Lesson Explainer: Intersection Point of Two Straight Lines on the Coordinate Plane Mathematics

In this explainer, we will learn how to find the intersection point between two straight lines on a coordinate system and use this concept to find equations of lines.

We begin by recalling what we mean by the intersection of two lines.

Definition: The Point of Intersection of Two Lines

The point of intersection of two distinct, nonparallel lines is the single point where they meet or cross. It is the ordered pair of the values of π‘₯ and 𝑦 where the lines meet on the graph and that satisfies the equations of both lines.

Distinct, parallel lines are lines in a plane that are always the same distance apart. They will have no points of intersection.

In the diagram below, 𝐴𝐡 and 𝑃𝑄 intersect at point 𝐹.

How To: Finding the Point of Intersection between Two Lines on the Coordinate Plane

We can find the point of intersection of two lines on the coordinate plane using two methods, a graphical or an algebraic method. For example, if we had the two lines represented by 𝑦=π‘₯+1 and 𝑦+2π‘₯=4, then we could graph these as follows.

The specific point of intersection can be found graphically by inspecting the graph to note the point (π‘₯,𝑦), where the lines meet or cross. However, for more complex equations, an algebraic solution may be preferable. For an algebraic solution, we recognize that a point lying on both lines must satisfy both equations. This is the same as solving a system of two linear equations in two unknowns.

In the first example, we will see how a graphical solution can be used to find the point of intersection between a horizontal and a vertical line.

Example 1: Finding the Point of Intersection of a Horizontal and a Vertical Line

At which point do the lines π‘₯=7 and 16𝑦=βˆ’1 intersect?

Answer

The intersection point of two distinct lines is the point where the lines cross. One way to answer this question is to sketch both lines. We begin by drawing the graph of π‘₯=7. Any equation that just has an π‘₯-variable, and no 𝑦-variable, will be a vertical line.

Next, we consider the graph of 16𝑦=βˆ’1. It may be helpful if we rearrange this to make 𝑦 the subject. So, by multiplying both sides of the equation, 16𝑦=βˆ’1, by 6, we have 𝑦=βˆ’6.

An equation with just a 𝑦-variable is a horizontal line. As 𝑦=βˆ’6, the horizontal line passes through βˆ’6 on the 𝑦-axis.

To find the intersection point of the two lines, π‘₯=7 and 16𝑦=βˆ’1, we look for the point where they cross or meet. Inspecting the graph, the intersection point occurs at the coordinate (7,βˆ’6).

In the next example, we will see how an algebraic method can be used to find the point of intersection between two lines.

Example 2: Finding the Point of Intersection of Two Straight Lines

Determine the point of intersection of the two straight lines represented by the equations π‘₯+3π‘¦βˆ’2=0 and βˆ’π‘¦+1=0.

Answer

We can recall that the intersection point of two distinct lines is the point where the lines cross. To find this point of intersection, or the point where the lines cross, we can use either an algebraic or a graphical approach.

A point of intersection lies on both lines, so it must satisfy the equations of both lines. Hence, we can find the intersection point by solving these equations as simultaneous equations, finding the values of π‘₯ and 𝑦, where (π‘₯,𝑦) is the point of intersection.

We can write our equations:

π‘₯+3π‘¦βˆ’2=0,βˆ’π‘¦+1=0.(1)(2)

To solve this using a substitution method, we rearrange the second equation, βˆ’π‘¦+1=0, by adding 𝑦 to both sides of the equation to give βˆ’π‘¦+1=01=𝑦𝑦=1.

Substituting 𝑦=1 into the equation (1), and rearranging, we have π‘₯+3(1)βˆ’2=0π‘₯+1=0π‘₯=βˆ’1.

We have the solution π‘₯=βˆ’1 and 𝑦=1. Therefore, the point of intersection can be given as (βˆ’1,1).

As an alternative method, or as a check on the algebraic method, we consider the graphs of the two lines.

We have established that the line βˆ’π‘¦+1=0 can be rearranged as 𝑦=1. Hence, the line will be a horizontal line passing through 1 on the 𝑦-axis.

We can sketch the equation π‘₯+3π‘¦βˆ’2=0 by finding two points on the line. For example, we could find the 𝑦-intercept by substituting π‘₯=0 and the π‘₯-intercept by substituting 𝑦=0.

Substituting 𝑦=0 into π‘₯+3π‘¦βˆ’2=0, and simplifying, we have π‘₯+3(0)βˆ’2=0π‘₯βˆ’2=0π‘₯=2.

Substituting π‘₯=0, and simplifying, gives 0+3π‘¦βˆ’2=03π‘¦βˆ’2=03𝑦=2𝑦=23.

We have now established that the line passes through the points (2,0) and ο€Ό0,23. It can be a little difficult to plot a set of coordinates with noninteger values, such as ο€Ό0,23; so, we may prefer to find another set of coordinates on the line π‘₯+3π‘¦βˆ’2=0.

Substituting π‘₯=βˆ’4 into π‘₯+3π‘¦βˆ’2=0, and simplifying, we have (βˆ’4)+3π‘¦βˆ’2=0βˆ’6+3𝑦=03𝑦=6𝑦=2.

This gives us a third coordinate, (βˆ’4,2), on the line π‘₯+3π‘¦βˆ’2=0. We can plot these three coordinates and sketch this line alongside the graph of βˆ’π‘¦+1=0 as shown.

Inspecting the graph, we can confirm the algebraic solution above, as the intersection point is at (βˆ’1,1).

In the previous example, we saw two different methods used, an algebraic and a graphical one. There are advantages to both methods, and often a graphical method is a good way to check on the result of an algebraic one. However, it is worth noting that a graphical method may not be fully accurate, particularly in the cases where the solution is a noninteger result. On the other hand, an algebraic solution will always give an accurate result.

In the next example, we will see how the intersection can be found between two lines, where one is given in vector form.

Example 3: Finding the Vector Form of a Line That Passes through the Intersection of Two Other Lines

Find the vector equation of the straight line that is parallel to the 𝑦-axis and passes through the point of intersection of the two straight lines βƒ‘π‘Ÿ=π‘˜(βˆ’6,βˆ’4) and βˆ’3π‘₯+5𝑦=βˆ’5.

Answer

We recall that the intersection of two lines is the point at which they meet or cross. In order to find the vector equation of a line, we need a point on the line and its direction. Since the line is parallel to the 𝑦-axis, it is a vertical line and has the direction vector (0,1). Therefore, we need to find a point on the line. This point will be the point of intersection.

We can write the line with vector equation βƒ‘π‘Ÿ=π‘˜(βˆ’6,βˆ’4) as one in Cartesian form. For any points 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦) in the coordinate system with a direction vector ⃑𝑑=(π‘Ž,𝑏), for any scalar π‘˜, (π‘₯,𝑦)=(π‘₯,𝑦)+π‘˜(π‘Ž,𝑏).

Hence, taking βƒ‘π‘Ÿ=(π‘₯,𝑦), we can write the equation βƒ‘π‘Ÿ=π‘˜(βˆ’6,βˆ’4) in parametric form as π‘₯=βˆ’6π‘˜,𝑦=βˆ’4π‘˜.

We can then eliminate π‘˜ by rearranging each equation above to make π‘˜ the subject. This gives us π‘₯βˆ’6=π‘˜,π‘¦βˆ’4=π‘˜.

Thus, we can set the left-hand side of each equation as equal, such that βˆ’π‘₯6=βˆ’π‘¦4.

Rearranging, we have 4π‘₯=6𝑦4π‘₯βˆ’6𝑦=02π‘₯βˆ’3𝑦=0.

Next, the line βˆ’3π‘₯+5𝑦=βˆ’5 can be rearranged as βˆ’3π‘₯+5𝑦+5=0, and we can find the point of intersection of the two lines in Cartesian form using a simultaneous method:

βˆ’3π‘₯+5𝑦+5=0,2π‘₯βˆ’3𝑦=0.(3)(4)

The second equation can be rearranged to make π‘₯ the subject, as follows: 2π‘₯=3𝑦π‘₯=32𝑦.

We then substitute this value into equation (3) to give βˆ’3ο€Ό32π‘¦οˆ+5𝑦+5=0βˆ’92𝑦+5𝑦+5=012𝑦+5=012𝑦=βˆ’5𝑦=βˆ’10.

Now, we can substitute this value of 𝑦=βˆ’10 into π‘₯=32𝑦 to give π‘₯=32(βˆ’10)=βˆ’15.

Thus, the point of intersection of the two lines is at (βˆ’15,βˆ’10).

We now have the vector direction, ⃑𝑑, as the unit vector (0,1) and the position vector of the point of intersection as (βˆ’15,βˆ’10).

Writing this equation in the form βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+π‘˜βƒ‘π‘‘οŠ¦, where π‘˜ as any scalar multiple, we have βƒ‘π‘Ÿ=(βˆ’15,βˆ’10)+π‘˜(0,1).

In the next example, we will use a given angle to find the direction of a line through the intersection of two other lines.

Example 4: Finding the Equation of a Straight Line That Passes through the Intersection of Two Other Lines

Determine the equation of the line passing through the point of intersection of the two lines whose equations are 5π‘₯+2𝑦=0 and 3π‘₯+7𝑦+13=0 while making an angle of 135∘ with the positive 𝑦-axis.

Answer

We recall that the intersection of two distinct lines is the point where they cross. We are given the information about the angle that the required line makes with the positive 𝑦-axis; however, we will also need to find a point on the line in order to write the equation of this line. As the point of intersection of the other lines lies on this line, then this will be an ideal point to choose.

We begin by finding the intersection of the lines 5π‘₯+2𝑦=0 and 3π‘₯+7𝑦+13=0. At the intersection point, the π‘₯- and 𝑦-values in each equation will be equal. We can find these values by using a substitution method.

Taking the equation 5π‘₯+2𝑦=0, and rearranging to make π‘₯ the subject, we have 5π‘₯=βˆ’2𝑦π‘₯=βˆ’25𝑦.

We can then substitute π‘₯=βˆ’25𝑦 into the equation 3π‘₯+7𝑦+13=0, and rearrange, to give 3ο€Όβˆ’25π‘¦οˆ+7𝑦+13=0βˆ’65𝑦+7𝑦+13=0295𝑦=βˆ’13𝑦=βˆ’6529.

To find the value of π‘₯, we substitute 𝑦=βˆ’6529 into our rearranged equation, π‘₯=βˆ’25𝑦. This gives us π‘₯=βˆ’25Γ—βˆ’6529π‘₯=2629.

The point of intersection, (π‘₯,𝑦), can be given as ο€Ό2629,βˆ’6529.

We need to find the equation of the line passing through this point ο€Ό2629,βˆ’6529 while making an angle of 135∘ with the positive 𝑦-axis. We can sketch a line with an angle of 135∘ to the 𝑦-axis. As the angle is positive, the measurement is taken counterclockwise.

To find the slope of the line, we need to calculate the positive angle that the line makes with the positive direction of the π‘₯-axis. We can use the angles on a line and the corresponding angles resulting from parallel lines and transversals.

We remember that the angles on a straight line sum to 180∘. Therefore, the line will make an angle of 180βˆ’135=45∘∘∘ to the positive 𝑦-axis, measured in a clockwise direction.

Hence, the line will make an angle of πœƒ=90βˆ’45=45∘∘∘ with the positive π‘₯-axis, measured in a counterclockwise direction.

If the line makes an angle πœƒ with the positive π‘₯-axis, then its slope is tanπœƒ.

The slope, π‘š, is π‘š=45=1.tan∘

We can use the point–slope form of a line to write the equation of this line. In this form, the equation of a line passing through (π‘₯,𝑦), with slope π‘š, is given by π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯).

So, the line passing through the point ο€Ό2629,βˆ’6529, with slope 1, is π‘¦βˆ’ο€Όβˆ’6529=1ο€Όπ‘₯βˆ’2629οˆπ‘¦+6529=π‘₯βˆ’2629.

Adding 2629 to both sides of this equation gives 𝑦+9129=π‘₯.

Then, multiplying the terms by 29, we have 29𝑦+91=29π‘₯.

Writing this equation in the form π‘Žπ‘₯+𝑏𝑦+𝑐=0, we subtract 29𝑦 and 91 from both sides of the equation to give 29π‘₯βˆ’29π‘¦βˆ’91=0.

We can now consider how we can write the general equation of the straight line passing through the point of intersection of two given lines.

We recognize that there is an infinite number of straight lines passing through any particular point. Hence, we can define the equation that represents all straight lines passing through an intersection point of two lines as follows.

Definition: The Equation of a Straight Line Passing through the Point of Intersection of Two Given Lines

The equation that represents all straight lines passing through an intersection point of two lines π‘Žπ‘₯+𝑏𝑦+𝑐=0 and π‘Žπ‘₯+𝑏𝑦+𝑐=0 is π‘š(π‘Žπ‘₯+𝑏𝑦+𝑐)+𝑙(π‘Žπ‘₯+𝑏𝑦+𝑐)=0, where π‘š,π‘™βˆˆβ„.

If π‘š=0, we have the equation of the second line.

If 𝑙=0, we have the equation of the first line.

When π‘šβ‰ 0 and 𝑙≠0, we have the equation of a straight line passing through the point of intersection, excluding the original lines. Hence, we can write the equation above in the form π‘Žπ‘₯+𝑏𝑦+𝑐+π‘˜(π‘Žπ‘₯+𝑏𝑦+𝑐)=0, for any π‘˜βˆˆβ„.

In the next example, we will see how we can apply this equation in an algebraic method to find the equation of a line passing through a given point and the intersection point of two lines.

Example 5: Finding the Equation of a Line That Passes through the Intersection of Two Other Lines

What is the equation of the line passing through 𝐴(βˆ’1,3) and the intersection of the lines 3π‘₯βˆ’π‘¦+5=0 and 5π‘₯+2𝑦+3=0?

  1. 23π‘₯+17𝑦+17=0
  2. 8π‘₯+𝑦+8=0
  3. 17π‘₯βˆ’2𝑦+23=0

Answer

We begin by recalling that the intersection point of two distinct lines is the point where they cross.

We can write the general equation of a line passing through the intersection point of two lines π‘Žπ‘₯+𝑏𝑦+𝑐=0 and π‘Žπ‘₯+𝑏𝑦+𝑐=0 as π‘Žπ‘₯+𝑏𝑦+𝑐+π‘˜(π‘Žπ‘₯+𝑏𝑦+𝑐)=0, for any π‘˜βˆˆβ„.

Substituting the values π‘ŽοŠ§, π‘οŠ§, and π‘οŠ§ as the values from the line 3π‘₯βˆ’π‘¦+5=0 and the π‘ŽοŠ¨, π‘οŠ¨, and π‘οŠ¨ values from the line 5π‘₯+2𝑦+3=0, we have

3π‘₯βˆ’π‘¦+5+π‘˜(5π‘₯+2𝑦+3)=0.(5)

As the line passes through the point with coordinates (βˆ’1,3), we can substitute π‘₯=βˆ’1 and 𝑦=3 into equation (5) above. This gives us 3(βˆ’1)βˆ’3+5+π‘˜[5(βˆ’1)+2(3)+3]=0βˆ’1+4π‘˜=04π‘˜=1π‘˜=14.

We can now substitute π‘˜=14 into equation (5). This gives us 3π‘₯βˆ’π‘¦+5+14(5π‘₯+2𝑦+3)=03π‘₯βˆ’π‘¦+5+54π‘₯+24𝑦+34=0174π‘₯βˆ’24𝑦+234=017π‘₯βˆ’2𝑦+23=0.

Therefore, the equation of the line passing through 𝐴(βˆ’1,3) and the intersection of the lines 3π‘₯βˆ’π‘¦+5=0 and 5π‘₯+2𝑦+3=0 is 17π‘₯βˆ’2𝑦+23=0.

As an alternative method, we can find the equation of a line by using two distinct points on the line. Therefore, we use the intersection point and point 𝐴 to find the equation of the line.

We can find the intersection of 3π‘₯βˆ’π‘¦+5=0 and 5π‘₯+2𝑦+3=0 by solving the equations simultaneously using the elimination method. Using an elimination method. We can number our equations as

3π‘₯βˆ’π‘¦+5=0,5π‘₯+2𝑦+3=0.(6)(7)

In order to eliminate either the π‘₯- or 𝑦-variables, their absolute values need to be the same in both equations. We observe that we can multiply equation (6) by 2 in order to have the same absolute value of 2𝑦 in each equation. Thus, we have

6π‘₯βˆ’2𝑦+10=0,5π‘₯+2𝑦+3=0.(8)(9)

We then eliminate 𝑦 by adding the two equations (8) and (9): 6π‘₯βˆ’2𝑦+10=0+5π‘₯+2𝑦+3=011π‘₯+13=0

Next, rearranging 11π‘₯+13=0 by subtracting 13 from both sides of the equation, and then dividing by 11, gives us 11π‘₯=βˆ’13π‘₯=βˆ’1311.

We have now found the π‘₯-coordinate of the point of intersection, so substituting this into either of the equations (6) or (7) would allow us to find the 𝑦-coordinate. Substituting π‘₯=βˆ’1311 into equation (6), and simplifying, we have 3ο€Όβˆ’1311οˆβˆ’π‘¦+5=0βˆ’3911βˆ’π‘¦+5=01611βˆ’π‘¦=0𝑦=1611.

This gives us the point of intersection ο€Όβˆ’1311,1611.

Now, we need to find the equation of the line passing through the points 𝐴(βˆ’1,3) and ο€Όβˆ’1311,1611.

We recall that a line containing a coordinate (π‘₯,𝑦), with slope π‘š, is given in point–slope form as π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯).

To use this form, we need to calculate the slope, π‘š, of a line joining two points (π‘₯,𝑦) and (π‘₯,𝑦), which is given by π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯.

We can designate ο€Όβˆ’1311,1611 as (π‘₯,𝑦) and (βˆ’1,3) as (π‘₯,𝑦). Substituting these in, we have π‘š=3βˆ’βˆ’1βˆ’ο€»ο‡=3βˆ’βˆ’1+=βˆ’+==172.

We now have the slope of the line being 172, along with two points that lie on the line. We only need one of these points to be able to use the point–slope form of an equation alongside the slope.

Thus, substituting 𝐴(βˆ’1,3) for the point (π‘₯,𝑦) and the slope π‘š=172 into the equation π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯), we have π‘¦βˆ’3=172(π‘₯βˆ’(βˆ’1))π‘¦βˆ’3=172(π‘₯+1).

Distributing 172 across the parentheses on the right-hand side, and then adding 3 to both sides of the equation, gives us π‘¦βˆ’3=172π‘₯+172𝑦=172π‘₯+232.

This is a valid equation for the line. However, we can also write this in the general form of the equation of a line, π‘Žπ‘₯+𝑏𝑦+𝑐=0, where π‘Ž,𝑏,π‘βˆˆβ„and.

We multiply all the terms by 2 and then subtract 2𝑦 from both sides of the equation, which gives 2𝑦=17π‘₯+230=17π‘₯βˆ’2𝑦+23.

This confirms the answer we found using the first method; the equation of the line is 17π‘₯βˆ’2𝑦+23=0.

We can now summarize the key points.

Key Points

  • The point of intersection between two distinct lines is the point where they meet or cross. It is the ordered pair of the values of π‘₯ and 𝑦 where the lines meet on the graph and that satisfies the equations of both lines.
  • Distinct, parallel lines are lines in a plane that are always the same distance apart. They will have no points of intersection.
  • We can find the intersection of two lines either graphically or algebraically. Algebraic solutions will always give an accurate result.
  • To find the point of intersection between two nonparallel lines, algebraically, we solve the system of two equations. The solution values of π‘₯ and 𝑦 form the intersection point (π‘₯,𝑦).
  • The equation that represents all straight lines passing through an intersection point of two lines π‘Žπ‘₯+𝑏𝑦+𝑐=0 and π‘Žπ‘₯+𝑏𝑦+𝑐=0 is π‘š(π‘Žπ‘₯+𝑏𝑦+𝑐)+𝑙(π‘Žπ‘₯+𝑏𝑦+𝑐)=0, where π‘š,π‘™βˆˆβ„. When π‘šβ‰ 0 and 𝑙≠0, then for any π‘˜βˆˆβ„; the equation of a straight line passing through the point of intersection can be written as π‘Žπ‘₯+𝑏𝑦+𝑐+π‘˜(π‘Žπ‘₯+𝑏𝑦+𝑐)=0.

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