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