### Video Transcript

For which values of π₯ and π¦ is π΄π· a perpendicular bisector of π΅πΆ?

Weβve been given a diagram of the line segment π΅πΆ and the line π΄π· and also the lines connecting the endpoints of π΅πΆ to the point π΄. Weβve been given expressions for the length of each line in terms of the variables π₯ and π¦. And weβre asked to determine the values of π₯ and π¦ that will make π΄π· a perpendicular bisector of π΅πΆ. Letβs consider how to approach this problem. If π΄π· is to be a perpendicular bisector of π΅πΆ, then this means that the line segments π΅π· and π·πΆ must be equal in length. We can therefore set the expressions for π΅π· and π·πΆ equal to one another, giving us the equation five π¦ minus one is equal to 10 minus three π₯.

I can manipulate this equation slightly by first adding one to both sides and then adding three π₯ to both sides, giving the equation five π¦ plus three π₯ is equal to 11. Now, this is just one equation with two unknown variables, π₯ and π¦. And so we canβt solve it. We need another equation. Letβs think about this point π΄ which clearly lies on the line π΄π· which we want to be a perpendicular bisector of π΅πΆ.

At this point, we need to recall the converse of the perpendicular bisector theorem which tells us that if a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment. π΅ and πΆ at the end points of the segment, so what this means is that if π΄ is the same distance from π΅ as it is from πΆ, then π΄π· will be the perpendicular bisector of π΅πΆ. We have expressions for π΄π΅ and π΄πΆ in terms of the variables π₯ and π¦. So we can form a second equation. Three π₯ plus two is equal to five π¦ plus three.

I can also manipulate this equation by first subtracting two from both sides and then subtracting five π¦ from both sides, to give the equation three π₯ minus five π¦ is equal to one. I now have two equations in two unknowns. Which means, I can solve the equation simultaneously in order to find the values of π₯ and π¦. If we look at the equations closely, weβll see that one has five π¦ and the other has negative five π¦. Which means, if we add the two equations together, the π¦ terms will be eliminated. Adding equations one and two together gives three π₯ plus three π₯ which is six π₯ is equal to 11 plus one which is 12.

To solve this equation for π₯, we just need to divide both sides by six, giving π₯ is equal to two. Now that we know the value of π₯, we can find the value of π¦ by substituting into either of the two equations. Iβm going to choose to substitute into equation one, giving five π¦ plus three multiplied by two is equal to 11. This simplifies to five π¦ plus six is equal to 11. And subtracting six from each side, we then have that five π¦ is equal to five. Dividing both sides of the equation by five gives the value of π¦. Itβs equal to one. So we have that π₯ is equal to two and π¦ is equal to one.

By the converse of the perpendicular bisector theorem, we know that when π₯ and π¦ take these values, the point π΄ will be equidistant from the points π΅ and πΆ. And therefore, the line π΄π· will be a perpendicular bisector of π΅πΆ.