Question

Four points A(6,3),B(-3,5) ,C(4,-2) and D(x,3x) are given such that area triangle DBC/area triangle ABC=1/2 find x

Answer 1

you can also use herons formula to calculate area but this method is easier
let me know if you find any problem.

Answer 2

Answer: 49/51

Step-by-step explanation:

Since, the area of a triangle having the vertices [tex[(x_1,y_1)[/tex], $(x_2,y_2)$ and $(x_3,y_3)$ is,

$A=1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Here, Four points A(6,3),B(-3,5) ,C(4,-2) and D(x,3x) are given,

Thus, by the above formula,

The area of triangle DBC =1/2( x(5-(-2))-3(-2-3x)+4(3x-5)) = 1/2(x(5+2)-3(-2-3x)+4(3x-5))

= 1/2(7x+10 +6+9x+12x-20)

= 14x-2 square unit

Now, the area of triangle ABC = 6(5-(-2)) -3(-2-3)+4(3-5) = 6(5+2)-3(-5)+4(-2)

= 1/2(42+15-8) = 49/2 square unit

Since, Area triangle DBC/Area triangle ABC = 1/2

$\frac{49/2}{14x-2}=\frac{1}{2}$

$\implies 49 = 14x - 2$

$\implies 51 = 14x\implies x = \frac{49}{51}$