Question

In each part that​ follows, you are given an equation of a line and a point. Find the equation of the line through the given point that is perpendicular to the given line.​ (The slope of the perpendicular line is the negative reciprocal of the slope of the given line if the given line is neither vertical nor​ horizontal.)
a) y=9x, P (0, 0)
b) y=3x+4, Q (1, 2)

Answer 1

Answer:

a) $y=-\frac{1}{9}x$

b)$y-2=-\frac{1}{3} (x-1)$

Step-by-step explanation:

We can write the equation of a line in 3 different forms including slope intercept, point-slope, and standard depending on the information we have. We have a point given and a slope from the equation. We will chose point-slope since we have a point and can find the slope.

Point slope:$y-y_1=m(x-x_1)$

a) $m\neq 9$ in our new equation because it is perpendicular to it. This means we will need to change it into its negative reciprocal which is $m=-\frac{1}{9}$.

We will substitute $m=-\frac{1}{9}$ and $x_1=0\\y_1=0$.

$y-0=-\frac{1}{9} (x-0)$

This simplifies to:

$y=-\frac{1}{9}x$

b) The equation y=3x+4 follows y=mx+b. For the perpendicular line, $m\neq 3$. We will need to change it into its negative reciprocal which is $m=-\frac{1}{3}$.

We will substitute $m=-\frac{1}{3}$ and $x_1=1\\y_1=2$.

$y-2=-\frac{1}{3} (x-1)$