If a ∆ ABC has vertices A (– 1, 7), B(– 7, 1) and C(5, – 5) , then...

Slope of line $BC$ is ${m}_{BC} = \dfrac{6}{- 12} = - \dfrac{1}{2}$

Slope of line $AD = \dfrac{1}{{m}_{BC}} = 2$

$∴$ Equation of line $AD$ is $y - 7 = 2 \left(x + 1\right)</p><p>$⇒ y = 2x + 9 . . . \left(1\right)$</p><p>Slope of line$AC$is${m}_{AC} = \dfrac{12}{- 6} = - 2$</p><p>Slope of line$BE = \dfrac{1}{{m}_{AC}} = \dfrac{1}{2}$</p><p>$∴$Equation of$BE$is$y - 1 = \dfrac{1}{2}\left(x + 7\right)

$⇒ y =$\dfrac{x}{2} + \dfrac{9}{2}$. . . \left(2\right)$

Comparing equations \left(1\right) \text{and }\left(2\right), we get</p><p>2x + 9 = $\dfrac{x + 9}{2}</p><p>$⇒ 4x + 18 = x + 9$</p><p>$⇒ 3x = - 9 ⇒ x = - 3$</p><p>$∴ y = 2x + 9 = 3

Therefore, orthocentre has co-ordinates \left(- 3, 3\right).