Question
The lines $L_{1}$ and $L_{2}$ denoted by $3x^{2} + 10xy + 8y^{2} + 14x + 22y + 15 = 0$ intersect at the point $P$ and have gradients $m_{1}$ and $m_{2}$ respectively. The acute angle between them is $\theta$. Which of following relations hold good:
Options
Solution
Sol. [Key Concept: The pair $3x^{2 + 10}xy + 8y^{2 + 14}x + 22y + 15 = 0$ factors as $(3x + 4y + 5)(x + 2y + 3)$; $m_{1}m_{2 =}\frac{38}{}$, $P = (1, - 2)$.]
$$\begin{matrix} (3x + 4y + 5)(x + 2y + 3) = 0 & \\ m_{1} & \ = \frac{- 3}{4},m_{2} = - \frac{1}{2} \Rightarrow m_{1}m_{2} = \frac{3}{8} \\ P & \ \equiv (1, - 2) \\ tan\theta & \ = \frac{- \frac{1}{2} + \frac{3}{4}}{1 + \frac{3}{8}} = \frac{2}{11} \\ sin\theta & \ = \frac{2}{5\sqrt{5}} \end{matrix}$$
✘ (a) actually $m_{1 +}m_{2 = -}\frac{5}{4}$; the option's $+ \frac{5}{4}$ has the wrong sign.
✔ (b) $m_{1}m_{2 =}\frac{38}{}$ from the factorisation.
✔ (c) $\theta = \sin^{- 1}\frac{2}{\sqrt{5}}$ — the acute angle between the lines.
✔ (d) at $P(1, - 2)$, abscissa + ordinate $= 1 + ( - 2) = - 1$.
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