The Question is :
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 $6x+13y=25$
Here is how I did it..
$L_1 = 3x + 2y -8 = 0$
$L_2 = 5x -11y +1 = 0$
$L_3= ?$
$L_4 = 6x + 13y -25= 0$
I found the point of intersection : $(-2, -1)$
Using formula: $L_1 + kL_2$ = 0
$(3x + 2y -8) + k(5x -11y +1)=0$
$(3(2) + 2(1) -88) + k(5(2) -11(1) +1) = 0$
$(6 +2 -8) + k(10 -11 +1) =0$
$8-8 + k(11-11) =0$
$0 +k(0) = 0$
What's wrong ?
Best Answer
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+65k = 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 :)