Question

Find the equation of the line passing through the points (3,-2) and (3, 4).The answer is x = 3. I'm just wondering how my textbook got to this solution.My work:y-y1=m(x-x1). m=y2-y1 / x2-x1. y=mx+bm=4--2 / 3-3 = 6/0 = 0. m=0.y--2=0(x-3) = y=0-2 y=-2 <<<

Answer 1

Given two points. we can find the equation of a line passing through the points

The formula to be used is:

$\frac{y_2-y_1}{x_2-x_!}=\frac{y-y_1}{x-x_!}$

where

$x_1=3,y_!=-2,x_2=3,y_2=4$$\frac{4-(-2)}{3-3}=\frac{y-(-2)}{x-3}$

=>

$\frac{6}{0}=\frac{y+2}{x-3}$

The next step is to cross multiply

$6(x-3)=0(y+2)$

=>

$6(x-3)=0$

Divide both sides by 6 and make x the subject

x=3