Four points A (6, 3), B (-3, 5), C (4, - 2) and D (x, 3x) are given in such a way that , find x.

Four points A (6, 3), B (−3, 5) C (4, −2) and D(x, 3x)


Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3)


= |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|


Area of ∆ABC


= |6(5 – (- 2)) - 3(-2 -3) + 4(3 – 5)|


= |42 + 15 -8|


= sq. units


Area of ∆DBC


= |x(5 – (-2)) + 3(-2 -3x) + 4(3x - 5))|


= |7x +6 + 9x + 12x - 20|


= | 28x -14|


= ± 7(2x -1 )


It is given that


∴2 × ∆DBC = ∆ABC


2 × (± 7(2x -1 )) =


∴ ± 4(2x -1 ) = 7


∴ 4(2x -1 ) = 7 or -4(2x -1 ) = 7


∴ 8x – 4 =7 or -8x + 4 =7


∴ 8x =11 or -8x =3


∴x= or x =


Hence, the value of x is or

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