A Straight line through p-15- -10- meets the straight line x - y -1 - 0- x - 2y - 5 and x - 3y - 7 respectively at A-B and C- If 12 PA - 40 PB - 52 PC - then prove that the straight line passes through origin-

Parametric form of straight line is. x x_1cosθ = y y_1sinθ = l Given (x_1 , y_1) = A( 15, 10) ∴ x = r cosθ 15 , y = r sinθ 10 for L_ ...

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Byju's Answer

Standard VIII

Mathematics

The Centroid

A Straight li...

Question

A Straight line through p(-15, -10), meets the straight line x - y -1 = 0, x + 2y = 5 and x + 3y = 7 respectively at A,B and C. If $\frac{12}{P A} + \frac{40}{P B} = \frac{52}{P C},$ then prove that the straight line passes through origin.

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Solution

Parametric form of straight line is.

$\frac{x - x_{1}}{cos θ} = \frac{y - y_{1}}{sin θ} = l$

Given $(x_{1}, y_{1}) = A (- 15, - 10)$

$∴ x = r cos θ - 15, y = r sin θ - 10$

for $L_{1} \Rightarrow x - y - 1 = 0$

$L_{1} = r_{1} cos θ - 15 - r_{1} sin θ + 10 - 1 = 0$

$r_{1} = \frac{6}{cos θ - sin θ} = A B$

Similarly for $L_{2} & L_{3}$

$L_{2} : r_{2} cos θ - 15 + 2 r_{2} sin θ - 20 = 5$

$r_{2} = \frac{40}{cos θ + 2 sin θ} = A C$

$L_{3} : r_{3} cos θ - 15 + 3 r_{3} sin θ - 30 = 7$

$r_{3} = \frac{52}{cos θ + 3 sin θ} = A D$

$\frac{12}{A B} + \frac{40}{A C} = \frac{52}{A D}$

$\frac{12}{r_{1}} + \frac{40}{r_{2}} = \frac{52}{r_{3}}$

$= 2 (cos θ - sin θ) + cos θ + 2 sin θ = cos θ + 3 sin θ$

$2 cos θ = 3 sin θ$

$tan θ = 2 / 3$

$m = 2 / 3$

Equation of the passing through $A (- 15, - 10)$ with slope $2 / 3$

$y + 10 = \frac{2}{3} (x + 15)$

$3 y + 30 = 2 x + 30$

$3 y = 2 x$

$∴$ Therefore it passes through origin.

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A Straight line through p(-15, -10), meets the straight line x - y -1 = 0, x + 2y = 5 and x + 3y = 7 respectively at A,B and C. If 12PA+40PB=52PC, then prove that the straight line passes through origin.

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