If a $\triangle \mathrm{ABC}$ have vertices $\mathrm{A}(-1,7), \mathrm{B}(-7,1)$ and $\mathrm{C}(5,-5)$, then its orthocenter has coordinates:

(A) $\left(\frac{3}{5},-\frac{3}{5}\right)$

(D) $\left(-\frac{3}{5}, \frac{3}{5}\right)$

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Straight Lines

If a

△ ABC

have vertices

A (- 1, 7), B (- 7, 1)

and

C (5, - 5)

, then its orthocenter has coordinates:

(A) $(\frac{3}{5}, - \frac{3}{5})$
(B) $(- 3, 3)$ ✓ Correct
(C) $(3, - 3)$
(D) $(- \frac{3}{5}, \frac{3}{5})$

Answer: (B)

Solution

The orthocentre is where two altitudes meet. Side BC has slope

\frac{- 5 - 1}{5 + 7} = - \frac{1}{2}

, so the altitude from A, perpendicular to it, has slope

2

y - 7 = 2 (x + 1)

, that is

y = 2 x + 9

. Side AC has slope

\frac{- 5 - 7}{5 + 1} = - 2

, so the altitude from B has slope

\frac{1}{2}

y - 1 = \frac{1}{2} (x + 7)

, that is

y = \frac{x}{2} + \frac{9}{2}

. Solving the two,

2 x + 9 = \frac{x}{2} + \frac{9}{2}

gives

x = - 3

and then

y = 3

. The orthocentre is

(- 3, 3)

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option

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