 Hi and welcome to the session. Let's work out the following question. The question says, using vectors proves that the midpoint of the hypotenuse of a right-angled triangle is equidistant from its vertices. Let's start with a solution to this question. Now, let the triangle OPQ be a right-angled triangle such that OP and OQ. They are taken as coordinate Xs respectively. Let R be the midpoint of hypotenuse PQ. Let position vector OP and Q be vector O, vector A and vector B respectively as vector OP is perpendicular to vector OQ. This implies vector A dot vector B is equal to 0 and this we call 1. Now, position vector of RB, vector A plus vector B divided by 2 because this is the midpoint of the hypotenuse PQ. Now, what we want to prove is that mod of vector OR is equal to mod of vector PR is equal to mod of vector QR. That is, the midpoint of the hypotenuse that is R is equidistant from the vertices OP and Q, so this is what we have to prove. Now, we know that vector OR is equal to vector A plus vector B divided by 2. This implies mod of vector OR is equal to mod of vector A plus vector B divided by 2. This implies mod of vector OR square is equal to 1 upon 4 into vector A plus vector B into mod of vector A plus vector B. This is equal to 1 by 4 into mod of vector A plus vector B square which can be written as 1 by 4 into vector A square plus vector B square plus 2 into vector A into vector B. But vector A dot vector B is equal to 0. This we have from equation 1. Therefore, vector OR square becomes equal to or mod of vector OR square becomes equal to 1 by 4 into mod of A square plus B square. This we call equation 2. Similarly, vector PR is equal to vector OR minus vector OP which is equal to half of vector A plus vector B minus vector A. So, mod of vector PR is equal to mod of vector B minus vector A divided by 2. This implies mod of vector PR square is equal to 1 by 4 into mod of vector B minus vector A square, which is equal to 1 by 4 into vector B minus vector A into vector B minus vector A which is also equal to 1 by 4 into vector B square plus vector A square minus 2 into vector A into vector B which is equal to half of vector A square plus vector B square because we know that vector A dot vector B is equal to 0. So, we call this 3. Now, again we say that vector QR is equal to vector OR minus vector OP which is equal to vector A plus vector B divided by 2 minus vector B. So, vector QR is equal to vector A minus vector B divided by 2 vector QR square is equal to vector A minus vector B divided by 2 the whole square which is equal to 1 by 4 into, here we have a mod. This is 1 by 4 into mod of vector A square plus vector B square minus 2 into vector A into vector B. Again this is equal to 0. So, vector QR square is 1 by 4 into vector A square plus vector B square and we call this equation 4. So, from 2, 3 and 4 we get mod of vector QR square is equal to mod of vector PR square is equal to mod of vector OR square each is equal to A square plus B square divided by 4. So, this is what you are supposed to prove in this question. I hope that you understood the solution and enjoyed the session. Have a good day.