Video: CBSE Class X • Pack 3 • 2016 • Question 13

CBSE Class X • Pack 3 • 2016 • Question 13

03:45

Video Transcript

If the point 𝑃 at 𝑥, 𝑦 is equidistant from the points 𝐴 at 𝑎 plus 𝑏, 𝑏 minus 𝑎 and 𝐵 at 𝑎 minus 𝑏, 𝑎 plus 𝑏, prove that 𝑏𝑥 is equal to 𝑎𝑦.

We’ll need to apply the distance formula here. The distance formula is derived from the Pythagorean theorem. And it says that for two points at 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, the distance between them is given by the square root of the square of the differences between the 𝑥-coordinates plus the square of the differences between the 𝑦-coordinates.

We can find the distance of the lines joining 𝐴 and 𝑃 and 𝐵 and 𝑃 by substituting the relevant values into this formula. For 𝐴𝑃, let’s call 𝑥, 𝑦 𝑥 one, 𝑦 one and the point at 𝐴 𝑥 two, 𝑦 two. We can then substitute these values into the formula as shown. The difference between the 𝑥-coordinates is 𝑎 plus 𝑏 minus 𝑥. And the difference between the 𝑦-coordinates is 𝑏 minus 𝑎 minus 𝑦.

We can repeat this process for the distance between the points 𝑃 and 𝐵. The difference between the 𝑥-coordinates is 𝑎 minus 𝑏 minus 𝑥. And the difference between the 𝑦-coordinates is 𝑎 plus 𝑏 minus 𝑦. Since the point 𝑃 is equidistant from the points 𝐴 and 𝐵, we know that the distance between 𝐴 and 𝑃 is equal to the distance between 𝐵 and 𝑃. This means we can equate the two expressions we formed for the distance between the lines.

At this point, we can square both sides of this equation. We will, however, need to expand each of these individual brackets. This is much like expanding products of binomials. We’ll need to make sure that each term in the first bracket multiplies by each term in the second. We can use a grid method though to do this and ensure we don’t lose any terms. Let’s multiply out this first pair of brackets.

Multiplying each term on the top of the grid by each term on the side, and we end up with these individual terms. We can repeat this for the second pair of brackets. And we get these terms. Remember, these are being added. So let’s add together everything we have in these two grids.

𝑎𝑏 minus 𝑎𝑏 is zero. So these 𝑎𝑏 terms cancel out. Adding these all together and we’re left with a rather nasty looking expression. But these should cancel out later on down the line. Multiplying each pair of brackets on the right-hand side of the equation, and we get this. Once again, we’re finding the sum of these expressions. So the 𝑎𝑏s cancel out. And when we simplify by collecting like terms, we get this.

Notice, we can subtract two 𝑎 squared and two 𝑏 squared from both sides. We can add two 𝑎𝑥 to both sides. We can add two 𝑏𝑦 to both sides. And we can subtract an 𝑥 squared and a 𝑦 squared. And this is what we’re left with. This looks much nicer at this point.

To simplify further, we’ll add two 𝑏𝑥 to both sides of the equation. We get two 𝑎𝑦 equals four 𝑏𝑥 minus two 𝑎𝑦. Next, we’ll add two 𝑎𝑦 to both sides of the equation, to leave us with four 𝑎𝑦 equals four 𝑏𝑥. We could divide everything by four. And changing the order to match the question, we’ve proved that 𝑏𝑥 is equal to 𝑎𝑦.

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