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# Solved Examples on Linear Functions

Pre-requisites:

(1.) Linear Equations

(2.) Literal Equations

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 any wrong answer.

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

Solve all questions.
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.) 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 $-\dfrac{4}{7}$
(13.) 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 $b = y-intercept$

$y = -\dfrac{15}{11}x + (-8) \\[5ex] y = -\dfrac{15}{11}x - 8$
(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 $(4, 3)$.
Write your answer in slope-intercept form and standard form.

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 $4x - y - 3 = 0$?

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

This question should take you less than ten seconds.
Just move it around...see what I explained in Expressions and Equations

$4x - y - 3 = 0 \\[3ex] 4x - 3 = y \\[3ex] y = 4x - 3$
(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 $(4, 3)$.

Write your answer in slope-intercept form and standard form.

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.) 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)

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

$x-intercept = (2, 0) \\[3ex] y-intercept = \left(0, \dfrac{2}{3}\right) \\[5ex] Point\:1\:(2, 0) \\[3ex] x_1 = 2 \\[3ex] y_1 = 0 \\[3ex] Point\:2\:\left(0, \dfrac{2}{3}\right) \\[5ex] x_2 = 0 \\[3ex] y_2 = \dfrac{2}{3} \\[5ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{\dfrac{2}{3} - 0}{0 - 2} \\[7ex] m = \dfrac{\dfrac{2}{3}}{-2} \\[7ex] m = \dfrac{2}{3} \div \dfrac{-2}{1} \\[5ex] m = \dfrac{2}{3} * \dfrac{-1}{2} \\[5ex] m = -\dfrac{1}{3} \\[5ex] y-intercept, b = \dfrac{2}{3} \\[5ex] y = mx + b ...Slope-Intercept\:\:Form \\[3ex] y = -\dfrac{1}{3}x + \dfrac{2}{3}$
(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 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

$y = 2x + 6 \\[3ex] y = 3x + 4 \\[3ex] y = y \\[3ex] \therefore 2x + 6 = 3x + 4 \\[3ex] 3x + 4 = 2x + 6 \\[3ex] 3x - 2x = 6 - 4 \\[3ex] x = 2$
(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.) ACT In the standard $(x, y)$ coordinate plane, what is the slope of the line through $(-6, 4)$ and $(1, 3)$?

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

$Point\:1 = (-6, 4) \\[3ex] x_1 = -6 \\[3ex] y_1 = 4 \\[3ex] Point\:2 = (1, 3) \\[3ex] x_2 = 1 \\[3ex] y_2 = 3 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{3 - 4}{1 - (-6)} \\[5ex] = \dfrac{-1}{1 + 6} \\[5ex] = -\dfrac{1}{7}$
(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.) 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: $4x + y = 11$

$\underline{Format\:\:of\:\:the\:\:Answer\:\:Options} \\[3ex] y = -4x + 11 \\[3ex] y + 4x - 11 = 0$
(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