The line through A(4,-2) and B(2,y) is: a.) parallel b.) perpendicular

Question

 to the line whose inclination is 135. 

Answer 1

Parallel line equation :

Points of the line (4,-2) and (2,y) .

Inclination of the line is 135° .

The slope of the line is tan ϴ  = tan 135 = -1 .

The slope of the lines using the points m = (y₂ – y₁)/(x₂ – x₁)

m = (y – (-2))/(2 – 4)

-1 = (y + 2)/( –2)

y+2 = 2

y = 0

So the points are  (4,-2) and (2,0) .

The equation of line is 

y – y₁ = m(x – x₁)

y – 4 = -1(x – (-2))

y – 4 = -1(x +2)

y – 4 = -x -2

x + y -2 = 0

So the line parallel to Inclination of the line is 135° is x + y -2 = 0 .

Answer 2

Perpendicular line equation :

Points of the line (4,-2) 

Inclination of the line is 135° .

The slope of the line is tan ϴ  = tan 135 = -1 .

So the slope of the line perpendicular to inclination is 1.                [since m1*m2 = -1]

The equation of line is 

y – y₁ = m(x – x₁)

y – 4 = 1(x – (-2))

y – 4 = 1(x +2)

y – 4 = x +2

x - y +6 = 0

So the line perpendicular to Inclination of the line is 135° is x - y +6 = 0.