Question: If the angle between the pair of lines represented by the equation 2x² + kxy +3y² = 0 is 45°, then find the positive value of k and also find the separate equations of the lines.

Solution:
Given,

Equation of a pair of straight line is 2x^2 + kxy +3y^2 = 0 - (i)
Angle between the lines (\theta) = 45°

Comparing equation (i) with ax^2 + 2hxy + y^2 = 0
a = 2, h = \frac{h}{2}, \; and \; b = 3

Angle between the lines is;
tan \theta = \pm \dfrac{2 \sqrt{h^2 - ab}}{a + b}

or, tan 45° = \pm \dfrac{2 \sqrt{(\frac{k}{2})^2 - 2×3}}{2+3}

or, 1 = \pm \dfrac{2 \sqrt{\frac{k^2}{4} - 6}}{5}

or, 5 = \pm 2 \sqrt{ \frac{k^2 - 24}{4}}

or, 5 = \pm \frac{2}{2} \sqrt{k^2 -24}

or, 5 = \pm \sqrt{k^2 -24}

Taking positive sign,
or, 5 = \sqrt{k^2 -24}
or, 5^2 = k^2 -24
or, 25 = k^2 -24
or, k^2 = 25+24
or, k^2 = 49
or, k^2 = 7^2
\therefore k = 7

Taking negative sign,
or, 5 = - \sqrt{k^2 -24}
[Squaring both sides]
or, 5^2 = (- \sqrt{k^2 -24})^2
or, 25 = k^2 -24
or, k^2 = 49
\therefore k = 7

So, the value of k is 7.

Our single equation of straight lines becomes 2x^2 +7xy +3y^2 = 0.

or, 2x^2+7xy +3y^2 = 0
or, 2x^2 +(6+1)xy +3y^2 = 0
or, 2x^2 +6xy +xy +3y^2 = 0
or, 2x(x +3y) + y(x +3y) = 0
or, (x +3y)(2x + y) = 0

Therefore, the required separate equations of straight lines are (x +3y = 0) and (2x + y = 0).

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