Question

If the area of the triangle formed by the points$(x,2x),(-2,6)$ and$(3,1)$ is $5$ square units , then $x$
(a)$23$
(b) $35$
(c)$3$
(d) $5$

Answer 1

We have the co-ordinates of the vertices of the triangle as$A(x,2x),B(-2,6),C(3,1)$ which has an area of 5 sq.units.

In general, if $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ are non-collinear points then area of the triangle formed is given by,
$ar\left(\Delta ABC\right)=\left|\frac{1}{2}{x}_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$

So,
$5=\left|\frac{1}{2}x\right(6-1)-2(1-2x)+3(2x-6\left)\right|$
$5=\frac{1}{2}|15x-20|$
$5=\frac{5}{2}|3x-4|$
Simplify the modulus function to get,
$3x-4=\pm 2$
$3x=4\pm 2$
$x=\frac{4\pm 2}{3}$
$x=\frac{4+2}{3}$ and $x=\frac{4-2}{3}$
$x=\frac{6}{3}$ and $x=\frac{2}{3}$

Therefore,
$x=2,\frac{2}{3}$
So, the answer is (a)