Question

The area of triangle formed by the points $(p,2-2p),(1-p,2p)$ and $(-4-p,6-2p)$ is $70$ sq. units. How many integral values of p are possible ?

• A. $2$
• B. $3$
• C. $4$
• D. None of these

Answer 1

Answer: (D)

None of these

Given the points of triangle, $A(p,2-2p),B(1-p,2p)$ and $C(-4-p,6-2p)$

Area =70sq.unit

$=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1\left)x_3\right(y_1-y_2\left)\right]=70\{p[2p-6+2p]+(1-p)[6-2p+2p]+(-4-p\left)[2-2p-2p]\right\}=140p(4p-6)+(1-p)\left(4\right)-\left[\right(4+p\left)\right(2-4p\left)\right]=1404p^2-6p+4-4p-[8-16p+2p-4p^2]=1404p^2-6p+4-4p-8+16p-2p+4p^2=1408p^2+4p-144=02p^2+p-36=0$
$2p^2+9p-8p-36=0$
$p(2p+9)-4(2p+9)=0$
$(2p+9)=0,(p-4)=0$

$p=-9/2,p=4$

$\therefore$ Only one integral value is possible.