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

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.

15:09

Video Transcript

In this video, we will learn how to find the intersection point of two straight lines on the coordinate plane and use this concept to find equations of lines. Let’s begin by thinking about what we mean by the intersection point. We say that the intersection of two lines is the single point where they meet or cross each other. So we could, for example, have this line segment 𝐴𝐡 and another line segment 𝑃𝑄. We could then mark on the point 𝐹 where these two line segments cross. Therefore, point 𝐹 is the intersection of line segment 𝐴𝐡 and line segment 𝑃𝑄. 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. So, the intersection point can be given as an ordered pair or coordinate. 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. We’ll look at a number of different methods of solving algebraically. But let’s have a look at the first question where we use a graphical method.

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 visualizing what these two lines would look like on the coordinate grid. The line π‘₯ equals seven indicates all the ordered pairs that have an π‘₯-value of seven. We’ll have a vertical line that goes through seven on the π‘₯-axis. For the second equation of a sixth 𝑦 equals negative one, sometimes it’s more easy to visualize if we have it 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 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. We can therefore give the answer of the coordinate seven, negative six. Usually when we find the intersection of two points, we could set the equations equal to each other. But as we had a horizontal and a vertical line here, the only points where these two have the same π‘₯- and 𝑦-values is when π‘₯ is equal to seven and 𝑦 is equal to negative six, giving us the coordinate seven, negative six.

In the next few questions, 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 to answer this question we’re not going to draw these lines to get a graphical solution. 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 will be the same. As we have two equations with the two unknowns of π‘₯ or 𝑦, then we’re going to need to solve this simultaneously or by using a substitution method. However, in our second equation, we don’t actually have an π‘₯-value. So perhaps, a substitution method here is the easiest. If we take our second equation of negative 𝑦 plus one equals zero and rearrange this to make 𝑦 the subject, then by adding 𝑦 to both sides, we would get one equals 𝑦 or 𝑦 equals one.

Now that we’ve established that 𝑦 is equal to one, we can plug this into the first equation. This gives us π‘₯ plus three times one subtract two equals zero. Evaluating this, we have π‘₯ plus one equals zero. Subtracting negative one, we have π‘₯ is equal to negative one. Now we know that at the point of intersection of these two equations, the π‘₯-value is negative one and the 𝑦-value is one, which means that we can give our answer as the coordinate negative one, one.

In the next question, we’ll find the equation of a line passing through a point and the intersection point of another two lines.

What is the equation of the line passing through 𝐴 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?

In this question, we’re really interested in finding two points. We’re given this point 𝐴 of negative one, three, and then we’ll find another point which is the intersection of these two lines. So, the first step to take here is to find the intersection of these lines. At the intersection point of two lines, the π‘₯- and the 𝑦-values are the same. You might observe that we have two unknowns to find, π‘₯ and 𝑦. But we do have two different equations to help us solve it. One way to solve it is by solving these two equations simultaneously. It’s helpful to number our equations equation one and equation two. And then we need to decide which value of π‘₯ or 𝑦 to eliminate.

In order to eliminate either the π‘₯- or 𝑦-value, their absolute values need to be the same. So if we choose to eliminate the 𝑦-value, then we’ll need to multiply all of equation one by two, which gives us six π‘₯ minus two 𝑦 plus 10 equals zero. We can then put the second equation underneath. In order to eliminate the two 𝑦’s as we have a negative two 𝑦 and a positive two 𝑦, we’ll be adding our two equations. Six π‘₯ plus five π‘₯ gives us 11π‘₯. Negative two 𝑦 plus two 𝑦 does indeed give us zero, which is what we wanted. 10 plus three gives us 13. And on the right-hand side, we have zero. Now that we have 11π‘₯ plus 13 equals zero, we can rearrange this to find the value of π‘₯. Subtracting 13 from both sides and then dividing through by 11 gives us that π‘₯ is equal to negative 13 over 11.

Now that we found the value of π‘₯, we can plug this into either of our equations to find the value of 𝑦. Choosing the first equation would give us three multiplied by negative 13 over 11 minus 𝑦 plus five equals zero. Simplifying the first value gives us negative 39 over 11 and then we can add five. Remembering that five is equal to fifty-five elevenths, we’ll get the equation sixteen elevenths minus 𝑦 equals zero. Adding 𝑦 to both sides gives us sixteen elevenths equals 𝑦 or 𝑦 equals sixteen elevenths. Now that we found these π‘₯- and 𝑦-values, it means that we know that the intersection point of our two lines is at the point negative thirteen elevenths, sixteen elevenths.

Next, we need to find the equation of the line that joins this point with the point 𝐴 negative one, three. Let’s start by finding the slope of this line. The slope or π‘š-value for any two points π‘₯ one, 𝑦 one and π‘₯ two, 𝑦 two is given by 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one. It doesn’t matter which coordinate we designate with π‘₯ one, 𝑦 one values. So let’s put those values with the point 𝐴. This gives us that the slope or π‘š is equal to sixteen elevenths subtract three over negative thirteen elevenths subtract negative one. Simplifying the numerator, sixteen elevenths subtract thirty-nine elevenths gives us negative seventeen elevenths. On the denominator, we have negative thirteen elevenths plus one, and one is equivalent to eleven elevenths. So we have the answer on the denominator of negative two elevenths.

Before we simplify this anymore, we can eliminate the negative from both the numerator and denominator. We can then recognize that this seventeen elevenths over two elevenths is equivalent to seventeen elevenths divided by two elevenths. When we’re dividing fractions, it’s equivalent to multiplying by the reciprocal of the second fraction. Taking out the common factor of 11 from the numerator and denominator before multiplying, we then get the answer that π‘š is equal to 17 over two. This means that the slope of the equation joining our two points is 17 over two. Let’s clear some space for the next part of our working.

Now that we have our two coordinates and the slope of the line, we can use the point-slope form to find the equation of the line. When we have the slope π‘š of a line and a coordinate π‘₯ one, 𝑦 one, then the equation is given by 𝑦 minus 𝑦 one equals π‘š times π‘₯ minus π‘₯ one. Given we have two points, we could use either of these in the equation, but let’s use the coordinate of 𝐴 at negative one, three with the π‘₯ one, 𝑦 one values. Plugging these into the formula, we have 𝑦 minus three, as that was our 𝑦 one value, equals 17 over two, as that was the π‘š-value, multiplied by π‘₯ minus negative one, as that was the π‘₯ one value. On the right-hand side within the parentheses, we have minus negative one, which is equivalent to plus one. So when we distribute the parentheses on the right-hand side, we have 17 over two π‘₯ plus 17 over two.

As we want to collect together our terms, then to collect the constant terms, we’ll want to add three to both sides. Three is equivalent to six over two, so when we add that to 17 over two, we get 23 over two. At this point, we have a perfectly valid answer for the equation of a line, but sometimes it’s nice to remove any fractions if we can. So let’s multiply this equation through by two. This gives us two 𝑦 equals 17π‘₯ plus 23. And then as an equation equal to zero, we’d have zero equals 17π‘₯ minus two 𝑦 plus 23, which means that we can write our answer in the slightly nicer format of 17π‘₯ minus two 𝑦 plus 23 equals zero. And that’s the equation of the line passing through negative one, three and the intersection of the two lines three π‘₯ minus 𝑦 plus five equals zero and five π‘₯ plus two 𝑦 plus three equals zero.

Let’s have a look at another question.

Find the equation of the straight line that passes through the point of intersection of the two straight lines π‘₯ minus eight 𝑦 equals two and negative six π‘₯ minus eight 𝑦 equals one and is parallel to the 𝑦-axis.

In this question, we need to find the equation of a line which passes through the intersection point of two other lines, and that’s the point where these two lines would cross or meet. That wouldn’t be enough information to find an equation, so we’re also told that this line is parallel to the 𝑦-axis. The first step to take then is to actually find the point of intersection between our two lines. At the point of intersection between two lines, their π‘₯-values will be equal and their 𝑦-values will be equal. We have two unknowns to find, the π‘₯- and the 𝑦-values, but we also have two equations to help us. We can solve this algebraically using substitution or by using simultaneous equations.

Using the simultaneous equations method, we’d need to eliminate either the π‘₯- or 𝑦-values. But we might notice here that we have negative eight 𝑦 in both of our equations. Therefore, if we subtracted our second equation from our first equation, we could eliminate the 𝑦-value. Doing so would give us π‘₯ subtract six π‘₯, which is seven π‘₯. Negative eight 𝑦 subtract negative eight 𝑦 would give us zero, which is what we wanted. And finally, two subtract one gives us one. Now that we know that seven π‘₯ is equal to one, we can divide through by seven, which would give us π‘₯ is equal to one-seventh. Now that we found the π‘₯-value, we can plug this into either of our equations to find the value of 𝑦.

So, when π‘₯ is equal to one-seventh, using the first equation, we’d have one-seventh subtract eight 𝑦 is equal to two. Subtracting one-seventh from both sides gives us negative eight 𝑦 equals two minus one-seventh. So on the right-hand side, we’ll have thirteen sevenths. Dividing both sides by negative eight, we have 𝑦 is equal to 13 over negative 56, or 𝑦 is equal to negative 13 over 56. We have now found the two values of π‘₯ and 𝑦, which means that the intersection point of the two straight lines we were given is one-seventh, negative thirteen fifty-sixths.

We now need to find the equation of the line which passes through this point and which is parallel to the 𝑦-axis. It might be useful to think about what this might look like on a coordinate grid. The point one-seventh, negative thirteen fifty-sixths would be approximately here. The line going through this point parallel to the 𝑦-axis will be a vertical line. But what exactly is the equation of this line? All the values on this line will have an π‘₯-value of one-seventh. Therefore, the equation of the line is π‘₯ equals one-seventh. And so we’ve answered this question by finding the intersection point of two lines and then working out of the equation of this line given it was parallel to the 𝑦-axis.

We can now summarize some of the key points of this video. Firstly, we saw that the intersection point of two lines is the single point where they meet or cross each other. We can find the intersection point of two lines on the coordinate plane graphically by drawing the lines on a coordinate grid or algebraically by solving simultaneously to find the π‘₯- and 𝑦-values of the intersection point. An algebraic method will always give us an exact solution, but it can be useful to incorporate graphical methods as a check on our answer.

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