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

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Question

Let the third vertex of $\Delta ABC$ be $(a,b)$.

or thocenter $=H(0,0)$

Let $A(5,-1)$ and $B(-2,3)$ be other two vertices of $\Delta A

Vedclass · Accepted Answer

$(-4, - 7)$

Vedclass · Answer

Let the third vertex of $\Delta ABC$ be $(a,b)$.

or thocenter $=H(0,0)$

Let $A(5,-1)$ and $B(-2,3)$ be other two vertices of $\Delta ABC$.

Now, (Slpoe of $AH$)$ 	imes $ (Slope of $BC$) $=-1$

$ \Rightarrow \left( {\frac{{ - 1 - 0}}{{5 - 0}}} ight)\left( {\frac{{b - 3}}{{a + 2}}} ight) =  - 1$

$ \Rightarrow b - 3 = 5\left( {a + 2} ight)\,\,\,\,\,\,\,\,\,....\left( 1 ight)$

Similariy,

(Slope of $BH$ )$ 	imes $(Slop of $AC$ )$=-1$

$ \Rightarrow  - \left( {\frac{3}{2}} ight) 	imes \left( {\frac{{b + 1}}{{a - 5}}} ight) =  - 1$

$ \Rightarrow 3b + 3 = 2a - 10$

$ \Rightarrow 3b - 2a + 13 = 0\,\,\,\,\,\,\,\,\,\,.....\left( 2 ight)$

On solving equation $(1)$ and $(2)$ we get $a=-4$, $b=-7$

Hence, 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

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