Question

If the area of the triangle with vertices (x, 0),(1,1) and (0,2) is 4 square units then the value of x is

Answer 1

Hi Mate !!


Given :- area of ∆ABC is 4


x1 = x
x2 = 1
x3 = 0


y1 = 0
y2 = 1
y3 = 2

• area of ∆ABC :-

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

$4 = \frac{1}{2} (x(1 - 2) + 1(2 - 0) + 0(0 - 1))$


$8 = x - 2x + 2$


$8 - 2 = - x$


$6 = - x$


$x = - 6$



So, the value of x is ( - 6 )

Answer 2

Answer:

x=-6

Step-by-step explanation:

Coordinates of vertices of triangle are :

$(x_1,y_1)=(x,0)\\(x_2,y_2)=(1,1)\\(x_3,y_3)=(0,2)$

Area of triangle = $\frac{1}{2} ( x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2))$

We are given that the area of triangle is 4 sq.units

Substitute the values

$4=\frac{1}{2} (x(1-2) +1(2-0) + 0(0-1))$

$x=-6$

Hence the value of x is -6