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

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(D) Let the third vertex be $(p, q)$. Since it lies on the line $y = x + 3$,we have $q = p + 3$ $(i)$.Given the area of the triangle is $5$,the area f

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

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(D) Let the third vertex be $(p, q)$. Since it lies on the line $y = x + 3$,we have $q = p + 3$ $(i)$.Given the area of the triangle is $5$,the area formula is:$\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 5$$\frac{1}{2} |2(-2 - q) + 3(q - 1) + p(1 - (-2))| = 5$$|2(-2 - q) + 3(q - 1) + 3p| = 10$$|-4 - 2q + 3q - 3 + 3p| = 10$$|q + 3p - 7| = 10$Substituting $q = p + 3$ into the equation:$|(p + 3) + 3p - 7| = 10$$|4p - 4| = 10$Case $1$: $4p - 4 = 10$ $\Rightarrow 4p = 14$ $\Rightarrow p = \frac{7}{2}$. Then $q = \frac{7}{2} + 3 = \frac{13}{2}$.Case $2$: $4p - 4 = -10$ $\Rightarrow 4p = -6$ $\Rightarrow p = -\frac{3}{2}$. Then $q = -\frac{3}{2} + 3 = \frac{3}{2}$.The possible vertices are $\left( \frac{7}{2}, \frac{13}{2} \right)$ or $\left( -\frac{3}{2}, \frac{3}{2} \right)$.Comparing with the given options,the correct option is $D$.

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

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