What is the equation of the line that passes through (-2, -3) and is perpendicular to 2x - 3y = 6?

Solution:

Given point (-2, -3) and the line 2x - 3y = 6

Consider 2x - 3y = 6

-3y = 6 - 2x

y= 2/3 x - 2

This is in the standard form y = mx + b

m₁ = 2/3

Also given that the line is perpendicular to the given line

So, the product of their slopes must be equal to -1.

So, m₂ = -3/2

We have point-slope formula y - y₁ = m(x - x₁)

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

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

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

2y + 6 = -3x - 6

3x + 2y + 12 = 0

What is the equation of the line that passes through (-2, -3) and is perpendicular to 2x - 3y = 6?

Summary:

The equation of the line that passes through (-2, -3) and is perpendicular to 2x - 3y = 6 is 3x + 2y + 12 = 0