Equation of the line passing through the intersection of two lines and is parallel to another line. 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 ?
 A: 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
15k+66k = 12-39
81k = -27
k= -1/3  
Put the value of k in equation (i)
(3x+2y-8) + (-1/3)(5x-11y+1)= 0
3(3x+2y-8) - (5x-11y+1)= 0
9x+6y-24 -5x+11y-1 =0
4x+17y-25 = 0   
This is not the answer in my book.
A: 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 :)
A: This will blow your mind. Create vectors from the coefficients of the lines $$\begin{array}{rlrlrl}
L_1 & = \begin{pmatrix} 3\\2\\-8\end{pmatrix} &
L_2 & = \begin{pmatrix} 5\\-11\\1\end{pmatrix} &
L_4 & = \begin{pmatrix} 6\\13\\-25\end{pmatrix} \end{array} $$
The intersection point $P_3$ between $L_1$ and $L_2$ is found with a vector cross product
$$ P_3 = \begin{pmatrix} 3\\2\\-8\end{pmatrix} \times \begin{pmatrix} 5\\-11\\1\end{pmatrix}   = \begin{pmatrix} -86\\-43\\-43\end{pmatrix} $$ with coordinates $$(x,y) = \left( \frac{-86}{-43}, \frac{-43}{-43} \right) = (2,1)$$
The coefficients of a line parallel to $L_4$ are $L_3 = (6,13,c_3)$. To make this line pass through $P_3$ set
$$ P_3 \cdot L_3 = 0 $$
$$ \begin{pmatrix} -86\\-43\\-43\end{pmatrix} \cdot \begin{pmatrix} 6\\13\\c_3\end{pmatrix} =0 $$
$$ \left. -43 c_3 -1075 =0  \right\} c_3 = -25 $$
So line $L_3$ is $L_3 = (6,13,-25)$ with equation
$$ \begin{pmatrix} x\\y\\1\end{pmatrix} \cdot \begin{pmatrix} 6\\13\\-25\end{pmatrix} =0 
\\ 6x+13y-25 = 0 $$
A: What;'s wrong is that $(-2, -1)$ is not on the line 
$$
5x - 11y + 1 = 0
$$
because
$$
5(-2) - 11(-1) + 1 = -10 + 11 + 1 = 2. 
$$
A: I don't know what's wrong with your reasoning.
You actually don't need to find the point of intersection. For any $k$, the line $L_1 + kL_2=0$ is a line that passes through the point of intersection of $L_1=0$ and $L_2=0$. This family of lines is called a "pencil" of lines.
But
$$
L_1 + kL_2 = 3x+2y−8 + k(5x−11y+1) = (3+5k)x + (2-11k)y -8+k
$$
This line will be parallel to $6x+13y−25=0$ if 
$$
\frac{3+5k}{2-11k} = \frac{6}{13}
$$
Solve for $k$ and plug into $L_1 + kL_2 = 0$.
A: dear friend nothing is wrong....
you just form a family of lines having a common point.
K which you used, would have many values which give you infinite lines passing through point.
So when you apply that point in family they will satisfy the equality independent of 'K' . Since they had that point on them.
So you get nothing but satisfying equality.
A: You used L1+ k L2 =0 , which means this eqaution shows all the possible equations of lines which would pass through intersection of L1 and L2 . 
Next you took the point of intersection by solving L1 and L2, 
And put the point in L1 + kL2 , which will obviously satisfy the eqation , as the eqaution L1 +  kL2  says , eqiation of lines passing through point of intersection.
Thus , you are revolving in the same circle and accidently verified the family of lines. 
Hope this helps :) 
