:x = 0 \\[3ex] AND \\[3ex] Solve\:\:for\:\:y \\[3ex] 6(0) - 2y = 6 \\[3ex] 0 - 2y = 6 \\[3ex] -2y = 6 \\[3ex] y = \dfrac{6}{-2} \\[5ex] y = -3 \\[3ex] y-intercept = (0, -3) \\[3ex] \underline{Second\:\:Method:\:\:Slope-Intercept\:\:Form} \\[3ex] y = mx + b \\[3ex] 6x - 2y = 6 \\[3ex] 6x - 6 = 2y \\[3ex] 2y = 6x - 6 \\[3ex] y = \dfrac{6x - 6}{2} \\[5ex] y = \dfrac{6x}{2} - \dfrac{6}{2} \\[5ex] y = 3x - 3 \\[3ex] b = -3 \\[3ex] y-intercept = (0, -3)$
(16.) ACT In the standard $(x, y)$ coordinate plane, if the $x-coordinate$ of each point on a line is $4$ less than twice its $y-coordinate$, the slope of the line is:

$F.\:\: -4 \\[3ex] G.\:\: -2 \\[3ex] H.\:\: \dfrac{1}{2} \\[5ex] J.\:\: 2 \\[3ex] K.\:\: 4 \\[3ex]$

$x = 2y - 4 \\[3ex] Put\:\:in\:\:Slope-Intercept\:\:Form \\[3ex] y = mx + b \\[3ex] 2y - 4 = x \\[3ex] 2y = x + 4 \\[3ex] y = \dfrac{x + 4}{2} \\[5ex] y = \dfrac{x}{2} + \dfrac{4}{2} \\[5ex] y = \dfrac{1}{2}x + 2 \\[5ex] m = \dfrac{1}{2}$
(17.) Determine the equation of a line that is parallel to $y = \dfrac{3}{4}x + 1$
and passes through $(4, 3)$.

Two lines are parallel to each other if their slopes are the same

$\underline{Line\:\: 1}$
$y = \dfrac{3}{4}x + 1$
Slope of Line $1$ = $m_1$
$m_1 = \dfrac{3}{4}$

$\underline{Line\:\: 2}$
$Line\:\: 2 \parallel Line\:\: 1$
Slope of Line $2$ = $m_2$

$m_2 = \dfrac{3}{4} \\[5ex]$ ${Line\:\: 2}$ should pass through the point $(4, 3)$
To find ${Line\:\: 2}$, we shall use the Point-Slope form because we have a slope and a point
Point-Slope Form: $y - y_1 = m(x - x_1)$

$m_2 = \dfrac{3}{4} \\[5ex] x_1 = 4, y_1 = 3 \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 3 = \dfrac{3}{4}(x - 4) \\[5ex] y = \dfrac{3}{4}(x - 4) + 3 \\[5ex] y = \dfrac{3}{4}x - 3 + 3 \\[5ex] y = \dfrac{3}{4}x \\[5ex]$ The equation of ${Line\:\: 2}$ in Slope-Intercept form is: $y = \dfrac{3}{4}x$

To write ${Line\:\: 2}$ in Standard form;
Multiply each term by the $LCD$
$LCD = 4$

$4 * y = 4 * \dfrac{3}{4}x \\[5ex] 4y = 3x \\[3ex] 3x = 4y \\[3ex] 3x - 4y = 0 \\[3ex]$ The equation of ${Line\:\: 2}$ in Standard form is: $3x - 4y = 0$

(18.) Determine the value of $p$ so that the line containing the points $(-3, p)$ and $(4, 8)$
is parallel to the line containing the points $(7, 3)$ and $(3, -6)$

Two lines are parallel to each other if their slopes are the same

$\underline{Line\:\: 1}$
Slope of Line $1$ = $m_1$
Point $1$ is $(-3, p)$
$x_1 = -3$
$y_1 = p$

Point $2$ is $(4, 8)$
$x_2 = 4$
$y_2 = 8$

$m_1 = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m_1 = \dfrac{8 - p}{4 - (-3)} \\[5ex] m_1 = \dfrac{8 - p}{4 + 3} \\[5ex] m_1 = \dfrac{8 - p}{7} \\[5ex]$ $\underline{Line\:\: 2}$
Slope of Line $1$ = $m_2$
Point $1$ is $(7, 3)$
$x_1 = 7$
$y_1 = 3$

Point $2$ is $(3, -6)$
$x_2 = 3$
$y_2 = -6$

$m_2 = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m_2 = \dfrac{-6 - 3}{3 - 7} \\[5ex] m_2 = \dfrac{-9}{-4} \\[5ex] m_2 = \dfrac{9}{4} \\[5ex]$ $Line\:\: 1 \parallel Line\:\: 2$
$m_1 = m_2 \\[3ex] \dfrac{8 - p}{7} = \dfrac{9}{4} \\[5ex]$ Multiply each term by the $LCD$
$LCD = 28$

$28 * \dfrac{8 - p}{7} = 28 * \dfrac{9}{4} \\[5ex] 4(8 - p) = 7(9) \\[3ex] 4(8) - 4(p) = 63 \\[3ex] 32 - 4p = 63 \\[3ex] 32 - 63 = 4p \\[3ex] -31 = 4p \\[3ex] 4p = -31 \\[3ex] p = -\dfrac{31}{4}$
(19.) CSEC (a) The equation of a straight line, $l$, is given as $3x - 4y = 5$.

(i) Write the equation of the line, $l$, in the form $y = mx + c$

(ii) Hence, determine the gradient of the line, $l$

(iii) The point, $P$ with coordinates $(r, 2)$ lies on the line $l$.
Determine the value of $r$

(iv) Find the equation of the straight line passing through the point $(6, 0)$ which is perpendicular to $l$.

$(i) \\[3ex] 3x - 4y = 5 \\[3ex] 3x - 5 = 4y \\[3ex] 4y = 3x - 5 \\[3ex] y = \dfrac{3x - 5}{4} \\[5ex] y = \dfrac{3}{4}x - \dfrac{5}{4} \\[5ex] (ii) \\[3ex] Gradient, m = \dfrac{3}{4} \\[5ex] (iii) \\[3ex] y = \dfrac{3}{4}x - \dfrac{5}{4} \\[5ex] For\:\:(r, 2) \\[3ex] x = r \\[3ex] y = 2 \\[3ex] \rightarrow 2 = \dfrac{3}{4}r - \dfrac{5}{4} \\[5ex] LCD = 4 \\[3ex] 4(2) = 4\left(\dfrac{3}{4}r\right) - 4\left(\dfrac{5}{4}\right) \\[5ex] 8 = 3r - 5 \\[3ex] 3r - 5 = 8 \\[3ex] 3r = 8 + 5 \\[3ex] 3r = 13 \\[3ex] r = \dfrac{13}{3} \\[5ex]$ Two lines are perpendicular to each other if the product of their slopes is $-1$

$\underline{Line\:\:1} \\[3ex] m_1 = \dfrac{3}{4} \\[5ex] \underline{Line\:\: 2} \\[3ex] Line\:\: 2 \perp Line\:\: 1 \\[3ex] Slope\:\:of\:\:Line\:\:2 = m_2 \\[3ex] m_2 * m_1 = -1 \\[3ex] m_2 = -1 \div m_1 \\[3ex] m_2 = -1 \div \dfrac{3}{4} \\[5ex] m_2 = -1 * \dfrac{4}{3} \\[5ex] m_2 = -\dfrac{4}{3} \\[5ex]$ ${Line\:\: 2}$ should pass through the point $(6, 0)$
To find ${Line\:\: 2}$, we shall use the Point-Slope form because we have a slope and a point
Point-Slope Form: $y - y_1 = m(x - x_1)$

$m_2 = -\dfrac{4}{3} \\[5ex] x_1 = 6, y_1 = 0 \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 0 = -\dfrac{4}{3}(x - 6) \\[5ex] y = -\dfrac{4}{3}x + \dfrac{4}{3}(6) \\[5ex] y = -\dfrac{4}{3}x + 4(2) \\[5ex] y = -\dfrac{4}{3}x + 8 \\[5ex]$ The equation of ${Line\:\: 2}$ in Slope-Intercept form is: $y = -\dfrac{4}{3}x + 8$

To write ${Line\:\: 2}$ in Standard form;
Multiply each term by the $LCD$
$LCD = 3$

$3 * y = 3 * -\dfrac{4}{3}x + 3 * 8 \\[5ex] 3y = -4x + 24 \\[3ex] 3y + 4x = 24 \\[3ex] 4x + 3y = 24 \\[3ex]$ The equation of ${Line\:\: 2}$ in Standard form is: $4x + 3y = 24$

(20.) ACT In the standard $(x, y)$ coordinate plane, what is the slope of the line that passes through the origin and the point $\left(\dfrac{1}{2}, \dfrac{2}{3}\right)$?

$F.\:\: \dfrac{1}{2} \\[5ex] G.\:\: \dfrac{1}{3} \\[5ex] H.\:\: \dfrac{2}{3} \\[5ex] J.\:\: \dfrac{3}{4} \\[5ex] K.\:\: \dfrac{4}{3} \\[5ex]$

Origin means $(0, 0)$

$Point\:1:(0, 0) \\[3ex] x_1 = 0 \\[3ex] y_1 = 0 \\[3ex] Point\:2:\left(\dfrac{1}{2}, \dfrac{2}{3}\right) \\[5ex] x_2 = \dfrac{1}{2} \\[5ex] y_2 = \dfrac{2}{3} \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] y_2 - y_1 = \dfrac{2}{3} - 0 = \dfrac{2}{3} \\[5ex] x_2 - x_1 = \dfrac{1}{2} - 0 = \dfrac{1}{2} \\[5ex] m = \dfrac{2}{3} \div \dfrac{1}{2} \\[5ex] m = \dfrac{2}{3} * \dfrac{2}{1} \\[5ex] m = \dfrac{4}{3}$

(21.) ACT What is the slope-intercept form of $4x - y - 3 = 0$?

$A.\:\: y = -4x - 3 \\[3ex] B.\:\: y = -4x + 3 \\[3ex] C.\:\: y = 3x - 4 \\[3ex] D.\:\: y = 4x - 3 \\[3ex] E.\:\: y = 4x + 3 \\[3ex]$

$4x - y - 3 = 0 \\[3ex] 4x - 3 = y \\[3ex] y = 4x - 3$
(22.) ACT What is the slope-intercept form of the equation $6x - 3y = 7$?

$A.\:\: y = 2x - 7 \\[3ex] B.\:\: y = 2x - \dfrac{7}{3} \\[5ex] C.\:\: y = 2x - \dfrac{3}{7} \\[5ex] D.\:\: y = 2x + \dfrac{7}{3} \\[5ex] E.\:\: y = 2x + 7 \\[3ex]$

$6x - 3y = 7 \\[3ex] 6x - 7 = 3y \\[3ex] 3y = 6x - 7 \\[3ex] y = \dfrac{6x - 7}{3} \\[5ex] y = \dfrac{6}{3}x - \dfrac{7}{3} \\[5ex] y = 2x - \dfrac{7}{3}$
(23.) ACT One of the following equations represents the line graphed in the standard $(x, y)$ coordinate plane below.
Which one?

$A.\:\: y = -2x + 2 \\[3ex] B.\:\: y = -2x + 4 \\[3ex] C.\:\: y = 2x + 4 \\[3ex] D.\:\: y = 4x - 2 \\[3ex] E.\:\: y = 4x + 2 \\[3ex]$

$\underline{First\:\:Method - Faster} \\[3ex] Recommended\:\:for\:\:ACT\:\:purposes \\[3ex] Slope, m = \dfrac{fall}{run}...Left\:\:to\:\:Right \\[5ex] fall = -4\:\:units \\[3ex] run = 2\:\:units \\[3ex] m = \dfrac{-4}{2} \\[5ex] m = -2 \\[3ex] y-intercept, b = 4 \\[3ex] y = mx + b \\[3ex] y = -2x + 4 \\[3ex] \underline{Second\:\:Method} \\[3ex] Point\:1 = (0, 4) \\[3ex] x_1 = 0 \\[3ex] y_1 = 4 \\[3ex] y-intercept, b = 4 \\[3ex] Point\:2 = (2, 0) \\[3ex] x_2 = 2 \\[3ex] y_2 = 0 \\[3ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{0 - 4}{2 - 0} \\[5ex] m = \dfrac{-4}{2} \\[5ex] m = -2 \\[3ex] y = mx + b \\[3ex] y = -2x + 4$
(24.) ACT In the standard $(x, y)$ coordinate plane, what is the slope of the line given by the equation $5x = 9y + 18?$

$F.\:\: -\dfrac{5}{9} \\[5ex] G.\:\: \dfrac{5}{9} \\[5ex] H.\:\: \dfrac{9}{5} \\[5ex] J.\:\: 5 \\[3ex] K.\:\: 9 \\[3ex]$

We need to write the equation in the Slope-Intercept form

$5x = 9y + 18 \\[3ex] 9y + 18 = 5x \\[3ex] 9y = 5x - 18 \\[3ex] y = \dfrac{5x - 18}{9} \\[5ex] y = \dfrac{5x}{9} - \dfrac{18}{9} \\[5ex] y = \dfrac{5}{9}x - 2 \\[5ex] Slope = \dfrac{5}{9}$
(25.) Determine the equation of a line that is perpendicular to $y = \dfrac{3}{4}x$

and passes through $(4, 3)$.

Two lines are perpendicular to each other if the product of their slopes is $-1$

$\underline{Line\:\: 1}$
$y = \dfrac{3}{4}x$
Slope of Line $1$ = $m_1$
$m_1 = \dfrac{3}{4}$

$\underline{Line\:\: 2}$
$Line\:\: 2 \perp Line\:\: 1$
Slope of Line $2$ = $m_2$

$m_2 * m_1 = -1 \\[3ex] m_2 = -1 \div m_1 \\[3ex] m_2 = -1 \div \dfrac{3}{4} \\[5ex] m_2 = -1 * \dfrac{4}{3} \\[5ex] m_2 = -\dfrac{4}{3} \\[5ex]$ ${Line\:\: 2}$ should pass through the point $(4, 3)$
To find ${Line\:\: 2}$, we shall use the Point-Slope form because we have a slope and a point
Point-Slope Form: $y - y_1 = m(x - x_1)$

$m_2 = -\dfrac{4}{3} \\[5ex] x_1 = 4, y_1 = 3 \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 3 = -\dfrac{4}{3}(x - 4) \\[5ex] y = -\dfrac{4}{3}(x - 4) + 3 \\[5ex] y = -\dfrac{4}{3}x + \dfrac{16}{3} + 3 \\[5ex] y = -\dfrac{4}{3}x + \dfrac{16}{3} + \dfrac{9}{3} \\[5ex] y = -\dfrac{4}{3}x + \dfrac{16 + 9}{3} \\[5ex] y = -\dfrac{4}{3}x + \dfrac{25}{3} \\[5ex]$ The equation of ${Line\:\: 2}$ in Slope-Intercept form is: $y = -\dfrac{4}{3}x + \dfrac{25}{3}$

To write ${Line\:\: 2}$ in Standard form;
Multiply each term by the $LCD$
$LCD = 3$

$3 * y = 3 * -\dfrac{4}{3}x + 3 * \dfrac{25}{3} \\[5ex] 3y = -4x + 25 \\[3ex] 3y + 4x = 25 \\[3ex] 4x + 3y = 25 \\[3ex]$ The equation of ${Line\:\: 2}$ in Standard form is: $4x + 3y = 25$
(26.) ACT For all $a \ne 0$, what is the slope of the line segment connecting $(a, b)$ and $(-a, b)$ in the usual $(x, y)$ co-ordinate plane?

$F.\:\: 0 \\[3ex] G.\:\: \dfrac{a}{b} \\[5ex] H.\:\: -\dfrac{b}{a} \\[5ex] J.\:\: \dfrac{b}{a} \\[5ex] K.\:\: 2a \\[3ex]$

$Point\:1: (a, b) \\[3ex] x_1 = a \\[3ex] y_1 = b \\[3ex] Point\:2: (-a, b) \\[3ex] x_2 = -a \\[3ex] y_2 = b \\[3ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{b - b}{-a - a} \\[5ex] m = \dfrac{0}{-2a} \\[5ex] m = 0$
(27.) JAMB Find the equation of the line through $(5, 7)$ parallel to the line $7x + 5y = 12$

$A.\:\: 5x + 7y = 20 \\[3ex] B.\:\: 7x + 5y = 70 \\[3ex] C.\:\: xy = 7 \\[3ex] D.\:\: 15x + 17y = 90 \\[3ex]$

Two lines are parallel to each other if their slopes are the same

$\underline{Line\:\: 1} \\[3ex] 7x + 5y = 12 \\[3ex] 5y = 12 - 7x \\[3ex] 5y = -7x + 12 \\[3ex] y = \dfrac{-7x + 12}{5} \\[5ex] y = -\dfrac{7}{5}x + \dfrac{12}{5} \\[5ex] Slope\:\:of\:\:Line\:\:1 = m_1 \\[3ex] m_1 = -\dfrac{7}{5} \\[5ex] \underline{Line\:\: 2} \\[3ex] Line\:\: 2 \parallel Line\:\: 1 \\[3ex] Slope\:\:of\:\:Line\:\:2 = m_2 \\[3ex] m_2 = -\dfrac{7}{5} \\[5ex]$ ${Line\:\: 2}$ should pass through the point $(5, 7)$

To find ${Line\:\: 2}$, we shall use the Point-Slope form because we have a slope and a point

$\underline{Point-Slope\:\:Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] m_2 = -\dfrac{7}{5} \\[5ex] x_1 = 5, y_1 = 7 \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 7 = -\dfrac{7}{5}(x - 5) \\[5ex] y = -\dfrac{7}{5}(x - 5) + 7 \\[5ex] y = -\dfrac{7}{5}x + 7 + 7 \\[5ex] y = -\dfrac{7}{5}x + 14 \\[5ex]$ The equation of ${Line\:\: 2}$ in Slope-Intercept form is: $y = -\dfrac{7}{5}x + 14$

To write ${Line\:\: 2}$ in Standard form;

Multiply each term by the $LCD$

$LCD = 5 \\[3ex] 5 * y = 5\left(-\dfrac{7}{5}x + 14\right) \\[5ex] 5y = -7x + 70 \\[3ex] 5y + 7x = 70 \\[3ex] 7x + 5y = 70 \\[3ex]$ The equation of ${Line\:\: 2}$ in Standard form is: $7x + 5y = 70$
(28.) ACT In the standard $(x, y)$ coordinate plane below, $\triangle AOB$ is formed by $\overleftrightarrow{AB}$, the $x-axis$, and the $y-axis$
Which of the following is an equation of $\overleftrightarrow{AB}$
$A.\:\: y = -x + 6 \\[3ex] B.\:\: y = x - 6 \\[3ex] C.\:\: y = x + 6 \\[3ex] D.\:\: y = -6x - 6 \\[3ex] E.\:\: y = 6x + 6 \\[3ex]$

$Point\:A = Point\:\:1 = (0, 6) \\[3ex] x_1 = 0 \\[3ex] y_1 = 6 \\[3ex] y-intercept, b = 6 \\[3ex] Point\:B = Point\:2 = (6, 0) \\[3ex] x_2 = 6 \\[3ex] y_2 = 0 \\[3ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{0 - 6}{6 - 0} \\[5ex] m = \dfrac{-6}{6} \\[5ex] m = -1 \\[3ex] y = mx + b \\[3ex] y = -x + 6$
(29.) ACT In the standard (x, y) coordinate plane, a line intersects the y-axis at (0, 2) and contains the point (8, 3).
What is the slope of the line?

$A.\:\: \dfrac{1}{8} \\[5ex] B.\:\: \dfrac{2}{5} \\[5ex] C.\:\: \dfrac{1}{2} \\[5ex] D.\:\: 2 \\[3ex] E.\:\: 8 \\[3ex]$

$Point\:\:1 = (0, 2) \\[3ex] x_1 = 0 \\[3ex] y_1 = 2 \\[3ex] y-intercept, b = 2 \\[3ex] Point\:2 = (8, 3) \\[3ex] x_2 = 8 \\[3ex] y_2 = 3 \\[3ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{3 - 2}{8 - 0} \\[5ex] m = \dfrac{1}{8}$
(30.) JAMB A variable point $P(x, y)$ traces a graph in a two-dimensional plane.
$(0, -3)$ is one position of $P$
If $x$ increases by $1$ unit, $y$ increases by $4$ units.
The equation of the graph is

$A.\:\: -3 = \dfrac{y + 4}{x + 1} \\[5ex] B.\:\: 4y = -3 + x \\[3ex] C.\:\: \dfrac{y}{x} = -\dfrac{3}{4} \\[5ex] D.\:\: y + 3 = 4x \\[3ex] E.\:\: 4y = x + 3 \\[3ex]$

If $x$ increases by $1$ unit, $y$ increases by $4$ units
This means that the slope is $4$

$Point\:\:1 = (0, -3) \\[3ex] x_1 = 0 \\[3ex] y_1 = -3 \\[3ex] y-intercept, b = -3 \\[3ex] Slope, m = 4 \\[3ex] Slope-Intercept\:\:Form \\[3ex] y = mx + b \\[3ex] y = 4x + (-3) \\[3ex] y = 4x - 3 \\[3ex] Similar\:\:to\:\:Option\:\:D:\:\: y + 3 = 4x$
(31.) WASSCE Find the equation of the line which passes through the points $(2, -3)$ and $(1, -3)$

The Slope-Intercept form is the most popular form of the equation of a straight line.
If the question did not specify the form it wants, then express it in the Slope-Intercept form.

Slope-Intercept form: $y = mx + b$
Point $1$ is $(2, -3)$
$x_1 = 2$
$y_1 = -3$

Point $2$ is $(1, -3)$
$x_2 = 1$
$y_2 = -3$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-3 - (-3)}{1 - 2} \\[5ex] m = \dfrac{-3 + 3}{-1} \\[5ex] m = \dfrac{0}{-1} \\[5ex] m = 0 \\[3ex]$ When the slope is $0$, this means that $y_1 = y_2$
When the slope is $0$, $y$ is a constant
$y$ is the same as $y_1$ or $y_2$
$y = -3$ is the equation.

Student: Why is $y = y_1$ when $m = 0$
Teacher: If $m \ne 0$, what should we have done next?
Student: We would use the Point-Slope form to find the $y-intercept$
Teacher: That is correct.
So, let's use the Point-Slope form

$m = 0 \\[3ex] Point = (2, -3) \\[3ex] x_1 = 2, y_1 = -3 \\[3ex] Point-Slope\:\: form: y - y_1 = m(x - x_1) \\[3ex] y - (-3) = 0(x - 2) \\[3ex] y + 3 = 0 \\[3ex] y = - 3 \\[3ex] y = y_1$
(32.) WASSCE Find the equation of a straight line which passes through the point $(2, -3)$ and is parallel to the line $2x + y = 6$

Two lines are parallel to each other if their slopes are the same

$\underline{Line\:\: 1} \\[3ex] 2x + y = 6 \\[3ex] y = 6 - 2x \\[3ex] y = -2x + 6 \\[3ex]$ Slope of Line $1$ = $m_1$
$m_1 = -2$

$\underline{Line\:\: 2}$
$Line\:\: 2 \parallel Line\:\: 1$
Slope of Line $2$ = $m_2$

$m_2 = -2 \\[3ex]$ ${Line\:\: 2}$ should pass through the point $(2, -3)$
To find ${Line\:\: 2}$, we shall use the Point-Slope form because we have a slope and a point
Point-Slope Form: $y - y_1 = m(x - x_1)$

$m_2 = -2 \\[3ex] x_1 = 2, y_1 = -3 \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - (-3) = -2(x - 2) \\[3ex] y + 3 = -2x + 4 \\[3ex] y = -2x + 4 - 3 \\[3ex] y = -2x + 1 \\[3ex]$ The equation of ${Line\:\: 2}$ in Slope-Intercept form is: $y = -2x + 1$

To write ${Line\:\: 2}$ in Standard form;
$y = -2x + 1 \\[3ex] y + 2x = 1 \\[3ex] 2x + y = 1 \\[3ex]$ The equation of ${Line\:\: 2}$ in Standard form is: $2x + y = 1$

(33.) ACT The equation of line $s$ below is $y = mx + b$.
Which of the following could be an equation for line $t$?

$F.\:\: y = 2mx \\[3ex] G.\:\: y = 2mx + b \\[3ex] H.\:\: y = 2mx - b \\[3ex] J.\:\: y = -2mx + b \\[3ex] K.\:\: y = -2mx - b \\[3ex]$

We do not have enough information to find the equation for line $t$
But we do know that:
(1.) Line $t$ has a negative slope
So, Options $F, G, H$ are eliminated
We are now left with Options $J$ and $K$

(2.) The $y-intercept$ for Line $t$ is $b$
Option $K$ is eliminated.

Therefore, the answer is Option $J$
(34.) ACT What is the $y-intercept$ of the line in the standard $(x, y)$ coordinate plane that goes through the points $(-3, 6)$ and $(3, 2)$?

$F.\:\: 0 \\[3ex] G.\:\: 2 \\[3ex] H.\:\: 4 \\[3ex] J.\:\: 6 \\[3ex] K.\:\: 8 \\[3ex]$

First: We need to find the slope
Second: We find the equation
Third: We find the $y-intercept$

$\underline{Slope} \\[3ex] Point\:\:1 = (-3, 6) \\[3ex] x_1 = -3 \\[3ex] y_1 = 6 \\[3ex] Point\:\:2 = (3, 2) \\[3ex] x_2 = 3 \\[3ex] y_2 = 2 \\[3ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{2 - 6}{3 - (-3)} \\[5ex] m = \dfrac{-4}{3 + 3} \\[5ex] m = \dfrac{-4}{6} \\[5ex] m = -\dfrac{2}{3} \\[5ex] \underline{Equation} \\[3ex] Point-Slope\:\:Form \\[3ex] You\:\:can\:\:use\:\:any\:\:point \\[3ex] Use\:\:same\:\:slope \\[3ex] Point\:\:1 = (-3, 6) \\[3ex] Slope, m = -\dfrac{2}{3} \\[5ex] y - y_1 = m(x - x_1) \\[3ex] x_1 = -3 \\[3ex] y_1 = 6 \\[3ex] y - 6 = -\dfrac{2}{3}(x - (-3)) \\[5ex] y - 6 = -\dfrac{2}{3}(x + 3) \\[5ex] y - 6 = -\dfrac{2}{3}x - \dfrac{2}{3} * 3 \\[5ex] y - 6 = -\dfrac{2}{3}x - 2 \\[5ex] y = -\dfrac{2}{3}x -2 + 6 \\[5ex] y = -\dfrac{2}{3}x + 4 \\[5ex] Slope-Intercept\:\:Form \\[3ex] y = mx + b \\[3ex] y = -\dfrac{2}{3}x + 4 \\[5ex] y-intercept = 4 \\[3ex] OR \\[3ex] \underline{Equation} \\[3ex] Point-Slope\:\:Form \\[3ex] You\:\:can\:\:use\:\:any\:\:point \\[3ex] Use\:\:same\:\:slope \\[3ex] Point\:\:2 = (3, 2) \\[3ex] Slope, m = -\dfrac{2}{3} \\[5ex] y - y_1 = m(x - x_1) \\[3ex] x_1 = 3 \\[3ex] y_1 = 2 \\[3ex] y - 2 = -\dfrac{2}{3}(x - 3) \\[5ex] y - 2 = -\dfrac{2}{3}(x - 3) \\[5ex] y - 2 = -\dfrac{2}{3}x - \dfrac{2}{3} * -3 \\[5ex] y - 2 = -\dfrac{2}{3}x + 2 \\[5ex] y = -\dfrac{2}{3}x + 2 + 2 \\[5ex] y = -\dfrac{2}{3}x + 4 \\[5ex] Slope-Intercept\:\:Form \\[3ex] y = mx + b \\[3ex] y = -\dfrac{2}{3}x + 4 \\[5ex] y-intercept = 4$
(35.) WASSCE Find the equation of the line which is perpendicular to the line $y = 2x - 1$ and passes through the point $(2, 5)$

Two lines are perpendicular to each other if the product of their slopes is $-1$

$\underline{Line\:\: 1} \\[3ex] y = 2x - 1 \\[3ex] Slope\:\:of\:\:Line\:1 = m_1 \\[3ex] m_1 = 2 \\[3ex] \underline{Line\:\: 2} \\[3ex] Line\:\: 2 \perp Line\:\: 1 \\[3ex] Slope\:\:of\:\:Line\:2 = m_2 \\[3ex] m_1 * m_2 = -1 \\[3ex] m_2 = \dfrac{-1}{m_1} \\[5ex] m_2 = -\dfrac{1}{2} \\[5ex]$ ${Line\:\: 2}$ should pass through the point $(2, 5)$
To find ${Line\:\: 2}$, we shall use the Point-Slope form because we have a slope and a point
Point-Slope Form: $y - y_1 = m(x - x_1)$

$m_2 = -\dfrac{1}{2} \\[5ex] Point(2, 5) \\[3ex] x_1 = 2, y_1 = 5 \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 5 = -\dfrac{1}{2}(x - 2) \\[5ex] y - 5 = -\dfrac{1}{2}x + \dfrac{1}{2} * 2 \\[5ex] y - 5 = -\dfrac{1}{2}x + 1 \\[5ex] y = -\dfrac{1}{2}x + 1 + 5 \\[5ex] y = -\dfrac{1}{2}x + 6 \\[5ex]$ The equation of ${Line\:\: 2}$ in Slope-Intercept form is: $y = -\dfrac{1}{2}x + 6$

To write ${Line\:\: 2}$ in Standard form;
$y = -\dfrac{1}{2}x + 6 \\[5ex] LCD = 2 \\[3ex] 2y = 2\left(-\dfrac{1}{2}x\right) + 2(6) \\[5ex] 2y = -x + 12 \\[3ex] 2y + x = 12 \\[3ex] x + 2y = 12 \\[3ex]$ The equation of ${Line\:\: 2}$ in Standard form is: $x + 2y = 12$

(36.) Determine the slope-intercept form for a line whose $x-intercept = 2$, and $y-intercept = \dfrac{2}{3}$

$x-intercept = (2, 0) \\[3ex] y-intercept = \left(0, \dfrac{2}{3}\right) \\[5ex] Point\:1\:(2, 0) \\[3ex] x_1 = 2 \\[3ex] y_1 = 0 \\[3ex] Point\:2\:\left(0, \dfrac{2}{3}\right) \\[5ex] x_2 = 0 \\[3ex] y_2 = \dfrac{2}{3} \\[5ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{\dfrac{2}{3} - 0}{0 - 2} \\[7ex] m = \dfrac{\dfrac{2}{3}}{-2} \\[7ex] m = \dfrac{2}{3} \div \dfrac{-2}{1} \\[5ex] m = \dfrac{2}{3} * \dfrac{-1}{2} \\[5ex] m = -\dfrac{1}{3} \\[5ex] y-intercept, b = \dfrac{2}{3} \\[5ex] y = mx + b ...Slope-Intercept\:\:Form \\[3ex] y = -\dfrac{1}{3}x + \dfrac{2}{3}$
(37.) ACT What is the $x-coordinate$ of the point in the standard $(x, y)$ coordinate plane at which the $2$ lines $y = 2x + 6$ and $y = 3x + 4$ intersect?

$A.\:\: 1 \\[3ex] B.\:\: 2 \\[3ex] C.\:\: 4 \\[3ex] D.\:\: 6 \\[3ex] E.\:\: 10 \\[3ex]$

Intersection of the lines implies that we should equate both lines

$y = 2x + 6 \\[3ex] y = 3x + 4 \\[3ex] y = y \\[3ex] \therefore 2x + 6 = 3x + 4 \\[3ex] 3x + 4 = 2x + 6 \\[3ex] 3x - 2x = 6 - 4 \\[3ex] x = 2$
(38.) ACT What is the slope of any line parallel to the line $7x + 9y = 6$?

$A.\:\: -7 \\[3ex] B.\:\: -\dfrac{7}{9} \\[5ex] C.\:\: \dfrac{7}{6} \\[5ex] D.\:\: 6 \\[3ex] E.\:\: 7 \\[3ex]$

Parallel lines have the same slope

$7x + 9y = 6 \\[3ex] 9y = 6 - 7x \\[3ex] 9y = -7x + 6 \\[3ex] y = \dfrac{-7x + 6}{9} \\[5ex] y = \dfrac{-7x}{9} + \dfrac{6}{9} \\[5ex] y = -\dfrac{7}{9}x + \dfrac{2}{3} \\[5ex] Compare:\:\: y = mx + b \\[3ex] m = -\dfrac{7}{9} \\[5ex]$ The slope of any line parallel to $7x + 9y = 6$ would also be $-\dfrac{7}{9}$
(39.) ACT The ordered pairs $(x, y)$ in one of the following tables belong to a linear function.
Which one?

There are several ways to solve this question.
Because it has an ACT question and you have to solve it in a minute, it is important to use a method that is fast, while accurate.

1st Method: By Definition
For a linear function:
For this question: the input, $x$ is increasing
these are the cases between the output, $y$ and the input, $x$:
$y$ increases as $x$ increases ...increasing function
$y$ decreases as $x$ increases ... decreasing function
$y$ is constant as $x$ increases ... constant function

Option $F$:
$y$ fluctuates.. Does not apply ... Not a linear function

Option $J$:
$y$ fluctuates.. Does not apply ... Not a linear function

Option $G$:
$y$ is not a linear function. It is a ramp function
It decreases, remains constant, and decreases again.. Does not apply ... Not a linear function

So, we are left with Options $H$ and $K$
This method alone was not sufficient to do this question.
However, it helps in eliminating the incorrect options so we save some time.
On a careful observation of Option $K$, we notice that $y = x^2$ because:

$0 = 0^2 \\[3ex] 1 = 1^2 \\[3ex] 4 = 2^2 \\[3ex] 9 = 3^2 \\[3ex]$ Option $K$ is a quadratic function.
It is not a linear function.

So, our remaining option is Option $H$
Observing Option $H$, we notice that the output decreases by $1$ unit (a unit) for each unit increase in the output.
This implies that the slope is one.
Option $H$ is the linear function.

If you have time and if you want to, you can calculate the slope.
That is the second method: determining the slope.
A linear function has the same slope between two consecutive points.
(40.) ACT The graph in the standard $(x, y)$ coordinate plane below is represented by one of the following equations.
Which equation?
$F.\:\: y = -\dfrac{3}{2}x + 2 \\[5ex] G.\:\: y = -\dfrac{3}{2}x + 3 \\[5ex] H.\:\: y = -\dfrac{2}{3}x + 2 \\[5ex] J.\:\: y = -\dfrac{2}{3}x + 3 \\[5ex] K.\:\: y = \dfrac{2}{3}x + 2 \\[5ex]$

$Point\:\:1 = (0, 2) \\[3ex] x_1 = 0 \\[3ex] y_1 = 2 \\[3ex] y-intercept, b = 2 \\[3ex] Point\:2 = (3, 0) \\[3ex] x_2 = 3 \\[3ex] y_2 = 0 \\[3ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{0 - 2}{3 - 0} \\[5ex] m = -\dfrac{2}{3} \\[5ex] y = mx + b \\[3ex] y = -\dfrac{2}{3}x + 2$

(41.) JAMB Find the value of $p$ if the line joining $(p, 4)$ and $(6, -2)$ is perpendicular to the line joining $(2, p)$ and $(-1, 3)$

$A.\:\: 0 \\[3ex] B.\:\: 3 \\[3ex] C.\:\: 4 \\[3ex] D.\:\: 6 \\[3ex]$

$\underline{Line\:\:1} \\[3ex] m_1 = slope\:\:of\:\:Line\:\:1 \\[3ex] Point\:1\:(p, 4) \\[3ex] x_1 = p \\[3ex] y_1 = 4 \\[3ex] Point\:2\:(6, -2) \\[3ex] x_2 = 6 \\[3ex] y_2 = -2 \\[3ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m_1 = \dfrac{-2 - 4}{6 - p} \\[5ex] m_1 = \dfrac{-6}{6 - p} \\[5ex] \underline{Line\:\:2} \\[3ex] m_2 = slope\:\:of\:\:Line\:\:2 \\[3ex] Point\:1\:(2, p) \\[3ex] x_1 = 2 \\[3ex] y_1 = p \\[3ex] Point\:2\:(-1, 3) \\[3ex] x_2 = -1 \\[3ex] y_2 = 3 \\[3ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m_2 = \dfrac{3 - p}{-1 - 2} \\[5ex] m_2 = \dfrac{3 - p}{-3} \\[5ex] For\:\:Line\:\:1\:\:to\:\:be\:\:\perp \:\:to\:\:Line\:\:2 \\[3ex] m_1 * m_2 = -1 \\[3ex] m_1 = \dfrac{-1}{m_2} \\[5ex] m_1 = -1 \div m_2 \\[3ex] \rightarrow \dfrac{-6}{6 - p} = -1 \div \dfrac{3 - p}{-3} \\[5ex] \dfrac{-6}{6 - p} = -1 * \dfrac{-3}{3 - p} \\[5ex] \dfrac{-6}{6 - p} = \dfrac{3}{3 - p} \\[5ex] Cross\:\:Multiply \\[3ex] -6(3 - p) = 3(6 - p) \\[3ex] -18 + 6p = 18 - 3p \\[3ex] 6p + 3p = 18 + 18 \\[3ex] 9p = 36 \\[3ex] p = \dfrac{36}{9} \\[5ex] p = 4$
(42.) ACT What is the slope-intercept form of $8x - y - 6 = 0$?

$F.\:\: y = -8x - 6 \\[3ex] G.\:\: y = -8x + 6 \\[3ex] H.\:\: y = 8x - 6 \\[5ex] J.\:\: y = 8x + 6 \\[3ex] K.\:\: y = 6x - 8 \\[3ex]$

$8x - y - 6 = 0 \\[3ex] 8x - 6 = 0 + y \\[3ex] 8x - 6 = y \\[3ex] y = 8x - 6$
(43.) JAMB $3y = 4x -1$ and $Ky = x + 3$ are equations of two straight lines.
If the two lines are perpendicular to each other, find $K$

$A.\:\: -\dfrac{4}{3} \\[5ex] B.\:\: -\dfrac{3}{4} \\[5ex] C.\:\: \dfrac{3}{4} \\[5ex] D.\:\: \dfrac{4}{3} \\[5ex]$

$y = mx + b ...Slope-Intercept\:\:Form \\[3ex] \underline{Line\:\:1} \\[3ex] m_1 = slope\:\:of\:\:Line\:\:1 \\[3ex] 3y = 4x - 1 \\[3ex] y = \dfrac{4x - 1}{3} \\[5ex] y = \dfrac{4}{3}x - \dfrac{1}{3} \\[5ex] m_1 = \dfrac{4}{3} \\[5ex] \underline{Line\:\:2} \\[3ex] m_2 = slope\:\:of\:\:Line\:\:2 \\[3ex] Ky = x + 3 \\[3ex] y = \dfrac{x + 3}{K} \\[5ex] y = \dfrac{x}{K} + \dfrac{3}{K} \\[5ex] y = \dfrac{1}{K}x + \dfrac{3}{K} \\[5ex] m_2 = \dfrac{1}{K} \\[5ex] For\:\:Line\:\:1\:\:to\:\:be\:\:\perp \:\:to\:\:Line\:\:2 \\[3ex] m_1 * m_2 = -1 \\[3ex] m_1 = \dfrac{-1}{m_2} \\[5ex] m_1 = -1 \div m_2 \\[3ex] \rightarrow \dfrac{4}{3} = -1 \div \dfrac{1}{K} \\[5ex] \dfrac{4}{3} = \dfrac{-1}{K} \\[5ex] Cross\:\:Multiply \\[3ex] 4(K) = 3(-1) \\[3ex] 4K = -3 \\[3ex] K = \dfrac{-3}{4} \\[5ex] K = -\dfrac{3}{4}$
(44.) ACT What is the slope of the line given by the equation $14x - 11y + 16 = 0$

$F.\:\: -11 \\[3ex] G.\:\: -\dfrac{14}{11} \\[5ex] H.\:\: -\dfrac{11}{14} \\[5ex] J.\:\: \dfrac{14}{11} \\[5ex] K.\:\: 14 \\[3ex]$

$14x - 11y + 16 = 0 \\[3ex] 14x + 16 = 0 + 11y \\[3ex] 14x + 16 = 11y \\[3ex] 11y = 14x + 16 \\[3ex] y = \dfrac{14x + 16}{11} \\[5ex] y = \dfrac{14}{11}x + \dfrac{16}{11} \\[5ex] Compare:\:\: y = mx + b \\[3ex] Slope, m = \dfrac{14}{11}$
(45.) CSEC The equation of line $l$ is $y = 4x + 5$

(i) State the gradient of any line that is parallel to $l$

(ii) Determine the equation of the line parallel to $l$ that passes through the point $(2, -6)$

Gradient of a line is the slope of the line
Parallel lines have the same slope

$(i) \\[3ex] m = 4 \\[3ex] (ii) \\[3ex] \underline{Line\:\: 1} \\[3ex] y = 4x + 5 \\[3ex] Slope\:\:of\:\:Line\:\:1 = m_1 \\[3ex] m_1 = 4 \\[3ex] \underline{Line\:\: 2} \\[3ex] Line\:\: 2 \parallel Line\:\: 1 \\[3ex] Slope\:\:of\:\:Line\:\:2 = m_2 \\[3ex] m_2 = 4 \\[3ex]$ ${Line\:\: 2}$ should pass through the point $(2, -6)$

To find ${Line\:\: 2}$, we shall use the Point-Slope form because we have a slope and a point

$Point-Slope\:\:Form:\:\: y - y_1 = m(x - x_1) \\[3ex] m_2 = 4 \\[3ex] x_1 = 2, y_1 = -6 \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - -6 = 4(x - 2) \\[3ex] y + 6 = 4x - 8 \\[3ex] y = 4x - 8 - 6 \\[3ex] y = 4x - 14 \\[3ex] \underline{Slope-Intercept\:\:Form} \\[3ex] y = 4x - 14 \\[3ex] \underline{Standard\:\:Form} \\[3ex] 4x - 14 = y \\[3ex] 4x - y = 14 \\[3ex]$ The equation of ${Line\:\: 2}$ in Slope-Intercept form is: $y = 4x - 14$

The equation of ${Line\:\: 2}$ in Standard form is: $4x - y = 14$

(46.) JAMB Find the equation of a line parallel to $y = -4x + 2$ passing through $(2, 3)$

$A.\:\: y + 4x + 11 = 0 \\[3ex] B.\:\: y - 4x - 11 = 0 \\[3ex] C.\:\: y + 4x - 11 = 0 \\[3ex] D.\:\: y - 4x + 11 = 0 \\[3ex]$

Two lines are parallel to each other if their slopes are the same

$\underline{Line\:\: 1} \\[3ex] y = -4x + 2 \\[3ex] Slope\:\:of\:\:Line\:\:1 = m_1 \\[3ex] m_1 = -4 \\[3ex] \underline{Line\:\: 2} \\[3ex] Line\:\: 2 \parallel Line\:\: 1 \\[3ex] Slope\:\:of\:\:Line\:\:2 = m_2 \\[3ex] m_2 = -4 \\[3ex]$ ${Line\:\: 2}$ should pass through the point $(2, 3)$

To find ${Line\:\: 2}$, we shall use the Point-Slope form because we have a slope and a point

$\underline{Point-Slope\:\:Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] m_2 = -4 \\[3ex] x_1 = 2, y_1 = 3 \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 3 = -4(x - 2) \\[3ex] y - 3 = -4x + 8 \\[3ex] y = -4x + 8 + 3 \\[3ex] y = -4x + 11 \\[3ex]$ The equation of ${Line\:\: 2}$ in Slope-Intercept form is: $y = -4x + 11$

To write ${Line\:\: 2}$ in Standard form;

$y = -4x + 11 \\[3ex] y + 4x = 11 \\[3ex] 4x + y = 11 \\[3ex]$ The equation of ${Line\:\: 2}$ in Standard form is: $4x + y = 11$

$\underline{Format\:\:of\:\:the\:\:Answer\:\:Options} \\[3ex] y = -4x + 11 \\[3ex] y + 4x - 11 = 0$
(47.) JAMB Find the value of $p$, if the line which passes through $(-1, -p)$ and $(-2p, 2)$ is parallel to the line $2y + 8x - 17 = 0$

$A.\:\: -\dfrac{6}{7} \\[5ex] B.\:\: -\dfrac{2}{7} \\[5ex] C.\:\: -\dfrac{6}{7} \\[5ex] D.\:\: \dfrac{7}{6} \\[5ex]$

$\underline{Line\:\:1} \\[3ex] m_1 = slope\:\:of\:\:Line\:\:1 \\[3ex] Point\:1\:(-1, -p) \\[3ex] x_1 = -1 \\[3ex] y_1 = -p \\[3ex] Point\:2\:(-2p, 2) \\[3ex] x_2 = -2p \\[3ex] y_2 = 2 \\[3ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m_1 = \dfrac{2 - -p}{-2p - -1} \\[5ex] m_1 = \dfrac{2 + p}{-2p + 1} \\[5ex] \underline{Line\:\:2} \\[3ex] 2y + 8x - 17 = 0 \\[3ex] 2y = 0 - 8x + 17 \\[3ex] 2y = -8x + 17 \\[3ex] y = \dfrac{-8x + 17}{2} \\[5ex] y = -\dfrac{8}{2}x + \dfrac{17}{2} \\[5ex] y = -4x + \dfrac{17}{2} \\[5ex] m_2 = slope\:\:of\:\:Line\:\:2 \\[3ex] m_2 = -4 \\[3ex] For\:\:Line\:\:1\:\:to\:\:be\:\: || \:\:to\:\:Line\:\:2 \\[3ex] m_1 = m_2 \\[3ex] \rightarrow \dfrac{2 + p}{-2p + 1} = -4 \\[5ex] 2 + p = -4(-2p+ 1) \\[3ex] 2 + p = 8p - 4 \\[3ex] 2 + 4 = 8p - p \\[3ex] 6 = 7p \\[3ex] 7p = 6 \\[3ex] p = \dfrac{6}{7}$
(48.) ACT What is the slope of the line through (−2, 1) and (2, −5) in the standard (x, y) coordinate plane?

$A.\:\: \dfrac{3}{2} \\[5ex] B.\:\: 1 \\[3ex] C.\:\: -1 \\[3ex] D.\:\: -\dfrac{3}{2} \\[5ex] E.\:\: -4 \\[3ex]$

$Point\:1 = (-2, 1) \\[3ex] x_1 = -2 \\[3ex] y_1 = 1 \\[3ex] Point\:2 = (2, -5) \\[3ex] x_2 = 2 \\[3ex] y_2 = -5 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-5 - 1}{2 - (-2)} \\[5ex] = \dfrac{-6}{2 + 2} \\[5ex] = -\dfrac{6}{4} \\[5ex] = -\dfrac{3}{2}$
(49.) ACT In the standard $(x, y)$ coordinate plane, what is the slope of the line given by the equation $4x = 7y + 5?$

$A.\:\: -\dfrac{4}{7} \\[5ex] B.\:\: \dfrac{4}{7} \\[5ex] C.\:\: \dfrac{7}{4} \\[5ex] D.\:\: 4 \\[3ex] E.\:\: 7 \\[3ex]$

$4x = 7y + 5 \\[3ex] 7y + 5 = 4x \\[3ex] 7y = 4x - 5 \\[3ex] y = \dfrac{4x - 5}{7} \\[5ex] y = \dfrac{4}{7}x - \dfrac{5}{7} \\[5ex] Compare\:\:to \\[3ex] y = mx + b \\[3ex] m = \dfrac{4}{7}$
(50.) JAMB Find the gradient of the line passing through the points $P(1, 1)$ and $Q(2, 5)$.

$A.\:\: 3 \\[3ex] B.\:\: 2 \\[3ex] C.\:\: 5 \\[3ex] D.\:\: 4 \\[3ex]$

$Point\:1 = (1, 1) \\[3ex] x_1 = 1 \\[3ex] y_1 = 1 \\[3ex] Point\:2 = (2, 5) \\[3ex] x_2 = 2 \\[3ex] y_2 = 5 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{5 - 1}{2 - 1} \\[5ex] = \dfrac{4}{1} \\[5ex] = 4$
(51.) (a.) Determine the slope of the line joining pair of points.

$(i.)\;\; (9, 3) \;\;\;and\;\;\; (19, -17) \\[3ex] (ii.)\;\; (2, 3) \;\;\;and\;\;\; (7, 5) \\[3ex] (iii.)\;\; (-4, 7) \;\;\;and\;\;\; (-6, -4) \\[3ex]$ (b.) Write the slope – intercept form of the equation of the line.
Then, determine the slope of the line.

$(i.)\;\; 6x + 5y = -15 \\[3ex] (ii.)\;\; 11x - 8y = -48 \\[3ex] (iii.)\;\; 2x + 3y = 9 \\[3ex] (iv.)\;\; 3x - 2y = -16 \\[3ex]$

$(a.) \\[3ex] slope = m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] (i.) \\[3ex] Point\;1 = (9, 3) \\[3ex] x_1 = 9 \\[3ex] y_1 = 3 \\[3ex] Point\;2 = (19, -17) \\[3ex] x_2 = 19 \\[3ex] y_2 = -17 \\[3ex] m = \dfrac{-17 - 3}{19 - 9} \\[5ex] = \dfrac{-20}{10} \\[5ex] = -2 \\[3ex] (ii.) \\[3ex] Point\;1 = (2, 3) \\[3ex] x_1 = 2 \\[3ex] y_1 = 3 \\[3ex] Point\;2 = (7, 5) \\[3ex] x_2 = 7 \\[3ex] y_2 = 5 \\[3ex] m = \dfrac{5 - 3}{7 - 2} \\[5ex] = \dfrac{2}{5} \\[5ex] (iii.) \\[3ex] Point\;1 = (-4, 7) \\[3ex] x_1 = -4 \\[3ex] y_1 = 7 \\[3ex] Point\;2 = (-6, -4) \\[3ex] x_2 = -6 \\[3ex] y_2 = -4 \\[3ex] m = \dfrac{-4 - 7}{-6 - -4} \\[5ex] = \dfrac{-11}{-6 + 4} \\[5ex] = \dfrac{-11}{-2} \\[5ex] = \dfrac{11}{2} \\[5ex] (b.) \\[3ex] Slope-Intercept\;\;Form:\;\; y = mx + b \\[3ex] slope = m \\[3ex] (i.) \\[3ex] 6x + 5y = -15 \\[3ex] 5y = -6x - 15 \\[3ex] y = \dfrac{-6x - 15}{5} \\[5ex] y = \dfrac{-6x}{5} - \dfrac{15}{5} \\[5ex] y = -\dfrac{6}{5}x - 3 \\[5ex] m = -\dfrac{6}{5} \\[5ex] (ii.) \\[3ex] 11x - 8y = -48 \\[3ex] 11x + 48 = 8y \\[3ex] 8y = 11x + 48 \\[3ex] y = \dfrac{11x + 48}{8} \\[5ex] y = \dfrac{11x}{8} + \dfrac{48}{8} \\[5ex] y = \dfrac{11}{8}x + 6 \\[5ex] m = \dfrac{11}{8} \\[5ex] (iii.) \\[3ex] 2x + 3y = 9 \\[3ex] 3y = -2x + 9 \\[3ex] y = \dfrac{-2x + 9}{3} \\[5ex] y = \dfrac{-2x}{3} + \dfrac{9}{3} \\[5ex] y = -\dfrac{2}{3}x + 3 \\[5ex] m = -\dfrac{2}{3} \\[5ex] (iv.) \\[3ex] 3x - 2y = -16 \\[3ex] 3x + 16 = 2y \\[3ex] 2y = 3x + 16 \\[3ex] y = \dfrac{3x + 16}{2} \\[5ex] y = \dfrac{3x}{2} + \dfrac{16}{2} \\[5ex] y = \dfrac{3}{2}x + 8 \\[5ex] m = \dfrac{3}{2}$
(52.) ACT One of the following equations, in slope-intercept form, is the equation of the line shown below in the standard $(x, y)$ coordinate plane. Which one?

$A.\:\: y = -\dfrac{3}{5}x + 2 \\[5ex] B.\:\: y = -\dfrac{3}{5}x - 2 \\[5ex] C.\:\: y = -\dfrac{5}{3}x - 2 \\[5ex] D.\:\: y = \dfrac{3}{5}x + 2 \\[5ex] E.\:\: y = \dfrac{5}{3}x + 2 \\[5ex]$

The graph has a negative slope: rises on the left and falls on the right
In that regard, Options $D$ and $E$ are eliminated

The graph has a $y-intercept$ of $2$...assume each line represents one unit
In that regard, Options $B$ and $C$ are eliminated.

Remaining Option $A$
Option $A$ is the answer
(53.) CSEC The diagram below represents the graph of a straight line which passes through the points $p$ and $Q$.

(a) State the coordinates of $P$ and $Q$

(b) Determine
(i) the gradient of the line segment $PQ$
(ii) the equation of the line segment $PQ$

(c) The point $(-8, t)$ lies on the line segment $PQ$. Determine the value of $t$.

The line segment, $AB$ is perpendicular to $PQ$, and passes through $(6, 2)$.

(d) Determine the equation of $AB$ in the form $y = mx + c$

$(a.) \\[3ex] Coordinates\:\:of\:\:Point\:\:P = (0, 3) \\[3ex] Coordinates\:\:of\:\:Point\:\:Q = (-2, 0) \\[3ex] (b.)\:\:(i) \\[3ex] \underline{Gradient\:\:of\:\:\overleftrightarrow{PQ}} \\[3ex] Two\:\:Methods...use\:\:any\:\:one\:\:you\:\:prefer \\[3ex] \underline{First\:\:Method} \\[3ex] Point\:\:P(0, 3) \rightarrow x_1 = 0, y_1 = 3 \\[3ex] Point\:\:Q(-2, 0) \rightarrow x_2 = -2, y_2 = 0 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{0 - 3}{-2 - 0} \\[3ex] = \dfrac{-3}{-2} \\[5ex] = \dfrac{3}{2} \\[5ex] \underline{Second\:\:Method} \\[3ex] m = \dfrac{rise}{run} \\[3ex] rise = 3\:units...going\:\:up \\[3ex] run = -(-2) units ...going\:\:left \\[3ex] run = 2\:\:units \\[3ex] m = \dfrac{3}{2}\:units \\[5ex] (b)\:\:(ii) \\[3ex] \underline{Equation\:\:of\:\:\overleftrightarrow{PQ}} \\[3ex] y-intercept = b = 3 \\[3ex] slope = m = \dfrac{3}{2} \\[5ex] y = mx + b \\[3ex] y = \dfrac{3}{2}x + 3 \\[5ex] (c) \\[3ex] Two\:\:Methods...use\:\:any\:\:one\:\:you\:\:prefer \\[3ex] \underline{First\:\:Method} \\[3ex] (-8,t)\rightarrow x = -8, y = t \\[3ex] y = \dfrac{3}{2}x + 3...equation\:\:of\:\:\overleftrightarrow{PQ} \\[5ex] t = \dfrac{3}{2} * -8 + 3 \\[5ex] t = 3(-4) + 3 \\[3ex] t = -12 + 3 \\[3ex] t = -9 \\[3ex] \underline{Second\:\:Method} \\[3ex] Point\:\:(-8,t)\:\:lies\:\:on \overleftrightarrow{PQ} \\[3ex] Use\:\:Point\:\:(-8,t)\:\:and\:\:Point\:\:Q \\[3ex] You\:\:can\:\:also\:\:use\:\:Point\:\:(-8,t)\:\:and\:\:Point\:\:P \\[3ex] Point\:\:(-8,t)\rightarrow x_1 = -8, y_1 = t \\[3ex] Point\:\:Q(-2, 0) \rightarrow x_2 = -2, y_2 = 0 \\[3ex] m = \dfrac{3}{2}...applies\:\:to\:\:all\:\:points\:\:on\:\:\overleftrightarrow{PQ} \\[3ex] Also: \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{0 - t}{-2 - -8} \\[3ex] = \dfrac{-t}{-2 + 8} \\[5ex] = \dfrac{-t}{6} \\[5ex] \rightarrow \dfrac{-t}{6} = \dfrac{3}{2} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:-6 \\[3ex] -6 * \dfrac{-t}{6} = -6 * \dfrac{3}{2} \\[5ex] t = -3(3) \\[3ex] t = -9 \\[3ex] (d) \\[3ex] \overleftrightarrow{AB} \perp \overleftrightarrow{PQ}\:\:and\:\:passes\:\:through\:\:Point(6, 2) \\[3ex]$ Two lines are perpendicular to each other if the product of their slopes is $-1$

$\overleftrightarrow{PQ} \\[3ex] m_1 = \dfrac{3}{2} \\[5ex] \overleftrightarrow{AB} \\[3ex] Slope\:\:of\:\:\overleftrightarrow{AB} \perp \overleftrightarrow{PQ} = m_2 \\[3ex] m_2 * m_1 = -1 \\[3ex] m_2 = -1 \div m_1 \\[3ex] m_2 = -1 \div \dfrac{3}{2} \\[5ex] m_2 = -1 * \dfrac{2}{3} \\[5ex] m_2 = -\dfrac{2}{3} \\[5ex]$ $\overleftrightarrow{AB}$ should pass through the point $(6, 2)$
To find $\overleftrightarrow{AB}$, we shall use the Point-Slope form because we have a slope and a point

$\underline{Point-Slope\:\:Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] m_2 = -\dfrac{2}{3} \\[5ex] x_1 = 6, y_1 = 2 \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 2 = -\dfrac{2}{3}(x - 6) \\[5ex] y - 2 = -\dfrac{2}{3}x + 4 \\[5ex] y = -\dfrac{2}{3}x + 4 + 2 \\[5ex] y = -\dfrac{2}{3}x + 6 \\[5ex]$ The equation of $\overleftrightarrow{AB}$ in Slope-Intercept form is: $y = -\dfrac{2}{3}x + 6$

To write $\overleftrightarrow{AB}$ in Standard form;
Multiply each term by the $LCD$
$LCD = 3$

$3 * y = 3 * -\dfrac{2}{3}x + 3 * 6 \\[5ex] 3y = -2x + 18 \\[3ex] 3y + 2x = 18 \\[3ex] 2x + 3y = 18 \\[3ex]$ The equation of $\overleftrightarrow{AB}$ in Standard form is: $2x + 3y = 18$

(54.) ACT Jayla plotted the data from her science project as a scatterplot in the standard $(x, y)$ coordinate plane.
She found the line containing $2$ of the points to be $y = 0.28x + 6$.
The scatterplot and the line are shown below.

Jayla decided that this line was not a good fit for her data.
To transform her line into the regression line for her data, Jayla must:

A.   increase both the slope and the y-intercept.
B.   increase the slope and decrease the y-intercept.
C.   decrease both the slope and the y-intercept.
D.   decrease the slope and the increase the y-intercept.
E.   use either a horizontal or vertical line.

Line in Slope-Intercept Form
The slope of the line is $0.28$

The y-intercept of the line is $6$

Regression Line
*The y-intercept of the regression line seems to be $2$.*
In that regard, it is important to note that the y-intercept of the line in slope-intercept form must be decreased.

This eliminates Options $A$ and $D$.

Option $E$ is also incorrect because the regression line is not a horizontal line or a vertical line.
Using a horizontal or vertical line is not a good fit for the regression line.

Let us analyze the remaining options:

Option $B:$ increase the slope and decrease the y-intercept.
The regression is steeper than the slope-intercept line
This implies that it has a greater slope than the slope-intercept line

Student: How? May you explain?
Teacher: Remember we discussed steepness here
We can also look at an example to prove what I said.
Let us find the slope of the regression line ...not by formula...but by picking any two points in that regression line
Student: Point $1$ = $(1, 3)$ and Point $2$ = $(7, 10)$
Teacher Okay. Find the slope
Student:

$x_1 = 1 \\[3ex] y_1 = 3 \\[3ex] x_2 = 7 \\[3ex] y_2 = 10 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{10 - 3}{7 - 1} \\[5ex] = \dfrac{7}{6} \\[5ex] = 1.16666667 \\[3ex]$ Teacher: $1.16666667 \gt 0.28$

Because the regression line is steeper than the slope-intercept line, it has a higher slope.
Therefore, Jayla needs to increase the slope of her slope-intercept line in order for the line to fit her regression line.
Option $B$ is correct.
Option $C$ is incorrect because decreasing the slope of the slope-intercept line would cause it to be less steep. It would not fit the regression line.
(55.)

(56.) ACT In the standard $(x, y)$ coordinate plane, what is the slope of the line through (−6, 4) and (1, 3)?

$F.\:\: -\dfrac{7}{5} \\[5ex] G.\:\: -\dfrac{1}{5} \\[5ex] H.\:\: -\dfrac{1}{7} \\[5ex] J.\:\: \dfrac{1}{7} \\[5ex] K.\:\: \dfrac{1}{5} \\[5ex]$

$Point\:1 = (-6, 4) \\[3ex] x_1 = -6 \\[3ex] y_1 = 4 \\[3ex] Point\:2 = (1, 3) \\[3ex] x_2 = 1 \\[3ex] y_2 = 3 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{3 - 4}{1 - (-6)} \\[5ex] = \dfrac{-1}{1 + 6} \\[5ex] = -\dfrac{1}{7}$
(57.) Determine the distance between these pairs of points.

$(a.)\;\; (-8, 10)\;\;\;and\;\;\;(-6, 7) \\[3ex] (b.)\;\; (6, 4)\;\;\;and\;\;\;(-5, -1) \\[3ex]$

$distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[3ex] (a.) \\[3ex] Point\;1 = (-8, 10) \\[3ex] x_1 = -8 \\[3ex] y_1 = 10 \\[3ex] Point\;2 = (-6, 7) \\[3ex] x_2 = -6 \\[3ex] y_2 = 7 \\[3ex] distance = \sqrt{(-6 - -8)^2 + (7 - 10)^2} \\[3ex] = \sqrt{(-6 + 8)^2 + (-3)^2} \\[3ex] = \sqrt{2^2 + 9} \\[3ex] = \sqrt{4 + 9} \\[3ex] = \sqrt{13}\;units \\[3ex] (b.) \\[3ex] Point\;1 = (6, 4) \\[3ex] x_1 = 6 \\[3ex] y_1 = 4 \\[3ex] Point\;2 = (-5, -1) \\[3ex] x_2 = -5 \\[3ex] y_2 = -1 \\[3ex] distance = \sqrt{(-5 - 6)^2 + (-1 - 4)^2} \\[3ex] = \sqrt{(-11)^2 + (-5)^2} \\[3ex] = \sqrt{121 + 25} \\[3ex] = \sqrt{146}\;units$
(58.) WASSCE Find the equation of the line joining the points $\left(-2, \dfrac{1}{2}\right)$ and $\left(1, -\dfrac{2}{3}\right)$

$Point\:1 = \left(-2, \dfrac{1}{2}\right) \\[5ex] x_1 = -2 \\[3ex] y_1 = \dfrac{1}{2} \\[5ex] Point\:2 = \left(1, -\dfrac{2}{3}\right) \\[5ex] x_2 = 1 \\[3ex] y_2 = -\dfrac{2}{3} \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-\dfrac{2}{3} - \dfrac{1}{2}}{1 - (-2)} \\[7ex] m = \dfrac{\dfrac{-4 - 3}{6}}{1 + 2} \\[7ex] m = \dfrac{\dfrac{-7}{6}}{3} \\[7ex] m = -\dfrac{7}{6} \div 3 \\[5ex] m = -\dfrac{7}{6} * \dfrac{1}{3} \\[5ex] m = -\dfrac{7}{18} \\[5ex] \underline{Point-Slope\;\;Form} \\[3ex] Use\;\;any\;\;of\;\;the\;\;two\;\;points \\[3ex] Let\;\;us\;\;use\;\;Point\;\;1 \\[3ex] Point\:1 = \left(-2, \dfrac{1}{2}\right) \\[5ex] x_1 = -2 \\[3ex] y_1 = \dfrac{1}{2} \\[5ex] y - y_1 = m(x - x_1) \\[3ex] y - \dfrac{1}{2} = -\dfrac{7}{18}(x - -2) \\[5ex] y - \dfrac{1}{2} = -\dfrac{7}{18}(x + 2) \\[5ex] y - \dfrac{1}{2} = -\dfrac{7x}{18} - \dfrac{7}{9} \\[5ex] y = -\dfrac{7x}{18} - \dfrac{7}{9} + \dfrac{1}{2} \\[5ex] y = -\dfrac{7}{18}x + \dfrac{-14 + 9}{18} \\[5ex] y = -\dfrac{7}{18}x + \dfrac{-5}{18} \\[5ex] y = -\dfrac{7}{18}x - \dfrac{5}{18}...Slope-Intercept\;\;Form \\[5ex] OR \\[3ex] y = \dfrac{-7x - 5}{18} \\[5ex] 18y = -7x - 5 \\[3ex] 18y + 7x = -5 \\[3ex] 7x + 18y = -5...Standard\;\;Form$
(59.) ACT A certain fraternity had its freshmen members keep a log of their hours speD! playing video gimes.
When midterm grades were known. the fraternity president plotted the data in the standard (x,y) coordinate plane with average hours per week spent playing video games on the x-axis and the midterm grade point average (GPA) on the y-axis as shown in the figure below.
He then performed a linear regression on the data.
Which of the following statements is true of the regression equation?

A. The slope and the y-intercept are both negative.
B. The slope and the y-intercept are both positive.
C. The slope is negative, and the y-intercept is positive.
D. The slope is positive, and the y-intercept is negative.
E. The slope is 0, and the y-intercept is positive.

The y-intercept is the point where the graph cuts the y-axis.
Based on the graph:
The slope is negative and the y-intercept is positive.
(60.) ACT When graphed in the standard $(x, y)$ coordinate plane, which of the following equations does NOT represent a line?

$A.\:\: x = 4 \\[3ex] B.\:\: 3y = 6 \\[3ex] C.\:\: x - y = 1 \\[3ex] D.\:\: y = \dfrac{3}{4}x - 2 \\[5ex] E.\:\: x^2 + y = 5 \\[3ex]$

The graph of a Linear Function is a straight line

The graph of a Quadratic Function is a parabola (a curve)

$A.\:\: x = 4 \\[3ex] Not\:\:a\:\:Linear\:\:Function \\[3ex] However,\:\:the\:\:graph\:\:is\:\:a\:\:straight\:\:line \\[3ex] It\:\:is\:\:a\:\:Vertical\:\:line...so\:\:Yes \\[3ex] B.\:\: 3y = 6 \\[3ex] Linear\:\:Function...Yes \\[3ex] The\:\:graph\:\:is\:\:a\:\:horizontal\:\:line \\[3ex] C.\:\: x - y = 1 \\[3ex] Linear\:\:Function...Yes \\[3ex] D.\:\: y = \dfrac{3}{4}x - 2 \\[5ex] Linear\:\:Function...Yes \\[3ex] E.\:\: x^2 + y = 5 \\[3ex] It\:\:is\:\:a\:\:Quadratic\:\:Function \\[3ex] The\:\:graph\:\:is\:\:NOT\:\:a\:\:straight\:\:line \\[3ex] It\:\:is\:\:a\:\:curve$

(61.)

(62.) ACT An on-demand movie service charges $\$5$per month, plus$\$2$ for each movie rented.
Which of the following equations models the relationship between $M$, the number of movies rented per month, and $T$, the total monthly cost, in dollars, for the service?

$F.\;\; M = 5 + 2t \\[3ex] G.\;\; M = 2 + 5T \\[3ex] H.\;\; T = 5 + 2M \\[3ex] J.\;\; T = 2 + 5M \\[3ex] K.\;\; T = (5 + 2)M \\[3ex]$

If you are confused about how to exactly solve this question, a good way to begin is to find the monthly cost to rent:
(1.) a movie
(2.) two movies
(3.) three movies

$\underline{Monthly\;\;Cost\;\;to\;\;rent\;\;1\;\;movie} \\[3ex] T = 5 + 2(1) \\[3ex] T = 5 + 2 \\[3ex] T = \$7.00 \\[3ex] \underline{Monthly\;\;Cost\;\;to\;\;rent\;\;2\;\;movies} \\[3ex] T = 5 + 2(2) \\[3ex] T = 5 + 4 \\[3ex] T = \$9.00 \\[3ex] \underline{Monthly\;\;Cost\;\;to\;\;rent\;\;3\;\;movies} \\[3ex] T = 5 + 2(3) \\[3ex] T = 5 + 6 \\[3ex] T = \$11.00 \\[3ex] \underline{Monthly\;\;Cost\;\;to\;\;rent\;\;M\;\;movies} \\[3ex] T = 5 + 2(M) \\[3ex] T = 5 + 2M \\[3ex] T = 2M + 5 \\[3ex] $(63.) (64.) ACT What is the slope of the line that passes through (1, 5) and (17, 7) in the standard (x, y) coordinate plane?$ F.\;\; \dfrac{1}{8} \\[5ex] G.\;\; \dfrac{2}{3} \\[5ex] H.\;\; \dfrac{3}{2} \\[5ex] J.\;\; \dfrac{5}{2} \\[5ex] K.\;\; 8 \\[3ex]  Point\;1:\;(1, 5) \implies x_1 = 1, \;\; y_1 = 5 \\[3ex] Point\;2:\;(17, 7) \implies x_2 = 17, \;\; y_2 = 7 \\[3ex] Slope, \; m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{7 - 5}{17 - 1} \\[5ex] = \dfrac{2}{16} \\[5ex] = \dfrac{1}{8} $(65.) (66.) ACT The line with equation$2x + 5y = 9$is graphed in the standard$(x, y)$coordinate plane. What is the slope of the line?$ A.\;\; -\dfrac{5}{2} \\[5ex] B.\;\; -\dfrac{2}{5} \\[5ex] C.\;\; \dfrac{2}{5} \\[5ex] D.\;\; \dfrac{5}{2} \\[5ex] E.\;\; 2 \\[3ex]  2x + 5y = 9 \\[3ex] 5y = 9 - 2x \\[3ex] 5y = -2x + 9 \\[3ex] y = \dfrac{-2x + 9}{5} \\[5ex] y = -\dfrac{2}{5}x + \dfrac{9}{5} \\[5ex] Compare\;\;to\;\; y = mx + b \\[3ex] m = -\dfrac{2}{5} $(67.) (68.) Write an equation inâ€‹ point-slope form of the line that passes through (−3, 1) and with the given slope m = 7$ (x_1, y_1) = (-3, 1) \\[3ex] x_1 = -3 \\[3ex] y_1 = 1 \\[3ex] m = 7 \\[3ex] y - y_1 = m(x - x_1) ...Point-Slope\;\;Form \\[3ex] y - 1 = 7(x - -3) \\[3ex] y - 1 = 7(x + 3) $(69.) (70.) (71.) (72.) Find an equation for the line with the given properties. Express the equation in slope-intercept form. Containing the points P = (−3, 1) and Q = (−2, −1)$ P = (-3, 1) \\[3ex] x_1 = -3 \\[3ex] y_1 = 1 \\[3ex] Q = (-2, -1) \\[3ex] x_2 = -2 \\[3ex] y_2 = -1 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-1 - 1}{-2 - (-3)} \\[5ex] m = \dfrac{-2}{-2 + 3} \\[5ex] m = \dfrac{-2}{1} \\[5ex] m = -2 \\[3ex] To\;\;find\;\;b \\[3ex] Let\;\;us\;\;focus\;\;on\;\;point\;\;P(-3, 1) \\[3ex] x = -3 \\[3ex] y = 1 \\[3ex] y = mx + b \\[3ex] 1 = -2(-3) + b \\[3ex] 1 = 6 + b \\[3ex] 6 + b = 1 \\[3ex] b = 1 - 6 \\[3ex] b = -5 \\[3ex] \implies \\[3ex] y = mx + b \\[3ex] y = -2x - 5 \$
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