The equation of the sides of a triangle are $x - 3 y = 0, 4 x + 3 y = 5$ and $3 x + y = 0$ . The line $3 x - 4 y = 0$ passes through:

the incentre

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the centroid

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the orthocentre

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the circumcentre

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Solution

The correct option is D the orthocentre
Given equation of line are,
$l_{1} : x - 3 y = 0$ ,Slope $m_{1} = \frac{1}{3}$
$l_{2} : 4 x + 3 y = 5$ slope $m_{2} = - \frac{4}{3}$
$l_{3} : 3 x + y = 0$ slope $m_{3} = - 3$

Here $m_{1} \times m_{3} = - 1$ Line $l_{1} ⊥ l_{3}$

Slope of given line: $3 x - 4 y = 0$ , $m = \frac{3}{4}$

$h e r e m \times m_{2} = - 1$
$∴$ Given line is perpendicular to Line $l_{2}$

Therefore The given line must pass throigh the orthocenter

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Similar questions

Equations to the sides of a triangle are $x - 3 y = 0$ , $4 x + 3 y = 5$ and $3 x + y = 0$ . The line $3 x - 4 y = 0$ passes through the

x - 3 y = 0, 4 x + 3 y = 5

3 x + y = 0

3 x - 4 y = 0

x y + 3 x + 3 y + 9 = 0

3 x + 4 y - 5 = 0

2 x + 3 y = 0; 2 x + 3 y - 5 = 0

3 x - 4 y = 0

3 x - 4 y = 3 .

x^{2} + y^{2} - 6 x + 4 y = 12

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