The equation x²-7 x y+12 y²=0 represents a

Correct Answer is : pair of non-perpendicular straight lines

pair of parallel straight lines

pair of perpendicular straight lines

pair of non-perpendicular straight lines

Correct Answer is : pair of non-perpendicular straight lines

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CGPET 2012 Solved Paper

Section: Mathematics

Question : 149 of 150

Marks: +1, -0

The equation

x^2-7 x y+12 y^2=0

Solution:

Given equation is

x^2-7 x y+12 y^2=0

On comparing with

a x^2+2 h x y+b y^2=0

\therefore \; a=1, h=-\frac{7}{2}, b=12

\therefore \; m_1+m_2=-\frac{2h}{b}

\text{ and } \; \; m_1 m_2 = \frac{a}{b}

\Rightarrow m_1+m_2 = \frac{7}{12}

. . . (i)

Now,

\; \; (m_1-m_2) = \sqrt{ (m_1+m_2)^2 - 4 m_1 m_2 }

\; = \sqrt{ \left( \frac{7}{12} \right)^2 - 4 \times \frac{1}{12} }

\; = \sqrt{ \frac{49}{144} - \frac{4}{12} }

\; = \sqrt{ \frac{49-48}{144} }

= \; \sqrt{ \frac{1}{144} } = \; \frac{1}{12}

\Rightarrow \; \; m_1-m_2 = \frac{1}{12}

. . . (ii)

On solving Eqs. (i) and (ii), we get

\; m_1 = \frac{1}{3}

\; m_2 = \frac{3}{12}

\; \text{ Now, } \; \tan \theta = \frac{m_1-m_2}{1+m_1 m_2} = \frac{ \frac{1}{3} - \frac{3}{12} }{ 1 + \frac{1}{3} \times \frac{3}{12} }

\tan \theta = \frac{ \frac{1}{12} }{ \frac{13}{12} } = \frac{1}{13} \neq 0

Hence, given equation is a non-perpendicular straight lines.

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