### Video Transcript

A quadrilateral has its vertices at the points π΄ negative five, three; π΅ zero, negative two; πΆ negative two, negative six; and π· negative eight, negative two. A point πΈ lies on π΄πΆ such that the lengths of π΄πΈ and πΆπΈ are in the ratio of one to two. And a point πΉ lies on π΅π· such that the lengths of π΅πΉ and π·πΉ are in the ratio of one to three. Find the coordinates of πΈ. Find the coordinates of πΉ. Find the slope of the line πΈπΉ. Find the equation of the line πΈπΉ, giving your answer in the form π¦ equals ππ₯ plus π.

It would seem sensible to start by plotting the four coordinates that weβre given for the quadrilateral. We can then join our four points to make the quadrilateral π΄π΅πΆπ·. We can now start by looking at our first question to find the coordinates of πΈ.

Weβre given that πΈ lies on the line segment π΄πΆ. We need to establish the coordinates of point πΈ, given that line segments π΄πΈ and πΆπΈ 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 πΈ, which divides the segment between π΄ and πΆ in the ratio one to two. It can be helpful to note down the values of π₯ one, π¦ one; π₯ two, π¦ two; and π one, π two, which we will substitute into the section formula.

Therefore, the π₯-value of our coordinate πΈ is given by one times negative two plus two times negative five over two plus one. And the π¦-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 π¦-value, zero over three, is simply zero. So now we know that our coordinate πΈ 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, weβre asked to find the coordinates of point πΉ, which lies on the line segment π΅π·. π΅πΉ and π·πΉ 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 π΅πΉ is the one part of the ratio and π·πΉ is the portion represented by three. Weβre going to apply the section formula again to answer this question to find the coordinates of πΉ.

We can assign our values of π₯ one, π¦ one; π₯ two, π¦ two; π one and π two to our coordinates π΅, π· and the ratio one to three. Substituting these into the section formula then, we have πΉ equals one times negative eight plus three times zero over three plus one. And the π¦-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 πΉ, which is negative two, negative two.

Weβll now clear some space to find the slope of the line πΈπΉ. We have established that coordinate πΈ is at negative four, zero. And coordinate πΉ is at negative two, negative two. We can recall that the slope or gradient of the line joining the coordinate π₯ one, π¦ one and the coordinate π₯ two, π¦ two is given by π¦ two minus π¦ one over π₯ two minus π₯ one.

When we are using this formula, it doesnβt 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 πΈπΉ is negative one.

In the final question part, weβre asked to find the equation of the line πΈπΉ in the form π¦ equals ππ₯ plus π. When an equation is in this form, the π represents the slope or gradient. And the π-value is the π¦-intercept. Since we have already started at the line πΈπΉ, we could in theory continue this line to see where it crosses the π¦-axis. It looks as though our π¦-intercept will be at negative four. But letβs see if we can show this using a more formal method.

We can start by writing the general form of the equation π¦ equals ππ₯ plus π. And since we know that the slope is equal to negative one, we can substitute in the value π equals one to give π¦ equals negative one π₯ plus π. Or more simply, π¦ equals negative π₯ plus π. At this point, we still donβt know our π, the π¦-intercept. But we can substitute in one of the coordinates that we do know.

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

Therefore, we have our answer for the final question part. The equation of the line πΈπΉ is π¦ equals negative π₯ plus four.