Question

A line passes through the origin and has a slope of -3 what is the equation of the line that is perpendicular to the first line and passes through the point (3,4)?

Answer 1

ANSWER

$y= \frac{1}{3} x + 3$

EXPLANATION

We want to find the equation of a line which is perpendicular to another line with slope -3 and passes through (3,4).

Our line of interest is has a slope that is the negative reciprocal of -3

$m = - \frac{1}{ - 3} = \frac{1}{3}$

The equation is given by

$y-y_1=m(x-x_1)$

We substitute the point and slope to get:

$y-4= \frac{1}{3} (x-3)$

Expand

$y-4= \frac{1}{3} x-1$

$y= \frac{1}{3} x-1 + 4$

$y= \frac{1}{3} x + 3$

Answer 2

Answer: $y=\frac{1}{3}x+3$

Step-by-step explanation:

The equation of the line is Slope-intercept form is:

$y=mx+b$

Where "m" is the slope and "b" the y-intercept.

The slopes of perpendicular lines are negative reciprocal.

Then, if the slope of the first line is -3, the slope of the other line must be:

$m=\frac{1}{3}$

Substitute the point (3,4) into the equation and solve for b:

$4=\frac{1}{3}(3)+b\\ 4-1=b\\b=3$

Then the equation of this line is:

$y=\frac{1}{3}x+3$