If the line l_1: 3y - 2x = 3 is the angular b | vedclass

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(A) The point of intersection of $l_1: 3y - 2x = 3$ and $l_2: x - y + 1 = 0$ is found by solving the system:$3y - 2x = 3$$x - y = -1 \Rightarrow x = y

Vedclass · Accepted Answer

$348$

Vedclass · Answer

(A) The point of intersection of $l_1: 3y - 2x = 3$ and $l_2: x - y + 1 = 0$ is found by solving the system:$3y - 2x = 3$$x - y = -1 \Rightarrow x = y - 1$Substituting $x$ in the first equation: $3y - 2(y - 1) = 3$ $\Rightarrow 3y - 2y + 2 = 3$ $\Rightarrow y = 1$.Then $x = 1 - 1 = 0$. So,the point of intersection $P$ is $(0, 1)$.Since $l_1$ is the angle bisector,$P$ must also lie on $l_3: \alpha x + \beta y + 17 = 0$.Substituting $(0, 1)$ into $l_3$: $\alpha(0) + \beta(1) + 17 = 0 \Rightarrow \beta = -17$.Now,consider a point $Q(x_0, y_0)$ on $l_2$. Let $Q = (-1, 0)$. The reflection $Q'$ of $Q$ about the line $l_1: 2x - 3y + 3 = 0$ must lie on $l_3$.The formula for reflection $(x', y')$ of $(x_0, y_0)$ about $ax + by + c = 0$ is $\frac{x' - x_0}{a} = \frac{y' - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$.For $Q(-1, 0)$ and $l_1: 2x - 3y + 3 = 0$:$\frac{x' + 1}{2} = \frac{y' - 0}{-3} = -2 \frac{2(-1) - 3(0) + 3}{2^2 + (-3)^2} = -2 \frac{1}{13} = -\frac{2}{13}$.$x' = -1 - \frac{4}{13} = -\frac{17}{13}$ and $y' = 0 + \frac{6}{13} = \frac{6}{13}$.Since $Q'(-\frac{17}{13}, \frac{6}{13})$ lies on $l_3: \alpha x - 17y + 17 = 0$:$\alpha(-\frac{17}{13}) - 17(\frac{6}{13}) + 17 = 0$.Dividing by $17$: $-\frac{\alpha}{13} - \frac{6}{13} + 1 = 0$ $\Rightarrow -\alpha - 6 + 13 = 0$ $\Rightarrow \alpha = 7$.Finally,$\alpha^2 + \beta^2 - \alpha - \beta = 7^2 + (-17)^2 - 7 - (-17) = 49 + 289 - 7 + 17 = 348$.

If the line $l_1: 3y - 2x = 3$ is the angular bisector of the lines $l_2: x - y + 1 = 0$ and $l_3: \alpha x + \beta y + 17 = 0$,then $\alpha^2 + \beta^2 - \alpha - \beta$ is equal to

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