Four points A (6, 3), B (-3, 5), C (4, - 2) and D (x, 3x) are given in such a way that △DBC/△ABC = 1/2 , find x. ​

Question

Four points A (6, 3), B (-3, 5), C (4, - 2) and D (x, 3x) are given in such a way that \(\frac{△DBC}{△ABC}\) = \(\frac{1}2\) , find x.

Answer 1

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) 

= \(\frac{1}2\) |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)| 

Area of ∆ABC 

= \(\frac{1}2\) |6(5 – (- 2)) - 3(-2 -3) + 4(3 – 5)| 

= \(\frac{1}2\) |42 + 15 -8| 

= \(\frac{49}2\) sq. units 

Area of ∆DBC 

= \(\frac{1}2\) |x(5 – (-2)) + 3(-2 -3x) + 4(3x - 5))| 

= \(\frac{1}2\) |7x +6 + 9x + 12x - 20| 

= \(\frac{1}2\) | 28x -14| 

= ± 7(2x -1 ) 

It is given that   \(\frac{△DBC}{△ABC}\) = \(\frac{1}2\)

∴2 × ∆DBC = ∆ABC 

2 × (± 7(2x -1 )) = \(\frac{49}2\)

∴ ± 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 = \(\frac{11}8\) or x = \(\frac{-3}8\)

Hence, the value of x is \(\frac{11}8\) or \(\frac{-3}8\)