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

`y = -(3)/(2)x -1`

Explanation:

Given

Perpendicular to
`y = (2)/(3)x + 2`

Pass through
`(-2,2)`

Required

Determine the line equation

First, we need to determine the slope of
`y = (2)/(3)x + 2`

An equation is of the form:

`y = mx + b`

Where

`m = slope`

In this case:

`m = (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 = (-1)/(m)`

`m_1 = (-1)/(2/3)`

`m_1 = -(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 = -(3)/(2)(x - (-2))`

`y - 2 = -(3)/(2)(x +2)`

`y - 2 = -(3)/(2)x -3`

Add 2 to both sides

`y - 2+2 = -(3)/(2)x -3 + 2`

`y = -(3)/(2)x -1`