Question

Street C is perpendicular to Street A and passes through

Answer 1

Given:

Street C is perpendicular to Street A and passes through (4, -6).

The equation of street A is:

$y=-2x+2$

To find:

The equation of street C.

Solution:

The equation of street A is:

$y=-2x+2$

On comparing this equation with slope intercept form $y=mx+b$, we get

$m_2=-2$

Slope of this line is -2.

We know that, the product of slopes of two perpendicular lines is always -1.

$m_1\times m_2=-1$

$m_1\times (-2)=-1$

$m_1=\dfrac{-1}{-2}$

$m_1=\dfrac{1}{2}$

The slope of street C is $m_1=\dfrac{1}{2}$ and it passes through the point (4,-6). So, the equation of street C is

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

$y-(-6)=\dfrac{1}{2}(x-4)$

$y+6=\dfrac{1}{2}(x-4)$

Therefore, the point slope form of the street C's equation is $y+6=\dfrac{1}{2}(x-4)$.