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

SATISH KUMAR SAHU

S1 & S2 are two points on AB of a triangle ABC with vertices (-2,3),(4,-6) and(1,1).CS1 and CS2 divide the triangle into three of equal area. the equation of the lines through the origin drawn parallel to CS1 and CS2 will be

rohit dhingra

the line AB passes thoough the origin ...the points lie on line AB there fore line ab is the required line

Nehal Wani

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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