 Hello and welcome to the session. Let us discuss the following question. It says, find the coordinates of the point equidistant from the three points A51, B-3-7 and C7-9. So we have to find the coordinates of the point which is equidistant from these three given points. So let us now move on to the solution and let Pxy be the point equidistant points A51, B-3-7 and C7-1. Now since this point Pxy is equidistant from these three given points we have PA is equal to PB is equal to PC that is distance between P and A is equal to distance between P and B is equal to distance between P and C and this is according to question. We have to find the distance between P and A and the distance formula the points Q1, X1, Y1 and Q2, X2, Y2 is given Y under the root X2 minus X1 whole square plus Y2 minus Y1 whole square. Now we find the distance between P and A this is Q1, Q2 so distance between P and A is given Y 5 minus X whole square that is X2 minus X1 whole square here X2 is 5, Y2 is 1 and X1 is X and Y1 is Y. So this becomes 5 minus X whole square plus 1 minus Y whole square similarly we find distance between P and B so this is given by minus 3 minus X whole square plus minus 7 minus Y whole square similarly PC is given by the formula 7 minus X whole square plus minus 1 minus Y whole square its mean this as 1 this as 2 and this as 3 now from the question we have P is equal to PB is equal to PC so from 1 and 3 we have P is equal to PC that means under the root 5 minus X whole square plus 1 minus Y whole square is equal to under the root 7 minus X whole square plus minus 1 minus Y whole square can be written as 1 plus Y whole square because if we take minus 1 common and since here we have square so minus 1 square becomes plus 1 now squaring both sides we have 5 minus X whole square is plus 1 minus Y whole square is equal to 7 minus X whole square plus 1 plus Y whole square now by the formula of A minus B whole square so we have 25 plus X square minus 10 X A minus B whole square is A square plus B square minus 2 AB so this becomes minus 10 X plus 1 plus Y square minus 2 Y is equal to 49 plus X square minus 14 X plus 1 plus Y square plus 2 Y here we have a counter formula of A plus B whole square now we see that X square gets cancelled with X square Y square gets cancelled with Y square 1 gets cancelled with 1 and we have 25 minus 10 X minus 2 Y minus 49 plus 14 X minus 2 Y is equal to 0 now 25 minus 49 is minus 24 minus 10 X plus 14 X is plus 4 X minus 2 Y minus 2 Y is minus 4 Y is equal to 0 so this implies 4 X minus 4 Y is equal to 24 so this implies X minus Y is equal to 6 taking 4 common now again from 2 and 3 we have PB is equal to PC so we have minus 3 minus X whole square plus minus 7 minus Y whole square equal to PC which is under the root 7 minus X whole square plus 1 plus Y whole square again squaring both sides we have minus 3 minus X whole square plus minus 7 minus Y whole square is equal to 7 minus X whole square plus 1 plus Y whole square now again taking minus 1 common from this we have 3 plus X whole square and again we are taking minus 1 common but since here we have square so this becomes 7 plus Y whole square is equal to 7 minus X whole square plus 1 plus Y whole square now we apply the formula of A plus B whole square so this becomes 9 plus X square plus 6 X plus 49 plus Y square plus 14 Y is equal to 49 plus X square minus 14 X plus 1 plus Y Y square plus 2 Y now we see that Y square gets cancelled with Y square 49 gets cancelled with 49 X square gets cancelled with X square and we have 9 plus 6 X 9 plus 6 X plus 14 Y plus 14 X minus 1 minus 2 Y is equal to 0 again this implies 6 X plus 14 X is 20 X plus 14 Y it should be 12 Y 14 Y minus 2 Y is 12 Y 9 minus 1 is 8 equal to 0 so taking 4 common we have 5 X plus 3 Y plus 2 is equal to 0 and this implies 5 X plus 3 Y is equal to minus 2 let us name this as A this as B equation A is X minus Y is equal to 6 and B is 5 X plus 3 Y is equal to minus 2 now we will solve this equation for X and Y by equating the coefficient of X or Y so here we will equate the coefficient of Y and we will multiply equation A Y 3 so we have 3 X minus 3 Y is equal to 18 and we will add equation A and B 5 X plus 3 Y is equal to minus 2 adding the 2 equations since we are adding sign will not change 3 X plus 5 X is 8 X minus 3 Y gets cancelled with plus 3 Y 18 minus 2 is 16 so this implies X is equal to 16 upon 8 that is 2 now from equation A we have X minus Y is equal to 6 now X is 2 minus Y is equal to 6 so we have Y is equal to 2 minus 6 and this implies Y is equal to minus 4 so X is equal to 2 and Y is equal to minus 4 and the coordinates of the point P X Y where X is 2 and Y is minus 4 thus the required point is which is equidistant from the 3 given points is 2 minus 4 so this completes the question and the session bye for now take care have a good day