### Video Transcript

An isosceles triangle has vertices at the points π΄ negative three, negative three; π΅ negative one, negative one; and πΆ negative five, one. Angle πΆπ΄π΅ is equal to angle πΆπ΅π΄. π is the midpoint of π΄π΅. Work out the equation of the line that passes through πΆ and π. Give your answer in the form ππ¦ plus ππ₯ plus π is equal to zero, where π, π, and π are integers.

It can be really helpful to start a question like this with a sketch to help us visualize the problem more clearly. As it is only a sketch, we are going to plot the coordinates of the three vertices accurately, but they should be roughly in the right place relative to each other.

The triangle looks something like this. π is the midpoint of π΄π΅. And weβre looking for the equation of the line that passes through the points πΆ and π. So thatβs the line that Iβve drawn on in orange. Now, weβre given some other information in the question. Weβre told that angle πΆπ΄π΅ is equal to angle πΆπ΅π΄ β those these two angles here. This tells us that the two sides of the isosceles triangle which equal in length are sides πΆπ΄ and πΆπ΅.

Now, as the triangle is isosceles with πΆπ΄ equal to πΆπ΅, this means that the line drawn from the point πΆ to the midpoint of the opposite side β so thatβs the line π΄π΅ β divides the triangle up into two identical right-angled triangles. Importantly, this means that the line through the points πΆ and π is perpendicular to the line π΄π΅.

Now, why is this important? Well, remember the general form of the equation of a straight line is π¦ equals ππ₯ plus π, where π represents the gradient of the line and π represents the π¦-intercept. This looks a little different from the form weβve been asked to give our answer in. But we can always rearrange later.

To find the equation of a straight line, we generally need to know two pieces of information: its gradient and the coordinates of any point on the line. We already know the coordinates of one point on our line because we know the point πΆ is on the line and it has the coordinates negative five, one. We, therefore, need to calculate the gradient of our line.

And the fact itβs perpendicular to the line π΄π΅ will be really useful here. Remember that if two lines are perpendicular, then the product of their gradients is equal to negative one. If we think of the gradient of one line as π one and the gradient of the perpendicular line as π two, we can write this as π one multiplied by π two is equal to negative one.

π one and π two are the negative reciprocals of one another. That means negative one divided by each other as we can divide this equation by π two to give π one equals negative one over π two or we can divide it by π one to give π two equals negative one over π one. This means that then if we can find the gradient of the line π΄π΅, we can find the gradient of the line through πΆ and π by finding the negative reciprocal of our value.

The gradient of a line is found by dividing the change in π¦ by the change in π₯, which is sometimes written as π¦ two minus π¦ one over π₯ two minus π₯ one. We can choose π΄ to be the point π₯ one, π¦ one and π΅ to be the point π₯ two, π¦ two. Substituting the values of π₯ one, π¦ one; π₯ two and π¦ two into our equation for the gradient gives negative one minus negative three over negative one minus negative three.

Now, negative one minus negative three is the same as negative one plus three. So both the numerator and denominator of this fraction simplify to two. We have two over two, which is equal to one. Now, notice that we could have chosen the point π΅ to be the point π₯ one, π¦ one and the point π΄ to be the point π₯ two, π¦ two. In this case, it would have given negative three minus negative one over negative three minus negative one. This would have simplified to negative two over negative two. But negative two divided by negative two is still one. So weβd get the same answer for our gradient, whichever way around we label the two points.

So we found that the gradient of the line π΄π΅ is one. But remember we need the gradient to the line through the points πΆ and π. So we find the negative reciprocal of one, which is negative one divided by one, which is negative one. This value makes sense as we can see from our sketch that the line through the points πΆ and π is sloping downwards. So it must have a negative gradient.

We can now substitute the gradient weβve calculated into the general form of the equation of a straight line, giving π¦ equals negative π₯ plus π. Remember we donβt write negative one π₯. We just write negative π₯. To find the value of this lowercase letter π, the constant term in the equation, we need to substitute the coordinates of the point πΆ into the equation of our line.

The π¦-coordinate of the point πΆ is one and the π₯-coordinate is negative five. So substituting these values for π¦ and π₯ gives the equation one is equal to negative one multiplied by negative five plus π. This is an equation that we can solve to find π. Negative one multiplied by negative five is positive five. So the equation becomes one equals five plus π.

To solve, we subtract five from each side of the equation, giving negative four is equal to π. Substituting this value π back into the equation of our line gives π¦ is equal to negative π₯ minus four.

Remember weβve been asked to give the equation of our line in a particular form: ππ¦ plus ππ₯ plus π is equal to zero. So this means we need all the terms on the same side of the equation. We need to add both π₯ and four to both sides of the equation, giving π¦ plus π₯ plus four is equal to zero. This is the equation of the line in the required form.

We havenβt been asked to explicitly state the values of π, π, and π. But comparing our equation to the requested form, we can see the π and π are both equal to one and π is equal to four.