Question

Find the equation of the line through the intersection of the lines 3x+2y−8=0,5x−11y+1=0 and parallel to the line 6x+13y=25

L1 = 3x + 2y -8 = 0,
L2 = 5x -11y +1 = 0,
L3= ? ,
L4 = 6x + 13y -25= 0

Answer 1

first solve the equations of L1 and L2 simultaneously to find the point of intersection

Answer 2

L1=3x+2y−8=0
L2=5x−11y+1=0
L3=?
L4=6x+13y−25=0

(3x+2y-8) + k(5x-11y+1)= 0 ------(i)
3x+2y-8+5kx-11ky+k =0
Arrange and take common:
(3+5k)x + (2 +11k) y - 8 +k =0
The slope from this equation is :
-(3+5k)/(2-11k)

Since L3 is parallel to L4 therefore:
Slope of L3 = Slope Of L4
-(3+5k)/(2-11k) = -6/13
39+15k = 12 -66k
65k+66k = 12-39
131k = -27
k= -27/131

Put the value of k in equation (i)
(3x+2y-8) + (-27/131)(5x-11y+1)= 0
393x+262y-1048 -135x+297y-27= 0
258x+559y-1075 =0
Divide by 43:
6x +13y-25 =0 ------(ANSWER)

ThankYOU :)

Answer 3

yes thats correct

Answer 4

its easier to solve the 2 equations first to find point of intersection though
then use the general form of the straight line

y - y1 = m(x -x1)

Answer 5

I tried to solve it that way but the answer didn't match.

Answer 6

Oh! mine did

did you get point of intersection to be (2,1)?

Answer 7

yes

Answer 8

so m = -6/13 , x1 = 2 and y1 = 1

y - 1 = (-6/13)( x - 2)

13y - 13 = -6x + 12
6x + 13y - 25 = 0