S1 & S2 are two points on AB of a triangle ABC with vertices (-2,3),(4,-6) and(1,1).CS1 and CS2 divi

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Hot goIITian

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The question is absolutely correct

Area of Triangle is 15/2 sq units.

Area formed with S2(x2,y2) , C(1.1) and B(4,-6) = 5/2

Therefore, 5 = |3y+7x-10| ......................(i)

The equation of AB is 3x+2y=0 ......................(ii)

Solving (i) and (ii)

(x2,y2) = (6,-9) or (2,-3)

(6,-9) can't be as it lies outside the triangle.

Therefore S2 is (2,-3)

Area of S1(x1,y1), S2(2,-3) and C(1,1) = 5/2

Therefore, 5=|-4x-y+5| .....................................(iii)

Solving (i) and (iii),

(x1,y1) = (4,-6) or (0,0)

(4,-6) is neglected!

Therefore S1 is (0,0)

Now equation of line passing through S1 parallel to CS1 : y=x

Now equation of line passing through S2 parallel to CS2 : y=-4x

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