First slide

Cartesian plane

Question

Two vertices of a triangle are $(4, - 3)$ and $(- 2, 5)$ . If the ortho centre of the triangle is at $(1, 2)$ , then the third vertex is

Easy

A

$(- 33, - 26)$
B

$(33, 26)$
C

$(26, 33)$
D

$(- 26, - 33)$

Solution

Let the third vertex be (h, k)
Now slope of AD is $\frac{(k - 2)}{(h - 1)}$ slope of BC is $\frac{(5 + 3)}{(- 2 - 4)} = - 4 / 3$ ,slope of BE is
$\frac{(- 3 - 2)}{(4 - 1)} = \frac{5}{3}$ And slope of AC is $\frac{(k - 5)}{(h + 2)}$ .
Since $A D ⊥ B C$ , so
$\frac{k - 2}{h - 1} \times \frac{- 4}{3} = - 1$
$\Rightarrow 3 h - 4 k + 5 = 0$ ……………(1)
Again since $B E ⊥ A C$ , so
$- \frac{5}{3} \times \frac{k - 5}{h + 2} = - 1$
$\Rightarrow 3 h - 5 k + 31 = 0$ ……………(2)
On solving (1) and (2) we get $h = 33, k = 26$ .
Hence the third vertex is $(33, 26)$

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