If two vertices of a triangle are (5, -1) and | vedclass

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Question

(B) Let the third vertex of $\Delta ABC$ be $C(a, b)$.Let $A(5, -1)$ and $B(-2, 3)$ be the other two vertices.The orthocentre $H$ is at $(0, 0)$.Since

Vedclass · Accepted Answer

$(-4, -7)$

Vedclass · Answer

(B) Let the third vertex of $\Delta ABC$ be $C(a, b)$.Let $A(5, -1)$ and $B(-2, 3)$ be the other two vertices.The orthocentre $H$ is at $(0, 0)$.Since $AH \perp BC$,the product of their slopes is $-1$:$\left( \frac{-1 - 0}{5 - 0} ight) 	imes \left( \frac{b - 3}{a - (-2)} ight) = -1$$\Rightarrow \left( \frac{-1}{5} ight) 	imes \left( \frac{b - 3}{a + 2} ight) = -1$$\Rightarrow b - 3 = 5(a + 2)$ $\Rightarrow b - 3 = 5a + 10$ $\Rightarrow 5a - b + 13 = 0 \dots (1)$Similarly,since $BH \perp AC$,the product of their slopes is $-1$:$\left( \frac{3 - 0}{-2 - 0} ight) 	imes \left( \frac{b - (-1)}{a - 5} ight) = -1$$\Rightarrow \left( \frac{3}{-2} ight) 	imes \left( \frac{b + 1}{a - 5} ight) = -1$$\Rightarrow 3(b + 1) = 2(a - 5)$ $\Rightarrow 3b + 3 = 2a - 10$ $\Rightarrow 2a - 3b - 13 = 0 \dots (2)$Multiplying equation $(1)$ by $3$:$15a - 3b + 39 = 0 \dots (3)$Subtracting equation $(2)$ from $(3)$:$(15a - 2a) + (-3b - (-3b)) + (39 - (-13)) = 0$$13a + 52 = 0 \Rightarrow a = -4$Substituting $a = -4$ into equation $(1)$:$5(-4) - b + 13 = 0$ $\Rightarrow -20 - b + 13 = 0$ $\Rightarrow b = -7$Thus,the third vertex is $(-4, -7)$.

If two vertices of a triangle are $(5, -1)$ and $(-2, 3)$ and its orthocentre is at $(0, 0)$,then the third vertex is:

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