# Video: CBSE Class X • Pack 2 • 2017 • Question 10

CBSE Class X • Pack 2 • 2017 • Question 10

05:55

### Video Transcript

If the distances of the point 𝑃 with coordinates 𝑥, 𝑦 from 𝐴 with coordinates five, one and 𝐵 with coordinates negative one, five are equal, prove that three 𝑥 equals two 𝑦.

We can draw a diagram to help us. Where is the point 𝐴? Well, it has coordinates five comma one. So it’s here. How about 𝐵? Well, it has coordinates negative one, five. And so it’s here. The point 𝑃 is an arbitrary point in the plane with coordinates 𝑥 and 𝑦. All we know is that the distances of 𝑃 from 𝐴 and 𝐵 are both the same. So let’s mark in our point somewhere. Let’s put it here, for example. Let’s mark the coordinates in. Now, we want these distances to be the same. This diagram is just to help us understand what we have to prove. It’s not needed as part of the proof itself. But, of course, it couldn’t hurt.

Okay, so what are these distances that we want to be equal? The distance of 𝑃 from 𝐴 is 𝐴𝑃. And the distance of 𝑃 from 𝐵 is 𝐵𝑃. To find these distances, we use the distance formula. The distance formula tells us that the distance between two points with coordinates 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two is the square root of 𝑥 two minus 𝑥 one squared plus 𝑦 two minus 𝑦 one squared.

To find the distance between the point with coordinates five, one and the point with coordinates 𝑥, 𝑦, we substitute five, one for 𝑥 one, 𝑦 one and 𝑥, 𝑦 for 𝑥 two, 𝑦 two in our formula. We get the square root of the difference of the 𝑥-coordinates squared plus the difference of the 𝑦-coordinates squared. It’s exactly the same for 𝐵𝑃. We have the difference of the 𝑥-coordinates squared. And, of course, 𝑥 minus negative one is just 𝑥 plus one, plus the difference of the 𝑦-coordinates squared.

Okay, we found the distances of 𝑃 from 𝐴 and 𝐵 using the distance formula. Now, we can set these two distances equal to find a relation between the coordinates 𝑥 and 𝑦 of 𝑃. Before we do this, let me just remark that had we forgotten the distance formula, we could find the distances of 𝑃 from 𝐴 and 𝐵 from first principles. To find the distance of 𝑃 from 𝐴, that’s the length of the line segment 𝐴𝑃, we can consider this right-angled triangle which makes 𝐴𝑃 the hypotenuse. Looking at the 𝑥 coordinates of 𝐴 and 𝑃, we can see that the base has a length of five minus 𝑥. And the height is the difference in the 𝑦-coordinates, 𝑦 minus one.

Applying Pythagoras’s theorem, we get the length of the hypotenuse 𝐴𝑃, as before. The only possible difference is that we might have five minus 𝑥 squared instead of 𝑥 minus five squared. But these are equal. Okay, so however we got the distances 𝐴𝑃 and 𝐵𝑃, we’re told that they’re equal. So we should set them equal to each other. 𝐴𝑃 equals 𝐵𝑃. We substitute the expressions we found for these distances.

We can square both sides and then expand the brackets. For example, 𝑥 minus five squared is 𝑥 squared minus two times five 𝑥 plus five squared. This is using the fact that 𝑥 minus 𝑎 squared is 𝑥 squared minus two 𝑎𝑥 plus 𝑎 squared. We can simplify this and move on to the next term. We expand 𝑦 minus one squared to 𝑦 squared minus two 𝑦 plus one. On the right-hand side, we expand 𝑥 plus one squared to 𝑥 squared plus two 𝑥 plus one and 𝑦 minus five squared to 𝑦 squared minus 10𝑦 plus 25.

Having done this, we see that there is some cancellation. An 𝑥 squared on the left-hand side cancels with an 𝑥 squared on the right. And similarly, the 𝑦 squareds on the left- and right-hand sides also cancel. Combining the constant terms 25 and one on the left-hand side, we get negative 10𝑥 minus two 𝑦 plus 26 on the left-hand side. And combining the constant terms on the right-hand side, we get two 𝑥 minus 10𝑦 plus 26. The constant terms cancel. And we’re left with only 𝑥- and 𝑦-terms. Adding 10𝑥 to both sides, we’re left with just negative two 𝑦 on the left-hand side. And the 10𝑥 combines with the two 𝑥 on the right-hand side to give 12𝑥 minus the 10𝑦.

We can also add 10𝑦 to both sides to get 10 minus two equals eight 𝑦 on the left-hand side and just 12𝑥 on the right-hand side. We can divide both sides by four, getting two 𝑦 on the left-hand side and three 𝑥 on the right-hand side. And finally, swapping both sides we get that three 𝑥 is equal to two 𝑦, which is what we were required to prove.

You might like to think about what this means graphically. The equation three 𝑥 equals two 𝑦 defines a line on the coordinate plane. We have proved that if a point 𝑃 is equidistant from points 𝐴 and 𝐵, then it lies on this line. You might recognize this line, whose equation we found, as the perpendicular bisector of the line segment 𝐴𝐵. This question used the distance formula which is really worth remembering. But you can, of course, rederive it if you’ve forgotten it using Pythagoras’s theorem. Once we applied the distance formula, the rest was just algebra.