The area of a triangle is 5 and two of its ve | vedclass

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(A) Let the third vertex be $C(h, k)$.Since $C$ lies on the line $y = x + 3$,we have $k = h + 3$  .........$(1)$The area of the triangle with vertices

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$\left( \frac{7}{2}, \frac{13}{2} \right)$

Vedclass · Answer

(A) Let the third vertex be $C(h, k)$.Since $C$ lies on the line $y = x + 3$,we have $k = h + 3$  .........$(1)$The area of the triangle with vertices $(h, k), (2, 1), (3, -2)$ is given by:$\frac{1}{2} |h(1 - (-2)) + 2(-2 - k) + 3(k - 1)| = 5$$|h(3) + 2(-2 - k) + 3(k - 1)| = 10$$|3h - 4 - 2k + 3k - 3| = 10$$|3h + k - 7| = 10$Case $1$: $3h + k - 7 = 10 \implies 3h + k = 17$  .........$(2)$Substitute $k = h + 3$ into $(2)$:$3h + (h + 3) = 17 \implies 4h = 14 \implies h = \frac{7}{2}$Then $k = \frac{7}{2} + 3 = \frac{13}{2}$. So,$C = \left( \frac{7}{2}, \frac{13}{2} \right)$.Case $2$: $3h + k - 7 = -10 \implies 3h + k = -3$  .........$(3)$Substitute $k = h + 3$ into $(3)$:$3h + (h + 3) = -3 \implies 4h = -6 \implies h = -\frac{3}{2}$Then $k = -\frac{3}{2} + 3 = \frac{3}{2}$. So,$C = \left( -\frac{3}{2}, \frac{3}{2} \right)$.

The area of a triangle is $5$ and two of its vertices are $A(2, 1)$ and $B(3, -2)$. The third vertex,which lies on the line $y = x + 3$,is:

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