Question

Find the equation of the line though the point (-3, -2) and perpendicular to the line y = 2/3x-2.Write your answer in the form y=mx+b.

Answer 1

For this question we know that we have a point (-3,-2) and we want to find an equation perpendicular to the line

y=2/3x-2.

Since both lines are perpendicular we need to satisfy this:

`m_1\cdot m_2=-1`

With m1= 2/3. If we solve for m2 we got:

`m_2=(-1)/(m_1)=(-1)/((2)/(3))=-(3)/(2)`

And then we can find the intercept for the new line using the point given with x=-3 and y=-2 and we got this:

`-2=-(3)/(2)(-3)+b`

And solving for b we got:

`b=-(9)/(2)-2=-(13)/(2)`

And then our final answer would be:

`y=-(3)/(2)x-(13)/(2)`