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Question 40 of 50
Find the equation of the straight line that passes through
(2,-3) and perpendicular to the line 3x-2y+4=0
A. 2y-3x=0
B. 3y-2x+5=0
C. 3y+2x+5=0
D. 2y-3x-5=0
Question 41 of 50

C

Step-by-step explanation:

We start with the second line;

3x -2y + 4 = 0

3x + 4 = 2y

divide through by 2

3x/2 + 2 = y

Comparing this with standard form of y = mx + c

where m is slope

This means 3/2 is slope of the line

Since this line is perpendicular to the other line, then the product of their slopes is -1

3/2 * slope of second line = -1

Slope of second line = -1 divided by 3/2 = -2/3

We use the point slope form to find the equation of the other line

y-y1 = m(x-x1)

where m = -2/3 and (x1,y1) = (2,-3)

y-(-3) = -2/3 (x-2)

y + 3 = -2/3 (x -2)

3(y + 3) = -2(x-2)

3y + 9 = -2x + 4

3y + 5 = -2x

3y + 2x + 5 = 0

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation: