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Chapter 14: Problem 1292

The equation of the bisector of acute angle between the lines \(3 x-4 y+7=0\) and \(-12 x-5 y+2=0\) is (a) \(11 \mathrm{x}-3 \mathrm{y}+9=0\) (b) \(3 x+11 y-13=0\) (c) \(3 x+11 y-3=0\) (d) \(11 x-3 y+2=0\)

Short Answer

Expert verified
The short answer to the problem is: \(11x - 3y + 9 = 0\).

Step by step solution

01

Find the angle between the lines

Given two lines \(L_1: 3x - 4y + 7 = 0\) and \(L_2: -12x - 5y + 2 = 0\), we find the angle between them. Let's first write these lines in the form of \(L_1: Ax + By + C_1 = 0\) and \(L_2: A'x + B'y + C_2 = 0\). We can find the angle \(\theta\) between the lines using the formula: \[\tan \theta = \frac{A B' - A' B}{A A' + B B'}\] Substitute the values of A, B, A', and B' in the formula: \[\tan \theta = \frac{(3)(-5) - (-12)(-4)}{(3)(-12) + (-4)(-5)}\]
02

Calculate the angle between the lines

Now, we will simplify and solve for \(\tan \theta\): \[\tan \theta = \frac{-15 - 48}{-36 + 20}\] \[\tan \theta = \frac{-63}{-16}\] \[\tan \theta = \frac{63}{16}\]
03

Find the angle bisector

Now that we have the angle \(\theta\) between the lines, let's find the bisector of this acute angle using the angle bisector formula. The equation of the angle bisector of the acute angle is given by: \[\frac{A x + B y + C_1}{\sqrt{A^2 + B^2}} = \pm \frac{A' x + B' y + C_2}{\sqrt{{A'}^2 + {B'}^2}}\] Since we are dealing with the bisector of the acute angle, we will use the positive sign. Substitute all the values into the formula: \[\frac{3 x - 4 y + 7}{\sqrt{3^2 + (-4)^2}} = \frac{-12 x - 5 y + 2}{\sqrt{(-12)^2 + (-5)^2}}\]
04

Simplify the equation of the angle bisector

Now, let's simplify the equation of the angle bisector: \[\frac{3 x - 4 y + 7}{\sqrt{9 + 16}} = \frac{-12 x - 5 y + 2}{\sqrt{144 + 25}}\] \[\frac{3 x - 4 y + 7}{\sqrt{25}} = \frac{-12 x - 5 y + 2}{\sqrt{169}}\] \[(3 x - 4 y + 7)(13) = (-12 x - 5 y + 2)(5)\]
05

Solve for the equation of the angle bisector

Next, we will multiply both sides of the equation and solve for the equation of the angle bisector: \[39 x - 52 y + 91 = -60 x - 25 y + 10\] Combine like terms: \[99 x - 27y + 81 = 0\] Divide by 9 to simplify: \[11 x - 3 y + 9 = 0\] So, the equation of the bisector of the acute angle between the given lines is \(11x - 3y + 9 = 0\). Comparing this with the given options, we find that the correct answer is: (a) \(11 x - 3 y + 9 = 0\)

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Most popular questions from this chapter

Three straight lines \(2 \mathrm{x}+11 \mathrm{y}-5=0,4 \mathrm{x}-3 \mathrm{y}-2=0\) and \(24 x+7 y-20=0\) (a) form a triangle (b) are only concurrent (c) are concurrent with one line bisecting the angle between the other two. (d) none of these

Let \(\mathrm{A}(2,-3)\) and \(\mathrm{B}(-2,1)\) be vertices of a triangle \(\mathrm{ABC}\). If the centroid of this triangle moves on the line \(2 x+3 y=1\), then locus of the vertex \(\mathrm{C}\) is the line (a) \(2 \mathrm{x}+3 \mathrm{y}=9\) (b) \(2 x-3 y=7\) (c) \(3 \mathrm{x}+2 \mathrm{y}=5\) (d) \(3 x-2 y=3\)

If \(5 \mathrm{x}+12 \mathrm{y}+13=0\) is transformed into \(x \cos \alpha+\mathrm{ysin} \alpha=\mathrm{p}\) then \(\alpha=? \alpha \in[-\pi, \pi]\) (a) \(\cos ^{-1}[-(5 / 13)]\) (b) \(\sin ^{-1}[-(12 / 13)]\) (c) \(\tan ^{-1}(12 / 5)-\pi\) (d) \(\tan ^{-1}(12 / 5)\)

If a line \(3 \mathrm{x}+4 \mathrm{y}=24\) intersects the axes at \(\mathrm{A}\) and \(\mathrm{B}\), then in radius of \(\Delta \mathrm{OAB}\) is \(\ldots \ldots\) (a) 1 (b) 2 (c) 3 (d) 4

The \(\mathrm{p}-\alpha\) form of the line \(\mathrm{x}+\sqrt{(3) \mathrm{y}-4}=0 \mathrm{is}\) (a) \(x \cos (\pi / 6)+\operatorname{ysin}(\pi / 6)=2\) (b) \(x \cos (\pi / 3)+y \sin (\pi / 3)=2\) (c) \(x \cos [-(\pi / 3)]+y \sin [-(\pi / 3)]=2\) (d) \(x \cos [-(\pi / 6)]+y \sin [-(\pi / 6)]=2\)
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