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

Step $1$

Find the slope of the given line

Let

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

slope mAB is equal to

$mAB=\frac{(y2-y1)}{(x2-x1)} \\ \\ mAB=\frac{(-1-2)}{(2+3)} \\ \\ mAB=-\frac{3}{5}$

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

$mAB*mCD=-1\\ mAB=-\frac{3}{5} \\ mCD=-\frac{1}{mAB} \\ mCD=\frac{5}{3}$

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

$y-y1=m(x-x1)\\\\ y-0=\frac{5}{3} *(x-3)\\\\ y=\frac{5}{3} x-5$

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 $5x-3y=15$

Answer 2

Answer:

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