Question

The vertices of a triangle ABC are $(\lambda ,2-\lambda ),(-\lambda +1,2\lambda )$ and $(-4-\lambda ,6-2\lambda )$. If its area is 70 units${}^2$, find the number of integral values of $\lambda$.

• A. $1$
• B. $2$
• C. $4$
• D. $10$

Answer 1

The correct option is A $1$

Given the vertices of $△$ABC

$A(\lambda ,2-\lambda ),B(-\lambda +1,2\lambda ),C(-4-\lambda ,6-2\lambda )$

Given total area $=70uni{t}^2$

Area

$=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]\Rightarrow \frac{1}{2}[[2\lambda -6+2\lambda ]+(1-\lambda )[6-2\lambda -2+\lambda ]-(4+\lambda \left)[2-\lambda -2\lambda ]\right]=70\lambda (4\lambda -6)+(1-\lambda )(4-\lambda )-(4+\lambda )(2-3\lambda )=2\times 704{\lambda }^2-6\lambda +4-x-4\lambda +{\lambda }^2-[8-12\lambda +2\lambda -3{\lambda }^2]=1405{\lambda }^2-10\lambda +4-8+12\lambda -2\lambda +3{\lambda }^2=1408{\lambda }^2+\lambda -4-2\lambda =140⟹8{\lambda }^2-\lambda =136\lambda =4.186,-4.061$

$\therefore$ No.of Intergral value of $\lambda =1[i.e,\lambda =4.186]$