Video: CBSE Class X • Pack 5 • 2014 • Question 18

CBSE Class X • Pack 5 • 2014 • Question 18

04:02

Video Transcript

If the point 𝐴 at zero, two is equidistant from the points 𝐵 at three, 𝑝 and 𝐶 at 𝑝, five, find 𝑝, and then find the length of 𝐴𝐵.

We are told that the point 𝐴 is equidistant from points 𝐵 and 𝐶. We can therefore use the distance formula to create an equation in terms of 𝑝. The distance formula says that for two points 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, the distance between them is found by the square root of the square of the differences between the 𝑥-coordinates plus the square of the differences between the 𝑦-coordinates.

Let’s begin by forming an expression for the length of the line segment 𝐴𝐵. Now it doesn’t really matter what we choose to be 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two as long as we’re consistent. In this case, choosing 𝐴 to be 𝑥 one, 𝑦 one and 𝐵 to be 𝑥 two, 𝑦 two gives us that the length or the distance between them is the square root of three minus zero squared plus 𝑝 minus two squared. Three minus zero squared is nine. So we get that the length 𝐴𝐵 is the square root of nine plus 𝑝 minus two squared.

We’ll repeat this process for the length of the line segment 𝐴𝐶. That gives us the square root of 𝑝 minus zero squared plus five minus two squared. 𝑝 minus zero squared is simply 𝑝 squared. And five minus two squared is three squared which is nine. So the length of the line segment 𝐴𝐶 is the square root of 𝑝 squared plus nine.

Since 𝐴𝐵 and 𝐴𝐶 are equidistant, this means that 𝐴𝐵 and 𝐴𝐶 are of equal length. We can therefore equate these two expressions to get an equation in terms of 𝑝. We can first find the square of both sides to get nine plus 𝑝 minus two all squared equals 𝑝 squared plus nine. We can then subtract nine from both sides. And we get 𝑝 minus two all squared equals 𝑝 squared. We’ll need to expand these brackets before we go any further.

Be careful, a common mistake is to think that we just square the 𝑝 and the two and we’re done. In fact, we’ll need to consider this as a product of two brackets and expand from there. That is to say we’ll need to write it as 𝑝 minus two multiplied by 𝑝 minus two. Multiplying the first term in the first bracket by the first term in the second bracket gives us 𝑝 squared. 𝑝 multiplied by negative two is negative two 𝑝. And negative two multiplied by 𝑝 is once again negative two 𝑝. In fact, we can write negative two 𝑝 minus two 𝑝 as negative four 𝑝. Finally, we multiply negative two by negative two to give us four. And we now have 𝑝 squared minus four 𝑝 plus four is equal to 𝑝 squared.

Notice that we can subtract 𝑝 squared from both sides. And that gives us negative four 𝑝 plus four is equal to zero. To solve this, we’ll first add four 𝑝 to both sides to give us four equals four 𝑝. And then we’ll divide through by four. That gives us that 𝑝 is equal to one. Remember, 𝑝 is a coordinate. So we don’t need to include any units. We do, however, still need to find the length of 𝐴𝐵.

Remember, we created an expression for the length of 𝐴𝐵 in terms of 𝑝. So we can now substitute one into this expression. Doing so, and we get that the length of 𝐴𝐵 is the square root of nine plus one minus two all squared. One minus two is negative one squared. And negative one squared is one. So the length becomes the square root of nine plus one. The length of 𝐴𝐵 is root 10 units.

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