The Straight lines represented by the equation $135 x^{2} - 136 x y + 33 y^{2} = 0$ are equally inclined to the line

x - 2 y = 7

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x + 2 y = 7

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x - 2 y = 4

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3 x + 2 y = 4

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Solution

The correct option is B $x + 2 y = 7$
Give pair of lines is $135 x^{2} - 136 x y + 33 y^{2} = 0$ ...(1)

The equation of bisector of angles between pair of lines (1) is

$\frac{x^{2} - y^{2}}{a - b} = \frac{x y}{h} \Rightarrow \frac{x^{2} - y^{2}}{135 - 33} = \frac{x y}{- 68}$

$\Rightarrow 2 x^{2} + 3 x y - 2 y^{2} = 0 \Rightarrow (x + 2 y) (2 x - y) = 0$

One of the bisectors is $x + 2 y = 0$ which is parallel to the line $x + 2 y = 7$ .

Hence, the line $x + 2 y = 7$ is equally inclined to the given lines.

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\frac{x}{2} = \frac{y}{3} = \frac{z}{4}

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Find the separate equation of the lines represented by the following equation:

$(x - 2)^{2} - 3 (x - 2) (y + 1) + 2 (y + 1)^{2} = 0$

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