Question

A triangle has

a x^{2} + 2 h x y + b y^{2} = 0

for two of its sides and the point

(α, β)

for the orthocentre . Prove that the third side is given by

(a + b) (α x + β y) = a β^{2} - 2 h α β + b α^{2}

.

Accepted Answer

Given, combined equation of two sides $O A & O B$ is $a x^{2} + 2 h x y + b y^{2} = 0$ and $H (α, β)$ be the orthocentre.

Let $y = m_{1} x & y = m_{2} x$ be the separate equation of $O A & O B$

$∴ m_{1} + m_{2} = - \frac{2 h}{b} & m_{1} \cdot m_{2} = \frac{a}{b}$

$∵ A H$ is perpendicular to $O B$ .

$∴ Slope of A H = - \frac{1}{m_{2}}$

$∴$ Equation of $A H$ is

$y - β = - \frac{1}{m_{2}} (x - α)$

Substituting $y = m_{1} x$

$m_{1} x - β = - \frac{1}{m_{2}} (x - α)$

$\Rightarrow m_{1} m_{2} x - m_{2} β = - x + α$

$\Rightarrow \frac{a}{b} x + x = α + m_{2} β$

$\Rightarrow x = \frac{b α + m_{2} b β}{a + b} & y = m_{1} x = \frac{m_{1} b α + m_{1} m_{2} b β}{a + b} = \frac{m_{1} b α + a β}{a + b}$

$∴$ Co-ordinate of $A \equiv (\frac{b α + m_{2} b β}{a + b}, \frac{m_{1} b α + a β}{a + b})$

$Slope of O H = \frac{β - 0}{α - 0} = \frac{β}{α}$

$∴ Slope of A B = \frac{- α}{β} [∵ O H ⊥ A B]$

Equation of line $A B$ is

$y - (\frac{m_{1} b α + a β}{a + b}) = \frac{- α}{β} (x - (\frac{m_{2} b β + b α}{a + b}))$

$\Rightarrow β y - (\frac{m_{1} b α β + a β^{2}}{a + b}) = - α x + \frac{m_{2} b α β + b α^{2}}{a + b}$

$\Rightarrow α x + β y = \frac{m_{2} b α β + b α^{2} + m_{1} b α β + a β^{2}}{a + b}$

$\Rightarrow (α x + β y) (a + b) = b α^{2} + (m_{1} + m_{2}) b α β + a β^{2}$

$\Rightarrow (α x + β y) (a + b) = b α^{2} - \frac{2 h}{b} \cdot b α β + a β^{2}$

$∴ (α x + β y) (a + b) = b α^{2} - 2 α β + a β^{2}$ .

Dr. SK Goyal Solutions for Chapter: Pair of Straight Lines, Exercise 6: EXERCISE ON LEVEL-I

Dr. SK Goyal Mathematics Solutions for Exercise - Dr. SK Goyal Solutions for Chapter: Pair of Straight Lines, Exercise 6: EXERCISE ON LEVEL-I

Questions from Dr. SK Goyal Solutions for Chapter: Pair of Straight Lines, Exercise 6: EXERCISE ON LEVEL-I with Hints & Solutions

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