Solved Examples on Linear Functions

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 $

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} $

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 $

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 $

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} $

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} $

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 = $(4, 0)$

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: $x = -\dfrac{7}{2}$

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 = $\left(\dfrac{19}{3}, 0\right)$

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) $

$f(x) = y$
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 $

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 $

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 $-\dfrac{4}{7}$

$ \dfrac{y}{2} = x \\[5ex] y = 2 * x \\[3ex] y = 2x \\[3ex] Compare:\:\: y = mx + b \\[3ex] m = 2 $

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) $

$ 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} $

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: $3x - 4y = 0$

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} $

$ (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: $4x + 3y = 24$

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} $

$ 4x - y - 3 = 0 \\[3ex] 4x - 3 = y \\[3ex] y = 4x - 3 $

$ 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} $

$ \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 $

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} $

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: $4x + 3y = 25$

$ 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 $

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: $7x + 5y = 70$

$ 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 $

$ 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} $

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 $

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 $

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: $2x + y = 1$

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 $J$

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 $

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: $x + 2y = 12$

$ 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} $

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 $

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 $-\dfrac{7}{9}$

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.

$ 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 $

$ \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 $

$ 8x - y - 6 = 0 \\[3ex] 8x - 6 = 0 + y \\[3ex] 8x - 6 = y \\[3ex] y = 8x - 6 $

$ 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} $

$ 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} $

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: $4x - y = 14$

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 $

$ \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} $

$ 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} $

$ 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} $

$ 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 $

$ (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} $

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

$ (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: $2x + 3y = 18$

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.

$ 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} $

$ 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 $

$ 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 $

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.

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 $

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] $

$ 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} $

$ 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} $

$ (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) $

$ 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 $