Question

What is the equation of a line in slope intercept form that is perpendicular to y=2/3x+2 and passes through the points (-2,2)? HELP ME PLEASE

Answer 1

Answer:

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

Step-by-step explanation:

Given

Perpendicular to $y = \frac{2}{3}x + 2$

Pass through $(-2,2)$

Required

Determine the line equation

First, we need to determine the slope of $y = \frac{2}{3}x + 2$

An equation is of the form:

$y = mx + b$

Where

$m = slope$

In this case:

$m = \frac{2}{3}$

Next, we determine the slope of the second line.

Since both lines are perpendicular, the second line has a slope of:

$m_1 = \frac{-1}{m}$

$m_1 = \frac{-1}{2/3}$

$m_1 = -\frac{3}{2}$

Since this line passes through (-2,2); The equation is calculated as thus:

$y - y_1 = m_1(x - x_1)$

Where

$(x_1,y_1) = (-2,2)$

This gives:

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

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

$y - 2 = -\frac{3}{2}x -3$

Add 2 to both sides

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

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