The point of intersection of the lines given by the equation $3 x^{2} + 10 x y + 3 y^{2} - 15 x - 21 y + 18 = 0$ is :

(\frac{15}{8}, \frac{3}{8})

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(\frac{15}{6}, \frac{1}{6})

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(\frac{115}{48}, \frac{1}{16})

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(\frac{1}{16}, \frac{115}{48})

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Solution

The correct option is A $(\frac{15}{8}, \frac{3}{8})$
$S : 3 x^{2} + 10 x y + 3 y^{2} - 15 x - 21 y + 18 = 0$
partial differentiation w.r.to x
$\Rightarrow 6 x + 10 y - 15 = 0 \to (i i)$
partial diffrentiation of equation
$\begin{matrix} (i) w . r . t o . x 10 x + 6 y - 21 = 0 \to (i i i) m u l t i p l y e q . (i i) b y 10 & e q . (3) b y 6 60 x + 100 y - 150 = 0 \to (i v) 60 x + 36 y - 126 = 0 \to (v) (v) - (i v) 36 x - 100 y - 126 + 150 = 0 64 y + - 24 y = \frac{24}{64} = \frac{3}{8} \end{matrix}$
put the value of y in equation (i)
$\begin{matrix} \Rightarrow 6 x + 10 \times \frac{3}{8} - 15 = 0 \Rightarrow 6 x = 15 - \frac{15}{4} \Rightarrow 6 x = \frac{45}{4} \Rightarrow x = \frac{45}{4 \times 6} = \frac{15}{8} p o i n t o f i n t e r sec t i o n = (\frac{15}{8}, \frac{3}{8}) \end{matrix}$

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