IIT-JEE 1978Question 1
The area of a triangle is $5$. Two of its vertices are $A(2,1)$ and $B(3,-2)$. The third vertex $C$ lies on $y=x+3$. Find $C$.
Points and Straight Lines
Answer$\left(-\dfrac32,\dfrac32\right)$ or $\left(\dfrac72,\dfrac{13}{2}\right)$
Solution
With $C=(t,t+3)$, area $=\frac12|4t-4|=2|t-1|=5$, so $t=\frac72$ or $-\frac32$: $C=\left(\frac72,\frac{13}{2}\right)$ or $\left(-\frac32,\frac32\right)$.