Question 2 of 9 · Pair of Straight Lines

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NIMCET 2010

If the two pairs of lines $x^{2} - 2 m x y - y^{2} = 0$ and $x^{2} - 2 n x y - y^{2} = 0$ are such that one of them represents the bisectors of the angles between the other, then

Answer (Detailed Solution Below)View answer ▾

Answer: Option A — $mn + 1 = 0$

Detailed Solution

The bisectors of the lines $a x^{2} + 2 h x y + b y^{2} = 0$ are $\frac{x ^{2} - y ^{2}}{a - b} = \frac{x y}{h}$ .

For $x^{2} - 2 n x y - y^{2} = 0$ : $a = 1$ , $b = - 1$ , $h = - n$ , so the bisectors are $\frac{x ^{2} - y ^{2}}{2} = \frac{x y}{- n}$ , i.e. $x^{2} + \frac{2}{n} x y - y^{2} = 0$ .

This must be $x^{2} - 2 m x y - y^{2} = 0$ , so $- 2 m = \frac{2}{n}$ , giving $mn = - 1$ , i.e. $mn + 1 = 0$ .

✓ Check

For

n = 1

the lines

x^{2} - 2 x y - y^{2} = 0

make angles

22.5^{\circ}

and

- 67.5^{\circ}

with the

x

-axis; their bisectors at

- 22.5^{\circ}

and

67.5^{\circ}

are

x^{2} + 2 x y - y^{2} = 0

, i.e.

m = - 1

, and

mn + 1 = 0

⚠ Common trap

(B)

mn - 1 = 0

comes from taking

h = n

instead of

h = - n

(the

x y

coefficient is

2 h = - 2 n

Quick tip

$\frac{x ^{2} - y ^{2}}{a - b} = \frac{x y}{h} \Rightarrow mn = - 1$

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Pair of Straight Lines — Preparation

This question tests your understanding of Pair of Straight Lines, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

A pair of straight lines can be represented by a single second-degree equation. This topic covers the homogeneous equation ax^2 + 2hxy + by^2 = 0, which represents two lines through the origin, and the general equation ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0. Exam questions test the condition for the general equation to represent a pair of lines, the angle between the lines (tan theta = 2 sqrt(h^2 - ab) / |a + b|), the conditions for the lines to be perpendicular (a + b = 0) or coincident (h^2 = ab), and the sum and product of the slopes (-2h/b and a/b). Knowing these results directly turns most questions into short calculations. Practise the pair of straight lines questions below, each with a step-by-step solution.

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If the two pairs of lines $x^{2} - 2 m x y - y^{2} = 0$ and $x^{2} - 2 n x y - y^{2} = 0$ are such that one of them represents the bisectors of the angles between the other, then

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If the two pairs of lines $x^{2} - 2 m x y - y^{2} = 0$ and $x^{2} - 2 n x y - y^{2} = 0$ are such that one of them represents the bisectors of the angles between the other, then