# Point of intersection of two lines

If two straight lines intersect, we have mentioned that they intersect at a single point, however no mention has been made about the nature of this point. Graphically, the point of intersection between these two lines is the point where the two are exactly the same. From this fact, we can calculate the value of the coordinates that define it, formally, if we consider two lines expressed as follows

$l_1 : y = m_1 x + b_1$
$l_2 : y = m_2 x + b_2$

The point $P_0 = (x_0,y_0)$ is the intersection point of $l_1$ and $l_2$, if the values of $x_0$ and $y_0$ satisfy both equations at the same time. This is known as a system of linear equations that consists of two equations and two unknowns, nevertheless, we will not elaborate on this topic since noticing that the lines are expressed in the form slope intercept, we will simply equalize the expressions that define them to later calculate the value of the unknowns.

Let’s see with some examples how to calculate the intersection point between two lines using this technique.

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## Examples

### Example 1

Calculate the intersection point between the lines $l_1 : y = 3x-3$ and $l_2 : y = -x + 1$.

For this we define our system of linear equations

$l_1 : y = 3x-3$
$l_2 : y = -x + 1$

We equal the two expressions that define these two lines, then we clear the variable $x$

$3x-3 = -x + 1$
$\Rightarrow 3x + x = 1 + 3$
$\Rightarrow 4x = 4$
$\Rightarrow x = \frac{4}{4}$
$\Rightarrow x = 1$

In this way, we can conclude that the coordinate value in the X-axis of the intersection point is $x=1$ and taking into account that this value is common in both lines, we can substitute it in the lines of our preference to calculate the value of $y$. Let’s substitute the value of $x=1$ in $l_1$:

$y = 3(1)-3 \Rightarrow y = 3-3 \Rightarrow y=0$

Note that if we substitute the value of $x=1$ in the $l_2$ line, we get the same value for $y$:

$y = -(1) + 1 \Rightarrow y = -1+1 \Rightarrow y=0$

Therefore, we conclude that the intersection point between the lines $l_1$ and $l_2$ is $P_0 = (1,0)$ and we can also, locate it in the Cartesian plane. We graph both lines making a table of values considering only the cutting points with the axes.

### Example 2

Calculate the intersection point between the lines $l_1 : y = -4x-2$ and $l_2 : y = \frac{1}{4}x + 3$.

For this we define our system of linear equations

$l_1 : y = -4x-2$
$l_2 : y = \frac{1}{4}x + 3$

We equal the two expressions that define these two lines, then we clear the variable $x$

$-4x-2 = \frac{1}{4}x + 3$
$\Rightarrow -4x - \frac{1}{4}x = 3 + 2$
$\Rightarrow -\frac{17}{4}x = 5$
$\Rightarrow x = -\frac{20}{17}$

In this way, we can conclude that the coordinate value in the X-axis of the intersection point is $x=-\frac{20}{17}$. Let us substitute this value in $l_1$:

$y = -4\left( -\frac{20}{17} \right)-2 \Rightarrow y = \frac{80}{17} -2 \Rightarrow y = \frac{46}{17}$

Therefore, we conclude that the intersection point between the lines $l_1$ and $l_2$ is $P_0 = \left( -\frac{20}{17} , \frac{46}{17} \right)$ and we can also locate it in the Cartesian plane. We graph both lines making a table of values considering only the cut-off points with the axes.

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### Example 3

Calculate the intersection point between the lines $l_1 : y = x+5$ and $l_2 : y = 2$.

In this case it is necessary to raise a system of equations, because as $l_2$ is a horizontal line, we simply substitute the value of $y$ that defines it in the line $l_1$ and from there, we calculate the value of $x$. Then, if $y=2$ we have to

$2 = x+5 \Rightarrow -x = 5-2 \Rightarrow -x = 3 \Rightarrow x = -3$

Therefore, we conclude that the point of intersection between the lines $l_1$ and $l_2$ is $P_0 = \left( -3 , 2 \right)$ and we can also locate it in the Cartesian plane. We graph both lines making a table of values considering only the cutting points with the axes.

### Example 4

Calculate the intersection point between the lines $l_1 : y = -\frac{1}{5}x+2$ and $l_2 : x = -1$.

In this case it is necessary to raise a system of equations, because as $l_2$ is a vertical line, we simply substitute the value of $x$ that defines it in the line $l_1$ and from there, we calculate the value of $y$. Then, if $x=-1$ we have to

$y = -\frac{1}{5}(-1)+2 \Rightarrow y = \frac{1}{5}+2 \Rightarrow y = \frac{11}{5}$

Therefore, we conclude that the point of intersection between the lines $l_1$ and $l_2$ is $P_0 = \left( -1 , \frac{11}{5} \right)$ and we can also locate it in the Cartesian plane. We graph both lines making a table of values considering only the cut-off points with the axes.

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### Example 5

Calculate the intersection point between the lines $l_1 : y = -3$ and $l_2 : x = 4$.

In this case it is necessary to raise a system of equations, because being $l_1$ a horizontal line and $l_2$ a vertical line, we can immediately conclude that the point of intersection between them is $(4,-3)$ and we can also locate it in the Cartesian plane.

We have seen the cases of intersections where the lines are expressed in the slope-ordered form, we now see the case in which we have lines expressed in a general way. formally, if we consider two lines expressed in the following way

$l_1 : a_1 x + b_1 y + c_1 = 0$
$l_2 : a_2 x + b_2 y + c_2 = 0$

Again, the point $P_0 = (x_0,y_0)$ is the point of intersection of $l_1$ and $l_2$, if the values of $x_0$ and $y_0$ satisfy both equations at the same time. However, the way to deal with this type of case is slightly different from a pending-ordered case.

In these cases it does not make sense to equalize the two expressions that define the lines, so the technique to find the solution consists in making operations between both equations to cancel one of the two variables. Let’s see with some examples how to calculate the solution of this type of systems of equations.

## Examples

### Example 6

Calculate the intersection point between the lines $l_1 : 2 x + 2 y - 1 = 0$ and $l_2 : - 2 x + y + 4 = 0$.

For this we define our system of linear equations

$l_1 : 2 x + 2 y - 1 = 0$
$l_2 : - 2 x + y + 4 = 0$

In this particular case, we can notice that in one equation is the expression $2x$ and in the other, the expression $-2x$, therefore, we can add both equations to obtain that

Considering the resulting equation, we can clear the variable $y$, and thus obtain the value $y_0$ of our point of intersection.

$0x + 3y + 3 = 0$
$\Rightarrow 3y + 3 = 0$
$\Rightarrow 3y = -3$
$\Rightarrow y = -\frac{3}{3}$
$\Rightarrow y = - 1$

In this way, we can conclude that the value of the Y-axis coordinate of the intersection point is $y=-1$ and taking into account that this value is common in both lines, we can substitute it in the lines of our preference to calculate the value of $x$. Let’s substitute the value of $y=-1$ in $l_1$:

$2 x + 2 (-1) - 1 = 0$
$\Rightarrow 2x - 2 - 1 = 0$
$\Rightarrow 2x - 3 = 0$
$\Rightarrow 2x = 3$
$\Rightarrow x = \frac{3}{2}$

Therefore, we conclude that the intersection point between the $l_1$ and $l_2$ lines is $P_0 = \left( \frac{3}{2}, -1 \right)$ and we can also locate it in the Cartesian plane. We graph both lines making a table of values considering only the cut-off points with the axes.

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### Example 6

Calculate the intersection point between the lines $l_1 : 2 x + 2 y - 1 = 0$ and $l_2 : - 2 x + y + 4 = 0$.

For this we define our system of linear equations

$l_1 : 2 x + 2 y - 1 = 0$
$l_2 : - 2 x + y + 4 = 0$

In this particular case, we can notice that in one equation is the expression $2x$ and in the other, the expression $-2x$, therefore, we can add both equations to obtain that

Considering the resulting equation, we can clear the variable $y$, and thus obtain the value $y_0$ of our point of intersection.

$0x + 3y + 3 = 0$
$\Rightarrow 3y + 3 = 0$
$\Rightarrow 3y = -3$
$\Rightarrow y = -\frac{3}{3}$
$\Rightarrow y = - 1$

In this way, we can conclude that the value of the Y-axis coordinate of the intersection point is $y=-1$ and taking into account that this value is common in both lines, we can substitute it in the lines of our preference to calculate the value of $x$. Let’s substitute the value of $y=-1$ in $l_1$:

$2 x + 2 (-1) - 1 = 0$
$\Rightarrow 2x - 2 - 1 = 0$
$\Rightarrow 2x - 3 = 0$
$\Rightarrow 2x = 3$
$\Rightarrow x = \frac{3}{2}$

Therefore, we conclude that the intersection point between the $l_1$ and $l_2$ lines is $P_0 = \left( \frac{3}{2}, -1 \right)$ and we can also locate it in the Cartesian plane. We graph both lines making a table of values considering only the cut-off points with the axes.

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### Example 7

Calculate the intersection point between the lines $l_1 : 3 x - 5 y + 2 = 0$ and $l_2 : x + y - 2 = 0$.

For this we define our system of linear equations

$l_1 : 3 x - 5 y + 2 = 0$
$l_2 : x + y - 2 = 0$

In the previous case we could cancel with relative simplicity the variable $x$ but in this particular case, we can notice that if we multiply the second equation by $5$ we obtain

$l_1 : 3 x - 5 y + 2 = 0$
$l_2 : 5x + 5y - 10 = 0$

Now, we can notice that in one equation is the expression $-5y$ and in the other, the expression $5y$, therefore, we can add both equations to obtain that

Considering the resulting equation, we can clear the variable $x$, and thus obtain the value $x_0$ of our point of intersection.

$8x + 0y - 8 = 0$
$\Rightarrow 8x - 8 = 0$
$\Rightarrow 8y = 8$
$\Rightarrow y = \frac{8}{8}$
$\Rightarrow y = 1$

In this way, we can conclude that the coordinate value in the X-axis of the intersection point is $x=1$ and taking into account that this value is common in both lines, we can substitute it in the lines of our preference to calculate the value of $y$. Let’s substitute the value of $x=1$ in $l_1$:

$3 (1) - 5 y + 2 = 0$
$\Rightarrow 3 - 5y + 2 = 0$
$\Rightarrow -5x + 5 = 0$
$\Rightarrow -5x = -5$
$\Rightarrow x = \frac{-5}{-5}$
$\Rightarrow x = 1$

Therefore, we conclude that the intersection point between the $l_1$ and $l_2$ lines is $P_0 = \left( 1, 1 \right)$ and we can also, locate it in the Cartesian plane. We graph both lines making a table of values considering only the cutting points with the axes.

### Example 8

Calculate the intersection point between the lines $l_1 : 6 x - 5 y + 4 = 0$ and $l_2 : 4 x + 3 y - 5 = 0$.

For this we define our system of linear equations

$l_1 : 6 x - 5 y + 4 = 0$
$l_2 : 4 x + 3 y - 5 = 0$

In this case we must notice that the variables are accompanied by different coefficients, so it is not enough to multiply only one equation to cancel terms. We must then, multiply both equations by numbers that help us to cancel summands. Let us multiply the first equation by $4$ and the second equation by $-6$.

$l_1 : 24 x - 20 y + 16 = 0$
$l_2 : -24 x - 18 y + 30 = 0$

Now, we can notice that in one equation is the expression $24x$ and in the other, the expression $-24y$, therefore, we can add both equations to obtain that

Considering the resulting equation, we can clear the variable $x$, and thus obtain the value $x_0$ of our point of intersection.

$0x - 38y + 36 = 0$
$\Rightarrow -38y + 36 = 0$
$\Rightarrow -38y = -36$
$\Rightarrow y = \frac{38}{36}$
$\Rightarrow y = \frac{19}{18}$

In this way, we can conclude that the value of the Y-axis coordinate of the intersection point is $y = \frac{19}{18}$ and taking into account that this value is common in both lines, we can substitute it in the lines of our preference to calculate the value of $x$. Let’s substitute the value of $y = \frac{19}{18}$ in $l_2$:

$4 x + 3 \left( \frac{19}{18} \right) - 5 = 0$
$\Rightarrow 4 x + \frac{19}{6} - 5 = 0$
$\Rightarrow 4 x - \frac{11}{6} = 0$
$\Rightarrow 4 x = \frac{11}{6}$
$\Rightarrow x = \frac{11}{24}$

Therefore, we conclude that the point of intersection between the lines $l_1$ and $l_2$ is $P_0 = \left(\frac{11}{24},\frac{19}{18}\right)$ and we can also locate it in the Cartesian plane. We graph both lines making a table of values considering only the cut-off points with the axes.

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