Question

A line includes the points (-39,49) and (0,0). What is its equation in slope-intercept form?

Answer 1

Answer:

The equation in the slope-intercept form will be:

$y=-\frac{49}{39}x+0$

Step-by-step explanation:

Given the points

• (-39,49)

• (0,0)

Finding the slope between the points using the formula

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(-39,\:49\right),\:\left(x_2,\:y_2\right)=\left(0,\:0\right)$

$m=\frac{0-49}{0-\left(-39\right)}$

$m=-\frac{49}{39}$

We know that the point-slope of the line equation is

$y-y_1=m\left(x-x_1\right)$

substituting $m=-\frac{49}{39}$ and (-39,49)  in the equation

$y-y_1=m\left(x-x_1\right)$

$y-49=\frac{-49}{39}\left(x-\left(-39\right)\right)$

Now writing the equation in slope-intercept form

$y=mx+b$

where m is the slope and b is the y-intercept

$y-49=\frac{-49}{39}\left(x-\left(-39\right)\right)$

$y-49=\frac{-49}{39}\left(x+39\right)$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$y-49=-\frac{49}{39}\left(x+39\right)$

$\mathrm{Add\:}49\mathrm{\:to\:both\:sides}$

$y-49+49=-\frac{49}{39}\left(x+39\right)+49$

$y=-\frac{49}{39}x+0$              ∵ $y=mx+b$

Where

$m=-\frac{49}{39}$ and the y-intercept i.e. $b=0$

Therefore, the equation in the slope-intercept form will be:

$y=-\frac{49}{39}x+0$