Lesson Video: Intersection Point of Two Straight Lines on the Coordinate Plane | Nagwa Lesson Video: Intersection Point of Two Straight Lines on the Coordinate Plane | Nagwa

Lesson Video: Intersection Point of Two Straight Lines on the Coordinate Plane Mathematics • First Year of Secondary School

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In this video, 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.

14:27

Video Transcript

In this video, we will learn how to find the intersection point of two straight lines on the coordinate plane. And we’ll use this concept to find equations of lines.

So, let’s begin by thinking about what we mean by the intersection point of lines. We say that the point of intersection of two distinct, nonparallel lines is the single point where they meet or cross. So we could, for example, have the line segment 𝐴𝐡 and another line segment 𝑃𝑄. We could label where these two line segments cross as the point 𝐹. Therefore, point 𝐹 is the intersection of line segment 𝐴𝐡 and line segment 𝑃𝑄. Notice that as part of this definition of the intersection of lines, we used the word β€œnonparallel.” That’s because if we have two distinct parallel lines, they are always the same distance apart and they will never intersect.

So, when it comes to finding the intersection of two lines on a coordinate plane, the same principle still applies. We’re looking to find the point where the two lines meet or cross. We’ll usually be given the equations of two lines like this. So, the intersection point 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. We can find this ordered pair graphically by drawing or algebraically by solving to find the π‘₯- and 𝑦-values. Notice that because we usually have two equations with two unknowns of π‘₯ and 𝑦, then we’ll often need to solve simultaneously or by using a substitution method.

In this video, we’ll look at different methods of solving algebraically. But first, let’s have a look at an example where we use a graphical method to find the intersection of two lines.

At which point do the lines π‘₯ equals seven and one-sixth 𝑦 equals negative one intersect?

In this question, we have the equations of two lines and we’re asked where these two lines intersect, which would be the point where they meet or cross. It might be useful to begin by visualizing what these two lines would look like on the coordinate plane. The line π‘₯ equals seven indicates all the ordered pairs that have an π‘₯-value of seven. And so we’ll have a vertical line that goes through seven on the π‘₯-axis.

For the second equation of one-sixth 𝑦 equals negative one, sometimes it’s easier to visualize the equation of a line like this if we write the equation with 𝑦 as the subject. Rearranging by multiplying through by six would give us the equation of 𝑦 equals negative six. So, the equation of the line 𝑦 equals negative six, or one-sixth 𝑦 equals negative one, will be a horizontal line passing through negative six on the 𝑦-axis.

The intersection point then is the ordered pair where these two lines intersect. Using the graph, we can see that this occurs at the point seven, negative six. And so this is the answer for the intersection point of the two lines.

If we wanted to consider an algebraic method here instead of drawing the lines, usually when we find the intersection of two lines, we could set the equations equal to each other. But as we had a horizontal line and a vertical line here, the only point where these two have the same π‘₯- and 𝑦-values is when π‘₯ is equal to seven and 𝑦 is equal to negative six, which would also give us the coordinates seven, negative six.

In the next examples, we’ll use an algebraic method to find the intersection of two lines.

Determine the point of intersection of the two straight lines represented by the equations π‘₯ plus three 𝑦 minus two equals zero and negative 𝑦 plus one equals zero.

Let’s say that to answer this question, we’re not going to draw these lines to get a graphical solution. But instead, we’re going to solve these algebraically. At the point of intersection, that’s the place where the two lines meet or cross, the π‘₯- and 𝑦-values must be the same. As we have two equations with the two unknowns of π‘₯ and 𝑦, then we can solve simultaneously or by using a substitution method. However, in our second equation, notice that we don’t actually have an π‘₯-value. So perhaps the more efficient way to solve for π‘₯ and 𝑦 is by using a substitution method. If we take this second equation of negative 𝑦 plus one equals zero and rearrange this to make 𝑦 the subject, by adding 𝑦 to both sides, we would get one equals 𝑦 or 𝑦 equals one.

And now that we have established that 𝑦 is equal to one, we can substitute this into the first equation. This gives us π‘₯ plus three times one minus two equals zero. Evaluating this, we have π‘₯ plus one equals zero. And so π‘₯ is equal to negative one. Now we know that at the point of intersection of these two lines, the π‘₯-value is negative one and the 𝑦-value is one, which means that we can give our answer as the coordinates negative one, one.

Let’s now consider how we could create the general equation of the straight lines passing through the point of intersection of two given lines. Let’s say we have two lines given as π‘Ž sub one π‘₯ plus 𝑏 sub one 𝑦 plus 𝑐 sub one equals zero and π‘Ž sub two π‘₯ plus 𝑏 sub two 𝑦 plus 𝑐 sub two equals zero. Then, we can write a general equation for any line which passes through the intersection of these two lines. We say that for the two given lines, then the equation of all straight lines passing through their intersection can be given as π‘š times π‘Ž sub one π‘₯ plus 𝑏 sub one 𝑦 plus 𝑐 sub one plus 𝑙 times π‘Ž sub two π‘₯ plus 𝑏 sub two 𝑦 plus 𝑐 sub two equals zero, where π‘š and 𝑙 are in the set of real numbers.

We can also note two important points about the values of π‘š and 𝑙. If π‘š is equal to zero, we would just have the equation of the second line. And if 𝑙 is equal to zero, we would simply have the equation of the first line. So, if π‘š is not equal to zero and 𝑙 is not equal to zero, we have the equation of a straight line passing through the point of intersection, excluding the original lines. And hence we can write the equation above in the form π‘Ž sub one π‘₯ plus 𝑏 sub one 𝑦 plus 𝑐 sub one plus π‘˜ times π‘Ž sub two π‘₯ plus 𝑏 sub two 𝑦 plus 𝑐 sub two equals zero, for any π‘˜ in the set of real numbers. And it is this form of the equation that we will see how we can apply in the following example.

What is the equation of the line passing through 𝐴 with coordinates negative one, three and the intersection of the lines three π‘₯ minus 𝑦 plus five equals zero and five π‘₯ plus two 𝑦 plus three equals zero?

We can begin by recalling that the intersection of two lines is the point where they meet or cross. We can also recall that there is a general equation of a line passing through the intersection point of two lines. If two lines have equations π‘Ž sub one π‘₯ plus 𝑏 sub one 𝑦 plus 𝑐 sub one equals zero and π‘Ž sub two π‘₯ plus 𝑏 sub two 𝑦 plus 𝑐 sub two equals zero, then we can write the equation as π‘Ž sub one π‘₯ plus 𝑏 sub one 𝑦 plus 𝑐 sub one plus π‘˜ times π‘Ž sub two π‘₯ plus 𝑏 sub two 𝑦 plus 𝑐 sub two equals zero, for any π‘˜ in the set of real numbers.

So, let’s take the equation three π‘₯ minus 𝑦 plus five equals zero to have the π‘Ž sub one, 𝑏 sub one, and 𝑐 sub one values. Meaning that π‘Ž sub one is three, 𝑏 sub one is negative one, and 𝑐 sub one is five. And in the same way, we can take the second equation for the π‘Ž sub two, 𝑏 sub two, and 𝑐 sub two values, which will be equal to five, two, and three, respectively. Then, we substitute these into the general equation of the line, which gives three π‘₯ minus 𝑦 plus five plus π‘˜ times five π‘₯ plus two 𝑦 plus three equals zero.

At this point, what we have worked out is the general equation of a line which passes through the intersection of the two lines three π‘₯ minus 𝑦 plus five equals zero and five π‘₯ plus two 𝑦 plus three equals zero. And there would in fact be an infinite number of lines that pass through their intersection point given by the infinite number of values we could take for π‘˜. However, we need to work out the equation of one specific line. It’s the line which passes through this intersection point and the point 𝐴 with coordinates negative one, three. And so we can substitute the values π‘₯ equals negative one and 𝑦 equals three into this equation. When we do this and simplify, we have that negative one plus π‘˜ times four equals zero. Rearranging, we have four π‘˜ equals one and π‘˜ equals one-quarter.

Now that we’ve worked out the value of π‘˜, we can substitute it into this general equation. And this gives three π‘₯ minus 𝑦 plus five plus one-quarter times five π‘₯ plus two 𝑦 plus three equals zero. To simplify, we distribute the one-quarter across the parentheses, and then we collect the like terms. This equation of 17 over four π‘₯ minus two-quarters 𝑦 plus 23 over four equals zero would be a valid equation of the line. But it would be a little bit nicer to write this with integer values for the coefficients of π‘₯ and 𝑦. So, by multiplying through by four, we have 17π‘₯ minus two 𝑦 plus 23 equals zero. And that’s the answer for the equation of the line which passes through the point 𝐴 and the intersection of the two given lines.

Notice that another way in which we could’ve solved this example without using this general equation form is by first solving the equations of the two lines simultaneously to find the intersection point. We would then have needed to calculate the slope of the line between the two points and used this along with the point–slope form of a line to obtain the equation that we found.

Let’s now summarize the key points of this video. The point of intersection between two distinct nonparallel 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. But note that 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. Then, the solution values of π‘₯ and 𝑦 form the intersection point with coordinates π‘₯, 𝑦. We also saw how we can create a general equation for any line which passes through the intersection point of two lines, which we can write as π‘š times π‘Ž sub one π‘₯ plus 𝑏 sub one 𝑦 plus 𝑐 sub one plus 𝑙 times π‘Ž sub two π‘₯ plus 𝑏 sub two 𝑦 plus 𝑐 sub two equals zero, where π‘š and 𝑙 are in the set of real numbers.

Importantly, by recognizing that when π‘š and 𝑙 are nonzero, we have the equation π‘Ž sub one π‘₯ plus 𝑏 sub one 𝑦 plus 𝑐 sub one plus π‘˜ times π‘Ž sub two π‘₯ plus 𝑏 sub two 𝑦 plus 𝑐 sub two equals zero, for any π‘˜ in the set of real numbers. And because π‘˜ can have an infinite set of values, then there are an infinite number of lines passing through the intersection point of two lines. As we saw in the previous example, if we are asked to find the equation of a specific line passing through an intersection point, then we input any additional information, such as a point on the line, to determine the exact equation.

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