Straight LineHardBloom L3
Question
If the lines $x + y + 1 = 0;4x + 3y + 4 = 0$ and $x + \alpha y + \beta = 0$, where $\alpha^{2} + \beta^{2} = 2$, are concurrent then:
Options
A.$\alpha = 1,\beta = - 1$
B.$\alpha = 1,\beta = \pm 1$
C.$\alpha = - 1,\beta = \pm 1$
D.$\alpha = \pm 1,\beta = 1$
Solution
Sol. [Key Concept: Three concurrent lines ⇒ determinant of coefficients $= 0$ fixes $\alpha,\beta$.]
Lines are
$$x + y + 1 = 0;4x + 3y + 4 = 0$$
and
$$x + \alpha y + \beta = 0$$
where, $\ \alpha^{2} + \beta^{2} = 2$
$$\begin{matrix} & \left| \begin{matrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{matrix} \right| = 0 \\ & 1(3\beta - 4\alpha) - 1(4\beta - 4) + 1(4\alpha - 3) \\ = & 3\beta - 4\alpha - 4\beta + 4 + 4\alpha - 3 \\ = & \ - \beta + 1 = 0\ \Rightarrow \ \beta = 1 \\ \therefore & \alpha = \pm 1 \end{matrix}$$
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