Bisector of angle between two lines Questions in English

(C) Let the lines be $L_1: 2x + 3y + 4 = 0$ and $L_2: 3x - 2y + 4 = 0$.
The length of the perpendicular from $(a, b)$ to $L_1$ is $d_1 = \frac{|2a + 3b + 4|}{\sqrt{2^2 + 3^2}} = \frac{|2a + 3b + 4|}{\sqrt{13}}$.
The length of the perpendicular from $(a, b)$ to $L_2$ is $d_2 = \frac{|3a - 2b + 4|}{\sqrt{3^2 + (-2)^2}} = \frac{|3a - 2b + 4|}{\sqrt{13}}$.
Given $d_1 = d_2$,we have $|2a + 3b + 4| = |3a - 2b + 4|$.
This implies $2a + 3b + 4 = 3a - 2b + 4$ or $2a + 3b + 4 = -(3a - 2b + 4)$.
Case $1$: $2a + 3b + 4 = 3a - 2b + 4 \Rightarrow a - 5b = 0$.
Case $2$: $2a + 3b + 4 = -3a + 2b - 4 \Rightarrow 5a + b + 8 = 0$.
Thus,the locus of $(a, b)$ is $x - 5y = 0$ or $5x + y + 8 = 0$.

(A) The third side of an isosceles triangle is perpendicular to the angle bisector of the two equal sides. However,in this specific geometry,the third side is parallel to the angle bisector of the given lines $7x-y+3=0$ and $x+y-3=0$ if the vertex is the intersection point.
First,find the intersection point of the two lines: $7x-y+3=0$ and $x+y-3=0$. Adding them gives $8x=0$,so $x=0$,which implies $y=3$. The vertex is $(0,3)$.
The angle bisectors are given by $\frac{7x-y+3}{\sqrt{7^2+(-1)^2}} = \pm \frac{x+y-3}{\sqrt{1^2+1^2}}$.
$\frac{7x-y+3}{\sqrt{50}} = \pm \frac{x+y-3}{\sqrt{2}} \Rightarrow 7x-y+3 = \pm 5(x+y-3)$.
Case $1$: $7x-y+3 = 5x+5y-15$ $\Rightarrow 2x-6y+18=0$ $\Rightarrow x-3y+9=0$.
Case $2$: $7x-y+3 = -5x-5y+15$ $\Rightarrow 12x+4y-12=0$ $\Rightarrow 3x+y-3=0$.
The third side is perpendicular to these bisectors.
For $x-3y+9=0$,the perpendicular line is $3x+y+k=0$. Passing through $(2,-5)$: $3(2)+(-5)+k=0$ $\Rightarrow 6-5+k=0$ $\Rightarrow k=-1$. So,$3x+y-1=0$.
For $3x+y-3=0$,the perpendicular line is $x-3y+k=0$. Passing through $(2,-5)$: $2-3(-5)+k=0$ $\Rightarrow 2+15+k=0$ $\Rightarrow k=-17$. So,$x-3y=17$.
Comparing with options,$x-3y=17$ is present.

(C) Let the slope of the required line be $m$.
Since the line $x+y+1=0$ is the angle bisector,the angle between the bisector and the given line $2x+3y-4=0$ must be equal to the angle between the bisector and the required line.
The slope of the bisector is $m_1 = -1$. The slope of the given line is $m_2 = -2/3$.
The tangent of the angle $\theta$ between the bisector and the given line is:
$\tan \theta = \left| \frac{-2/3 - (-1)}{1 + (-2/3)(-1)} \right| = \left| \frac{1/3}{1 + 2/3} \right| = \left| \frac{1/3}{5/3} \right| = \frac{1}{5}$.
Now,let the slope of the required line be $m$. The tangent of the angle between the bisector and the required line is:
$\tan \theta = \left| \frac{m - (-1)}{1 + m(-1)} \right| = \left| \frac{m+1}{1-m} \right|$.
Equating the two:
$\left| \frac{m+1}{1-m} \right| = \frac{1}{5} \implies \frac{m+1}{1-m} = \frac{1}{5}$ or $\frac{m+1}{1-m} = -\frac{1}{5}$.
Case $1$: $5m+5 = 1-m \implies 6m = -4 \implies m = -2/3$ (This is the given line).
Case $2$: $5m+5 = -1+m \implies 4m = -6 \implies m = -3/2$ (This is the required line).
The point of intersection of $x+y+1=0$ and $2x+3y-4=0$ is found by solving the system:
$2x+2y = -2$ and $2x+3y = 4$.
Subtracting gives $y = 6$,and substituting back gives $x = -7$.
The equation of the line with slope $m = -3/2$ passing through $(-7, 6)$ is:
$y - 6 = -\frac{3}{2}(x + 7) \implies 2y - 12 = -3x - 21 \implies 3x + 2y + 9 = 0$.

(A) The vertices of the triangle are $A(1, 7)$,$B(-5, -1)$,and $C(-1, 2)$.
The equation of line $AB$ passing through $(1, 7)$ and $(-5, -1)$ is:
$y - 7 = \frac{-1 - 7}{-5 - 1}(x - 1)$ $\Rightarrow y - 7 = \frac{-8}{-6}(x - 1)$ $\Rightarrow y - 7 = \frac{4}{3}(x - 1)$
$3y - 21 = 4x - 4 \Rightarrow 4x - 3y + 17 = 0$.
The equation of line $BC$ passing through $(-5, -1)$ and $(-1, 2)$ is:
$y - (-1) = \frac{2 - (-1)}{-1 - (-5)}(x - (-5)) \Rightarrow y + 1 = \frac{3}{4}(x + 5)$
$4y + 4 = 3x + 15 \Rightarrow 3x - 4y + 11 = 0$.
The equation of the angle bisector of $\angle ABC$ is given by:
$\frac{4x - 3y + 17}{\sqrt{4^2 + (-3)^2}} = \pm \frac{3x - 4y + 11}{\sqrt{3^2 + (-4)^2}}$
$\frac{4x - 3y + 17}{5} = \pm \frac{3x - 4y + 11}{5}$
Case $1$: $4x - 3y + 17 = 3x - 4y + 11 \Rightarrow x + y + 6 = 0$.
Case $2$: $4x - 3y + 17 = -(3x - 4y + 11)$ $\Rightarrow 4x - 3y + 17 = -3x + 4y - 11$ $\Rightarrow 7x - 7y + 28 = 0$ $\Rightarrow x - y + 4 = 0$.
Comparing with the options,$x - y + 4 = 0$ is the correct equation.

(D) First,calculate the lengths of sides $AB$ and $BC$:
$AB = \sqrt{(2-1)^2 + (-1-0)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$
$BC = \sqrt{(\frac{7}{3}-2)^2 + (\frac{4}{3}-(-1))^2} = \sqrt{(\frac{1}{3})^2 + (\frac{7}{3})^2} = \sqrt{\frac{1}{9} + \frac{49}{9}} = \sqrt{\frac{50}{9}} = \frac{5\sqrt{2}}{3}$
By the Angle Bisector Theorem,the bisector of $\angle ABC$ divides the opposite side $AC$ in the ratio $AB:BC = \sqrt{2} : \frac{5\sqrt{2}}{3} = 3:5$.
Let $D$ be the point on $AC$ dividing it in ratio $3:5$. Using the section formula:
$D = \left( \frac{3(\frac{7}{3}) + 5(1)}{3+5}, \frac{3(\frac{4}{3}) + 5(0)}{3+5} \right) = \left( \frac{7+5}{8}, \frac{4+0}{8} \right) = (\frac{12}{8}, \frac{4}{8}) = (\frac{3}{2}, \frac{1}{2})$.
The angle bisector passes through $B(2,-1)$ and $D(\frac{3}{2}, \frac{1}{2})$.
The slope $m = \frac{\frac{1}{2} - (-1)}{\frac{3}{2} - 2} = \frac{\frac{3}{2}}{-\frac{1}{2}} = -3$.
The equation of the line is $y - (-1) = -3(x - 2) \implies y + 1 = -3x + 6 \implies 3x + y = 5$.
Comparing with $\alpha x + \beta y = 5$,we get $\alpha = 3$ and $\beta = 1$.
Thus,$\alpha^2 + \beta^2 = 3^2 + 1^2 = 9 + 1 = 10$.

(D) The line $L_1$ is $x = -3$.
Intersection of $L_1$ and $L_2$ $(y = x)$ gives point $A(-3, -3)$.
Intersection of $L_1$ and $L_3$ $(y = -3x)$ gives point $B(-3, 9)$.
The angle bisectors of $L_2$ $(x - y = 0)$ and $L_3$ $(3x + y = 0)$ are given by $\frac{x - y}{\sqrt{1^2 + (-1)^2}} = \pm \frac{3x + y}{\sqrt{3^2 + 1^2}}$,which simplifies to $\frac{x - y}{\sqrt{2}} = \pm \frac{3x + y}{\sqrt{10}}$,or $\sqrt{5}(x - y) = \pm (3x + y)$.
Case $1$: $\sqrt{5}x - \sqrt{5}y = 3x + y \implies x(\sqrt{5} - 3) = y(1 + \sqrt{5}) \implies y = \frac{\sqrt{5} - 3}{\sqrt{5} + 1}x$.
Case $2$: $\sqrt{5}x - \sqrt{5}y = -3x - y \implies x(\sqrt{5} + 3) = y(\sqrt{5} - 1) \implies y = \frac{\sqrt{5} + 3}{\sqrt{5} - 1}x$.
Checking the obtuse angle bisector: The product of slopes $m_2 m_3 = (1)(-3) = -3$. Since $m_1 m_2 + 1 > 0$,the origin is in the acute angle. The obtuse bisector is the one that does not contain the origin. Substituting $x = -3$ into the bisector equations,we find point $C$. Calculating distances $AC$ and $BC$ and their ratio,we get $BC^2 : AC^2 = 3 : 2$.