# 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 ?

• All three lines intersect at the same point. I know that from the triple product $$(6,13,-25)\cdot(3,2,-8)\times(5,-11,1)=0$$ – ja72 Sep 25 '14 at 14:32

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:

ThankYOU :)

• $39+15k$ should be $39+65k$ – bubba Sep 21 '14 at 12:38

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

$$\begin{pmatrix} x\\y\\1\end{pmatrix} \cdot \begin{pmatrix} 6\\13\\-25\end{pmatrix} =0 \\ 6x+13y-25 = 0$$

• how did you multiply L1 and L2 ? – It'sMe Sep 25 '14 at 15:36
• $\cdot$ is the vector inner (dot) product, and $\times$ is the vector cross product. – ja72 Sep 25 '14 at 15:37
• I am using homogeneous coordinates for planar points and lines where the geometry meet and join operations are simply $\cdot$ and $\times$. – ja72 Sep 25 '14 at 15:41
• I still don't get it. can you help me in this please ? math.stackexchange.com/questions/945808/… – It'sMe Sep 25 '14 at 15:46

What;'s wrong is that $(-2, -1)$ is not on the line $$5x - 11y + 1 = 0$$ because $$5(-2) - 11(-1) + 1 = -10 + 11 + 1 = 2.$$

• I tried doing it another way.. I'll post it. Can you please check it ? – It'sMe Sep 21 '14 at 10:58

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.

• I see the error here: Just after "Slope of $L_3$ = Slope of $L_4$", you multiply through by $13$, but $5 \cdot 13$ is not $15$. – John Hughes Sep 21 '14 at 11:08
• 39+65k = 12 -66k, 65k+66k = 12-39, 131k=27, k=27/131? Does this seem right? – It'sMe Sep 21 '14 at 11:12
• It seems right, but to help us check your work, please format your mathematics better; for a start, just put dollar signs around everything math, changing "13k+11" to $13k + 11$, for instance. – John Hughes Sep 21 '14 at 11:19
• Okay :) I did it Thanks :) – It'sMe Sep 21 '14 at 11:30

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 $L_1 + kL_2 = 0$.

• shouldn't it be 3+5k/2-11k? – It'sMe Sep 21 '14 at 11:07
• Yes, you're right. Sorry. I'll fix it. – bubba Sep 21 '14 at 11:50

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.

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 :)