consider the line y=2/3x-7

Question

fine the equation of the line that is parallel to this line and passes through the point (9,-6)

 

find the equation of the line that that is perpendicular to this line and passes through the point (9,-6)

Answer 1

Given line is y = (2/3)x - 7

This is in the form of slope intercept form y = mx + b

Slope(m1) = 2/3

If  two lines are parllel they have same slope but the y-intercept will changes

Given that point (9 , -6) and slope(m2) = m1 = 2/3

y = mx + b

-6 = 2/3(9) + b

-6 = 2(3) + b

-6 = 6 + b

-6 - 6 = b

b = -12

Substitute  b = -12 and m = 2/3 in slope intercept form

y = mx + b

y = 2/3x - 12

If  two lines are perpedicular the slope and y-intercept will changes

Given that point (9 , -6) and slope(m2) = -1/m1 = -1/(2/3) = -3/2

y = mx + b

-6 = -3/2(9) + b

-6 = -27/2 + b

-6 +27/2 = b

(-12 + 27)/ 2 = b

b = 15/2

Substitute  b = 15/2  and m = -3/2 in slope intercept form

y = mx + b

y = -3/2 x + 15/2

Therefore equation of line parallel to the given line is y = 2/3x - 12

Therefore equation of line perpendicular to the given line is y = -3/2 x + 15/2

Answer 2

Given equation of line , y=2/3x-7

This is in the form of y = mx+c

Therefore slope(m) = 2/3

The parallel equation slope is m1=m=2/3

The perpendicular equation slope is m2=-1/m=-1/(2/3)=-3/2

The equation of line parallel to the given equation and passing through (9,-6) is,

=> y-y1=m1(x-x1)

=> y-(-6)=(2/3)(x-9)

=> y+6=(2/3)(x-9)

=> 3y+18=2x-18

=> 2x-3y -36=0

Parallel line equation is 2x-3y-36=0

The equation of line perpendicular to the given equation and passing through (9,-6)

=> y-(-6)=(-3/2)(x-9)

=> y+6=-3/2(x-9)

=> 2y+12=-3x+27

=> 3x+2y+12-27=0

=> 3x+2y-15=0

Perpendicular line equation is 3x+2y-15=0