Video: Pack 5 β€’ Paper 1 β€’ Question 22

Pack 5 β€’ Paper 1 β€’ Question 22

06:41

Video Transcript

An isosceles triangle has vertices at the points 𝐴 negative three, negative three; 𝐡 negative one, negative one; and 𝐢 negative five, one. Angle 𝐢𝐴𝐡 is equal to angle 𝐢𝐡𝐴. 𝑀 is the midpoint of 𝐴𝐡. Work out the equation of the line that passes through 𝐢 and 𝑀. Give your answer in the form π‘Žπ‘¦ plus 𝑏π‘₯ plus 𝑐 is equal to zero, where π‘Ž, 𝑏, and 𝑐 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. 𝑀 is the midpoint of 𝐴𝐡. And we’re looking for the equation of the line that passes through the points 𝐢 and 𝑀. So that’s the line that I’ve drawn on in orange. Now, we’re given some other information in the question. We’re told that angle 𝐢𝐴𝐡 is equal to angle 𝐢𝐡𝐴 β€” those these two angles here. This tells us that the two sides of the isosceles triangle which equal in length are sides 𝐢𝐴 and 𝐢𝐡.

Now, as the triangle is isosceles with 𝐢𝐴 equal to 𝐢𝐡, this means that the line drawn from the point 𝐢 to the midpoint of the opposite side β€” so that’s the line 𝐴𝐡 β€” divides the triangle up into two identical right-angled triangles. Importantly, this means that the line through the points 𝐢 and 𝑀 is perpendicular to the line 𝐴𝐡.

Now, why is this important? Well, remember the general form of the equation of a straight line is 𝑦 equals π‘šπ‘₯ plus 𝑐, where π‘š represents the gradient of the line and 𝑐 represents the 𝑦-intercept. This looks a little different from the form we’ve 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 𝐢 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 it’s perpendicular to the line 𝐴𝐡 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‘š one and the gradient of the perpendicular line as π‘š two, we can write this as π‘š one multiplied by π‘š two is equal to negative one.

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

The gradient of a line is found by dividing the change in 𝑦 by the change in π‘₯, which is sometimes written as 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one. We can choose 𝐴 to be the point π‘₯ one, 𝑦 one and 𝐡 to be the point π‘₯ two, 𝑦 two. Substituting the values of π‘₯ one, 𝑦 one; π‘₯ two and 𝑦 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 𝐡 to be the point π‘₯ one, 𝑦 one and the point 𝐴 to be the point π‘₯ two, 𝑦 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 we’d get the same answer for our gradient, whichever way around we label the two points.

So we found that the gradient of the line 𝐴𝐡 is one. But remember we need the gradient to the line through the points 𝐢 and 𝑀. 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 𝐢 and 𝑀 is sloping downwards. So it must have a negative gradient.

We can now substitute the gradient we’ve calculated into the general form of the equation of a straight line, giving 𝑦 equals negative π‘₯ plus 𝑐. Remember we don’t write negative one π‘₯. We just write negative π‘₯. To find the value of this lowercase letter 𝑐, the constant term in the equation, we need to substitute the coordinates of the point 𝐢 into the equation of our line.

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

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

Remember we’ve been asked to give the equation of our line in a particular form: π‘Žπ‘¦ plus 𝑏π‘₯ plus 𝑐 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 𝑦 plus π‘₯ plus four is equal to zero. This is the equation of the line in the required form.

We haven’t been asked to explicitly state the values of π‘Ž, 𝑏, and 𝑐. But comparing our equation to the requested form, we can see the π‘Ž and 𝑏 are both equal to one and 𝑐 is equal to four.

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