 Hello and welcome to the session. Let us discuss the following question. It says, find a point on the x-axis which is equiliscent from the points 7, 6 and 3, 4. Let us now understand the formula of distance between two points. If px1, y1 and qx2, y2 are any two points, then distance between pq is given by under root of x2-x1 whole square plus y2-y1 whole square. So this knowledge will work as key idea. Let us now move on to the solution. We have to find a point on the x-axis. So if we have to find a point on the x-axis, the coordinate of y will be 0. So let the point on the x-axis and let us call this point as p. And we are given 2.7 and 6 and 3 and 4. Let us call them as a and b respectively. Now we are given that point on the x-axis is equiliscent from these two points. That is, pA is equal to pB. Now distance between p and A is given by root of 7-x whole square plus 6-0 whole square. And distance between p and B is given by root of 3-x whole square plus 4-0 whole square. Now taking square on both sides, we have 7-x whole square plus 36 is equal to 3-x whole square plus 16. This implies 7-x whole square minus 3-x whole square is equal to 16-36. And this again implies 7-x whole square minus 3-x whole square is equal to minus 20. Now we apply the formula of A square minus B square here. So it is equal to 7-x plus 3-x into 7-x minus 3-x is equal to minus 20 as we know that A square minus B square is equal to A plus B into A minus B. This implies 10-2x into 7-x minus 3 plus x is equal to minus 20. And this again implies 10-2x into 4 is equal to minus 20. Now dividing by 4 on both sides we have 10-2x is equal to minus 5. And this implies minus 10 is equal to 5. And this implies 2x is equal to 15. And this again implies x is equal to 15 upon 2. Hence the point on the x-x's is 15 upon 2-0. So this completes the question. Hope you enjoyed the session. Goodbye and take care.