Find the equations of the bisectors of the angles between the lines 4x 3y + 1 = 0 and 12x 5y + 7 = 0 . Also prove that the bisectors are at right angles to each other. OR Find the angle between the…

Solution to the First Question: Angle Bisectors of Two Lines

Step 1: Find the Slopes of the Given Lines

We are given the two lines:

1. $L_1:4x-3y+1=0$
2. $L_2:12x-5y+7=0$

First, rewrite them in slope-intercept form $y=mx+c$:

For $L_1$: $$4x-3y+1=0⟹3y=4x+1⟹y=\frac{4}{3}x+\frac{1}{3}$$ So, slope $m_1=\frac{4}{3}$.

For $L_2$: $$12x-5y+7=0⟹5y=12x+7⟹y=\frac{12}{5}x+\frac{7}{5}$$ So, slope $m_2=\frac{12}{5}$.

Step 2: Find the Angle Bisectors Using the Angle Bisector Formula

The angle bisectors of two lines $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ are given by: $$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm \frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}$$

For our lines: $$\frac{4x-3y+1}{\sqrt{4^2+(-3{)}^2}}=\pm \frac{12x-5y+7}{\sqrt{12^2+(-5{)}^2}}$$ Simplify the denominators: $$\sqrt{4^2+(-3{)}^2}=\sqrt{16+9}=5$$ $$\sqrt{12^2+(-5{)}^2}=\sqrt{144+25}=13$$

Thus, the two angle bisectors are: $$\frac{4x-3y+1}{5}=\frac{12x-5y+7}{13}\text{(Positive sign)}$$ $$\frac{4x-3y+1}{5}=-\frac{12x-5y+7}{13}\text{(Negative sign)}$$

Step 3: Solve for the Two Bisectors

First Bisector (Positive Sign): $$13(4x-3y+1)=5(12x-5y+7)$$ $$52x-39y+13=60x-25y+35$$ $$52x-60x-39y+25y+13-35=0$$ $$-8x-14y-22=0⟹8x+14y+22=0$$ Simplify by dividing by 2: $$4x+7y+11=0\text{(Bisector 1)}$$

Second Bisector (Negative Sign): $$13(4x-3y+1)=-5(12x-5y+7)$$ $$52x-39y+13=-60x+25y-35$$ $$52x+60x-39y-25y+13+35=0$$ $$112x-64y+48=0$$ Simplify by dividing by 8: $$14x-8y+6=0\text{(Bisector 2)}$$

Step 4: Prove That the Bisectors Are Perpendicular

To prove that the two bisectors are perpendicular, we find their slopes and check if the product of the slopes is $-1$.

Slope of Bisector 1 ($4x+7y+11=0$): $$7y=-4x-11⟹y=-\frac{4}{7}x-\frac{11}{7}$$ Slope $m_3=-\frac{4}{7}$.

Slope of Bisector 2 ($14x-8y+6=0$): $$-8y=-14x-6⟹y=\frac{14}{8}x+\frac{6}{8}=\frac{7}{4}x+\frac{3}{4}$$ Slope $m_4=\frac{7}{4}$.

Now, check the product of the slopes: $$m_3\cdot m_4=\left(-\frac{4}{7}\right)\cdot \left(\frac{7}{4}\right)=-1$$ Since the product of the slopes is $-1$, the two bisectors are perpendicular to each other.

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Solution to the Second Question: Angle Between Two Lines Represented by $a{x}^2+2hxy+b{y}^2=0$

Step 1: Find the Slopes of the Two Lines

The given equation represents a pair of straight lines passing through the origin: $$a{x}^2+2hxy+b{y}^2=0$$

Divide by $y^2$ (assuming $y\ne 0$): $$a{\left(\frac{x}{y}\right)}^2+2h\left(\frac{x}{y}\right)+b=0$$ Let $m=\frac{x}{y}$, then: $$a{m}^2+2hm+b=0$$

This is a quadratic in $m$, whose roots $m_1$ and $m_2$ are the slopes of the two lines. By Vieta’s formulas: $$m_1+m_2=-\frac{2h}{a},m_1{m}_2=\frac{b}{a}$$

Step 2: Find the Angle Between the Two Lines

The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by: $$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$$

We know: $$(m_1-m_2{)}^2=(m_1+m_2{)}^2-4m_1{m}_2$$ Substitute Vieta’s results: $$(m_1-m_2{)}^2={\left(-\frac{2h}{a}\right)}^2-4\left(\frac{b}{a}\right)=\frac{4h^2}{a^2}-\frac{4b}{a}=\frac{4(h^2-ab)}{a^2}$$ Thus: $$|m_1-m_2|=\frac{2\sqrt{h^2-ab}}{a}$$

Now, substitute into the angle formula: $$\tan \theta =\left|\frac{\frac{2\sqrt{h^2-ab}}{a}}{1+\frac{b}{a}}\right|=\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$$ So, the angle $\theta$ between the two lines is: $$\theta ={\tan }^{-1}\left(\frac{2\sqrt{h^2-ab}}{a+b}\right)$$

Step 3: Conditions for Perpendicular and Coincident Lines

a) Perpendicular Lines: Two lines are perpendicular if the product of their slopes is $-1$: $$m_1{m}_2=-1$$ From Vieta’s formula: $$\frac{b}{a}=-1⟹b=-a$$ Thus, the condition for perpendicularity is: $$a+b=0$$

b) Coincident Lines: Two lines are coincident if they have the same slope, i.e., $m_1=m_2$. This happens when the discriminant of the quadratic in $m$ is zero: $$(2h{)}^2-4ab=0⟹4h^2-4ab=0⟹h^2=ab$$ Thus, the condition for coincident lines is: $$h^2=ab$$