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hoa [83]
3 years ago
12

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

Mathematics
2 answers:
lozanna [386]3 years ago
6 0

Answer:

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

olga2289 [7]3 years ago
4 0

Step 1

<u>Find the slope of the given line</u>

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

<u>Find the slope of the line that is perpendicular to the given line</u>

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

<u>Find the equation of the line with mCD and the point (3,0)</u>

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

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