Question

What is the equation of the line that is perpendicular to the given line and passes through the point (3, 0)?

a
5x − 3y = 15

b
5x − 3y = −15

c
3x + 5y = −9

d
3x + 5y = 9

Answer 1

Answer choice (B) for this question: 5x - 3y = 15

Answer 2

Step
`1`

Find the slope of the given line

Let

`A(-3,2)\ B(2,-1)`

slope mAB is equal to

`mAB=((y2-y1))/((x2-x1)) \\ \\ mAB=((-1-2))/((2+3)) \\ \\ mAB=-(3)/(5)`

Step
`2`

Find the slope of the line that is perpendicular to the given line

Let

CD ------> the line that is perpendicular to the given line

we know that

If two lines are perpendicular, then the product of their slopes is equal to
`-1`

so

`mAB*mCD=-1\\ mAB=-(3)/(5) \\ mCD=-(1)/(mAB) \\ mCD=(5)/(3)`

Step
`3`

Find the equation of the line with mCD and the point (3,0)

we know that

the equation of the line in the form point-slope is equal to

`y-y1=m(x-x1)\\\\ y-0=(5)/(3) *(x-3)\\\\ y=(5)/(3) x-5`

Multiply by
`3` both sides

`3y=5x-15`

`5x-3y=15`

therefore

the answer is

the equation of the line that is perpendicular to the given line is the equation
`5x-3y=15`