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

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 to the line whose inclination is 135. 

asked Nov 12, 2014 in PRECALCULUS by anonymous

2 Answers

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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 .

answered Nov 12, 2014 by yamin_math Mentor
reshown Nov 12, 2014 by bradely
0 votes

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.

answered Nov 12, 2014 by yamin_math Mentor

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