Question #5616

find the slope of the line bisecting the angle from the line through (-2,2) and (2,0) to the line through (3,-15) and (10,9)

Expert's answer

Let's find the equations of the lines.

Line through the points (-2,2) and (2,0):

(x+2)/(2-(-2)) = (y-2)/(0-2)

x = -2y+2

y = -0.5x+1

Line through the points (3,-15) and (10,9):

(x-3)/(10-3) = (y-(-15))/(9-(-15))

24x-72 = 7y+105

y = 24x/7 - 177/7.

So, the slope of the first line is k1 = -1/2 and the slope of the second one is 24/7.

The angles of this lines are

a1 = arctan(k1) = arctan(-1/2);

a2 = arctan(k2) = arctan(24/7).

There are two bisecting lines with the slopes

k3 = tan((a1+a2)/2) = tan((arctan(-1/2)+arctan(24/7))/2) ≈ 0.4367,

k4 = tan((a1-a2)/2) = tan((arctan(-1/2)-arctan(24/7))/2) ≈ -1.1982,

respectively.

Line through the points (-2,2) and (2,0):

(x+2)/(2-(-2)) = (y-2)/(0-2)

x = -2y+2

y = -0.5x+1

Line through the points (3,-15) and (10,9):

(x-3)/(10-3) = (y-(-15))/(9-(-15))

24x-72 = 7y+105

y = 24x/7 - 177/7.

So, the slope of the first line is k1 = -1/2 and the slope of the second one is 24/7.

The angles of this lines are

a1 = arctan(k1) = arctan(-1/2);

a2 = arctan(k2) = arctan(24/7).

There are two bisecting lines with the slopes

k3 = tan((a1+a2)/2) = tan((arctan(-1/2)+arctan(24/7))/2) ≈ 0.4367,

k4 = tan((a1-a2)/2) = tan((arctan(-1/2)-arctan(24/7))/2) ≈ -1.1982,

respectively.

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