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


3x + 5y = −9


3x + 5y = 9


5x − 3y = −15


5x − 3y = 15

What is the equation of the line that is perpendicular to the given line and passes through the point (3, 0)? 3x + 5y = 9 3x + 5y = 9 5x 3y = 15 5x 3y = 15 class=

Verified Answer :

I did this last year if i remember it was 5x − 3y = 15

Answer:

5x - 3y = 15

Step-by-step explanation:

Given line has two points (-3,2) and (3,0)

Now we find the slope of given line

\(slope m = \frac{y_2-y_2}{x_2-x_1} =\frac{-1-2}{2+3} =\frac{-3}{5}\)

Slope of given line is -3/5

Slope of perpendicular line is the negative reciprocal of the slope of given line

slope of perpendicular line = \(\frac{5}{3}\)

It passes through the point (3, 0)

We know the slope and the point (3,0), so we use point slope form

\(y-y1= m(x-x1)\)

\(y - 0 = \frac{5}{3}(x-3)\)

\(y= \frac{5}{3}x-5\)

Now we multiply the whole equation by 3

3y = 5x - 15

Subtract 5x on both sides

-5x + 3y = -15

Divide the whole equation by -1

5x - 3y = 15