Home
Class 11
MATHS
Find the equations to the straight lines...

Find the equations to the straight lines bisecting the angles between the following pairs of straight lines, placing first the bisector of the angle in which the origin lies.
`4x+ 3y- 7=0 and 24x+ 7y- 31= 0`

Text Solution

AI Generated Solution

The correct Answer is:
To find the equations of the straight lines bisecting the angles between the given pairs of straight lines, we will follow these steps: ### Step 1: Identify the equations of the lines The equations of the lines are: - \( L_1: 4x + 3y - 7 = 0 \) - \( L_2: 24x + 7y - 31 = 0 \) ### Step 2: Evaluate the lines at the origin We will check the values of \( L_1 \) and \( L_2 \) at the origin (0, 0): - For \( L_1 \): \[ L_1(0, 0) = 4(0) + 3(0) - 7 = -7 \] - For \( L_2 \): \[ L_2(0, 0) = 24(0) + 7(0) - 31 = -31 \] ### Step 3: Determine the product of the values Now, we calculate the product of the values obtained: \[ L_1(0, 0) \times L_2(0, 0) = (-7) \times (-31) = 217 \] Since the product is positive, the origin lies between the two lines, and we can find the first angle bisector. ### Step 4: Use the angle bisector formula The formula for the angle bisector is given by: \[ \frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}} \] Where: - For \( L_1 \): \( A_1 = 4, B_1 = 3, C_1 = -7 \) - For \( L_2 \): \( A_2 = 24, B_2 = 7, C_2 = -31 \) ### Step 5: Calculate the left-hand and right-hand sides First, we calculate the denominators: \[ \sqrt{A_1^2 + B_1^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \] \[ \sqrt{A_2^2 + B_2^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \] ### Step 6: Substitute into the angle bisector formula Substituting into the angle bisector formula: \[ \frac{4x + 3y - 7}{5} = \frac{24x + 7y - 31}{25} \] ### Step 7: Cross-multiply to eliminate the fractions Cross-multiplying gives: \[ 25(4x + 3y - 7) = 5(24x + 7y - 31) \] Expanding both sides: \[ 100x + 75y - 175 = 120x + 35y - 155 \] ### Step 8: Rearranging the equation Rearranging the equation gives: \[ 100x + 75y - 120x - 35y = -155 + 175 \] \[ -20x + 40y = 20 \] Dividing through by -20: \[ x - 2y + 1 = 0 \] ### Step 9: Find the second angle bisector For the second angle bisector, we use the negative sign in the angle bisector formula: \[ \frac{4x + 3y - 7}{5} = -\frac{24x + 7y - 31}{25} \] Cross-multiplying gives: \[ 25(4x + 3y - 7) = -5(24x + 7y - 31) \] Expanding both sides: \[ 100x + 75y - 175 = -120x - 35y + 155 \] ### Step 10: Rearranging the second equation Rearranging gives: \[ 100x + 75y + 120x + 35y = 155 + 175 \] \[ 220x + 110y = 330 \] Dividing through by 110: \[ 2x + y - 3 = 0 \] ### Final Result The equations of the angle bisectors are: 1. \( x - 2y + 1 = 0 \) (the bisector in which the origin lies) 2. \( 2x + y - 3 = 0 \)
Promotional Banner

Similar Questions

Explore conceptually related problems

Find the equation of the lines bisecting the angles between the pair of lines 3x^(2)+xy-2y^(2)=0

The equation to the straight line bisecting the angle between straight lines 12x+5y-2=0 and 3x-4y+7=0is:

Find the angles between the pairs of straight line x-y sqrt(3)=5 and sqrt(3)x+y=7

The angle between the pair of straight lines x^(2)+4y^(2)-7xy=0 is

Findthe angle between the pair of straight lines x^2 -4y^2+3x - 4 = 0

Find the angles between each of the following pairs of straight line: x-4y=3\ a n d\ 6x-y=11.

The angle between the pair of straight lines x^(2)-y^(2)-2y-1=0 is

Find the angle between the pair of straight lines x^(2) - 3xy +2y^(2) = 0

Find the equations of the bisectors of the angles formed by the following pairs of lines x+2y+3=0 and 2x+y-2=0