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Analytical Geometry

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11 Mar 2010 11:20:46 IST

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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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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

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rohit dhingra

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Joined: 5 Mar 2010

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11 Mar 2010 19:17:02 IST

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

Nehal Wani

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Joined: 19 Jul 2009

Posts: 166

18 Mar 2010 15:31:15 IST

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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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