Question

Find the equation of the line through (3,2) that is perpendicular to the line through (-3,1) and (4,-2).

Answer 1

Let us solve for the slope(m)

The slope formula is,

`m=(y_2-y_1)/(x_2-x_1)`

Given:

`\begin{gathered} (x_1,y_1)=(-3,1) \\ (x_2,y_2)=(4,-2) \end{gathered}`

Therefore,

`m=(-2-1)/(4--3)=(-3)/(4+3)=-(3)/(7)`

Perpendicular law

`\begin{gathered} m_1m_2=-1 \\ \therefore m_2=-(1)/(m_1) \end{gathered}`

Hence,

`\begin{gathered} m_2=-(1)/(-(3)/(7))=-(1/-(3)/(7))=-(1*-(7)/(3))=(7)/(3) \\ \therefore m_2=(7)/(3) \end{gathered}`

Therefore, the formula for the equation of the line given a point is,

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

Point

`(3,2)`

Therefore,

`\begin{gathered} y-2=(7)/(3)(x-3) \\ y=(7)/(3)(x-3)+2 \\ y=(7)/(3)x-7+2 \\ y=(7)/(3)x-5 \end{gathered}`

Hence, the answer is

`y=(7)/(3)x-5`