A (1, 3) and C−25,−25 are the vertices of a triangle ABC and the equation of the angle bisector of ABC is x + y = 2Equation of BC isCoordinates of vertex BEquation of side AB is

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7x+3y+4=0

3x+7y+4=0

13x+7y+8=0

x+9y+4=0

310,1710

1710,310

-52,92

(1,1)

13x−7y+8=0

13x+7y−34=0

3x+7y−24=0

3x+7y+24=0

answer is , , .

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Detailed Solution

Let B=(α,2−α)Lies on x+y=2tanθ=m1−m21+m1m22+25−αα+25+11−2+25−αα+25=1+α1−α+11−1+α1−α⇒α=−52∴B=−52,92Equation of BC is 7x+3y+4=0Equation of AB is 3x + 7y - 24 = 0Let B=(α,2−α)Lies on x+y=2tanθ=m1−m21+m1m22+25−αα+25+11−2+25−αα+25=1+α1−α+11−1+α1−α⇒α=−52∴B=−52,92Equation of BC is 7x+3y+4=0Equation of AB is 3x + 7y - 24 = 0Let B=(α,2−α)Lies on x+y=2tanθ=m1−m21+m1m22+25−αα+25+11−2+25−αα+25=1+α1−α+11−1+α1−α⇒α=−52∴B=−52,92Equation of BC is 7x+3y+4=0Equation of AB is 3x + 7y - 24 = 0

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