Video: Finding the Coordinates of Points on a Line Segment and Finding the Equation of a Line

A quadrilateral has its vertices at the points 𝐴(βˆ’5, 3), 𝐡(0, βˆ’2), 𝐢(βˆ’2, βˆ’6), and 𝐷(βˆ’8, βˆ’2). A point 𝐸 lies on 𝐴𝐢 such that the lengths of 𝐴𝐸 and 𝐢𝐸 are in the ratio of 1 : 2, and a point 𝐹 lies on 𝐡𝐷 such that the lengths of 𝐡𝐹 and 𝐷𝐹 are in the ratio of 1 : 3. Find the coordinates of 𝐸. Find the coordinates of 𝐹. Find the slope of the line 𝐸𝐹. Find the equation of the line 𝐸𝐹, giving your answer in the form 𝑦 = π‘šπ‘₯ + 𝑐.

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Video Transcript

A quadrilateral has its vertices at the points 𝐴 negative five, three; 𝐡 zero, negative two; 𝐢 negative two, negative six; and 𝐷 negative eight, negative two. A point 𝐸 lies on 𝐴𝐢 such that the lengths of 𝐴𝐸 and 𝐢𝐸 are in the ratio of one to two. And a point 𝐹 lies on 𝐡𝐷 such that the lengths of 𝐡𝐹 and 𝐷𝐹 are in the ratio of one to three. Find the coordinates of 𝐸. Find the coordinates of 𝐹. Find the slope of the line 𝐸𝐹. Find the equation of the line 𝐸𝐹, giving your answer in the form 𝑦 equals π‘šπ‘₯ plus 𝑐.

It would seem sensible to start by plotting the four coordinates that we’re given for the quadrilateral. We can then join our four points to make the quadrilateral 𝐴𝐡𝐢𝐷. We can now start by looking at our first question to find the coordinates of 𝐸.

We’re given that 𝐸 lies on the line segment 𝐴𝐢. We need to establish the coordinates of point 𝐸, given that line segments 𝐴𝐸 and 𝐢𝐸 are in the ratio of one to two. To do this, we can use the section formula. The section formula allows us to find the coordinates of a point which divides a line segment into a given ratio. So here we have coordinate 𝐸, which divides the segment between 𝐴 and 𝐢 in the ratio one to two. It can be helpful to note down the values of π‘₯ one, 𝑦 one; π‘₯ two, 𝑦 two; and π‘š one, π‘š two, which we will substitute into the section formula.

Therefore, the π‘₯-value of our coordinate 𝐸 is given by one times negative two plus two times negative five over two plus one. And the 𝑦-value will be given by one times negative six plus two times three over two plus one. Simplifying, we have the coordinate negative two minus 10 over three, negative six plus six over three. This gives us negative 12 over three, which is four. And the 𝑦-value, zero over three, is simply zero. So now we know that our coordinate 𝐸 lies on the π‘₯-axis. And we found the answer for question one.

We will now clear some space to look at answering question two. In this part of the question, we’re asked to find the coordinates of point 𝐹, which lies on the line segment 𝐡𝐷. 𝐡𝐹 and 𝐷𝐹 are in the ratio of one to three. Notice that if we were doing it on the diagram, it would appear to be a ratio of three to one since 𝐡𝐹 is the one part of the ratio and 𝐷𝐹 is the portion represented by three. We’re going to apply the section formula again to answer this question to find the coordinates of 𝐹.

We can assign our values of π‘₯ one, 𝑦 one; π‘₯ two, 𝑦 two; π‘š one and π‘š two to our coordinates 𝐡, 𝐷 and the ratio one to three. Substituting these into the section formula then, we have 𝐹 equals one times negative eight plus three times zero over three plus one. And the 𝑦-coordinate is one times negative two plus three times negative two over three plus one. So we have the coordinate negative eight plus zero over four, negative two minus six over four. And then, since negative eight over four simplifies to negative two, we will have the coordinate negative two, negative two. And therefore, we have answered question two to find the coordinates of 𝐹, which is negative two, negative two.

We’ll now clear some space to find the slope of the line 𝐸𝐹. We have established that coordinate 𝐸 is at negative four, zero. And coordinate 𝐹 is at negative two, negative two. We can recall that the slope or gradient of the line joining the coordinate π‘₯ one, 𝑦 one and the coordinate π‘₯ two, 𝑦 two is given by 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one.

When we are using this formula, it doesn’t matter which coordinate we designate as the first coordinate and which is the second coordinate. Substituting our values then, we have the slope equals negative two minus zero over negative two minus negative four. On the denominator, the minus negative four is equivalent to adding four. So we have the fraction negative two over two, which is equal to negative one. And therefore, the slope of the line 𝐸𝐹 is negative one.

In the final question part, we’re asked to find the equation of the line 𝐸𝐹 in the form 𝑦 equals π‘šπ‘₯ plus 𝑐. When an equation is in this form, the π‘š represents the slope or gradient. And the 𝑐-value is the 𝑦-intercept. Since we have already started at the line 𝐸𝐹, we could in theory continue this line to see where it crosses the 𝑦-axis. It looks as though our 𝑦-intercept will be at negative four. But let’s see if we can show this using a more formal method.

We can start by writing the general form of the equation 𝑦 equals π‘šπ‘₯ plus 𝑐. And since we know that the slope is equal to negative one, we can substitute in the value π‘š equals one to give 𝑦 equals negative one π‘₯ plus 𝑐. Or more simply, 𝑦 equals negative π‘₯ plus 𝑐. At this point, we still don’t know our 𝑐, the 𝑦-intercept. But we can substitute in one of the coordinates that we do know.

Taking the π‘₯- and 𝑦-values in coordinate 𝐸, we can substitute π‘₯ equals negative four and 𝑦 equals zero into our equation, giving us zero equals minus negative four plus 𝑐. So zero equals four plus 𝑐. And subtracting four from both sides will give us negative four equals 𝑐. So 𝑐 equals negative four. We can then substitute this into the equation 𝑦 equals negative π‘₯ plus 𝑐, giving us 𝑦 equals negative π‘₯ minus four. And finally, we can factor out the negative to give us 𝑦 equals negative π‘₯ plus four.

Therefore, we have our answer for the final question part. The equation of the line 𝐸𝐹 is 𝑦 equals negative π‘₯ plus four.

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