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If two vertices of a triangle are $(5, -1)$ and $( - 2, 3)$ and its orthocentre is at $(0, 0)$, then the third vertex
$\left( {4, - 7} \right)$
$(-4, - 7)$
$(-4, 7)$
$(4, 7)$
Solution
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 $AH$)$ \times $ (Slope of $BC$) $=-1$
$ \Rightarrow \left( {\frac{{ – 1 – 0}}{{5 – 0}}} \right)\left( {\frac{{b – 3}}{{a + 2}}} \right) = – 1$
$ \Rightarrow b – 3 = 5\left( {a + 2} \right)\,\,\,\,\,\,\,\,\,….\left( 1 \right)$
Similariy,
(Slope of $BH$ )$ \times $(Slop of $AC$ )$=-1$
$ \Rightarrow – \left( {\frac{3}{2}} \right) \times \left( {\frac{{b + 1}}{{a – 5}}} \right) = – 1$
$ \Rightarrow 3b + 3 = 2a – 10$
$ \Rightarrow 3b – 2a + 13 = 0\,\,\,\,\,\,\,\,\,\,…..\left( 2 \right)$
On solving equation $(1)$ and $(2)$ we get $a=-4$, $b=-7$
Hence, third vertex is $(-4,-7)$
