Question

If A(-2, 3), B(6,5), C(x,-5) and D(-4,-3) are the vertices of a quadrilateral ABCD of area 80 sq units, then find positive value of x. coordinate geometry
please elaborate the answer

Answer 1

Answer:

Step-by-step explanation:

My answer is in the picture

Answer 2

Answer:

The value of x is 50.

Step-by-step explanation:

Given vertices of the quadrilateral ABCD are,

A(-2, 3), B(6,5), C(x,-5) and D(-4,-3),

Thus, the area of quadrilateral ABCD = area of ΔABC + area of ΔADC

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

$A=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Thus, the area of the ΔABC,

$A_1=\frac{1}{2}[-2(5+5)+6(-5-3)+x(3-5)]$

$=\frac{1}{2}[-2(10)+6(-8)+x(-2)]$

$=\frac{1}{2}(-20-48-2x)$

$=\frac{1}{2}(-68-2x)$

$=-34-x\text{ square unit}$

Similarly, the area of the ΔADC,

$A_2=\frac{1}{2}[-2(-3+5)-4(-5-3)+x(3+3)]$

$=\frac{1}{2}(-4+32+6x)$

$=\frac{1}{2}(28+6x)$

$=14+3x\text{ square unit}$

Hence, the area of the quadrilateral ABCD,

$A=A_1+A_2=-34-x+14+3x=-20+2x$

According to the question,

A = 80 square unit,

$\implies -20+2x= 80$

$\implies 2x = 100$

$\implies x = 50$

Hence, the value of x would be 50.