 Hello and welcome to the session. In this session we discussed the following question that says, would that the points minus three zero, one minus three and four one are the vertices of an isosceles right angle triangle. So we have given three points. We need to show that these three points are the vertices of isosceles right angle triangle. For that we will need distance formula according to which we have that if we have a point say p with coordinates x1, y1 and q with coordinates x2, y2 then the distance between the two points is given by pq and this is equal to square root of x2 minus x1 whole square plus y2 minus y1 whole square. This is the key idea to be used in this question. Now we move on to the solution. We take let point A has coordinates minus three zero, B has coordinates one minus three and C has coordinates four one then we need to show that ABC are the vertices of an isosceles right angle triangle. Let's find out AB that is the distance between the points A and B. This is equal to square root x2 that is the x-coordinate of point B that is one minus of minus three that is x-coordinate of point A whole square plus minus three that is y-coordinate of point B minus y-coordinate of point A that is zero whole square. So from here we have AB is equal to square root of one plus three whole square plus minus three whole square. This gives us AB is equal to square root of 16 plus 9 that is AB is equal to square root 25 and this is equal to 5. So thus we get the distance between the points A and B that is AB is equal to five units. Now let's find out BC that is the distance between the points B and C. This is equal to square root x2 that is x-coordinate of point C four minus one that is x-coordinate of point B whole square plus y2 that is the y-coordinate of point C that is one minus y-coordinate of point B that is minus three whole square. So from here we get BC is equal to square root of three square plus four square. Now BC is equal to square root of 9 plus 16 and this is equal to square root of 25 and that is equal to 5. Thus we have BC is equal to 5 units. Next we find out AC that is the distance between the points A and C. This is equal to square root of x2 that is the x-coordinate of point C four minus x1 that is x-coordinate of point A that is minus three whole square plus y2 that is the y-coordinate of point C which is one minus y-coordinate of point A that is zero whole square. So from here we get AC is equal to square root of four plus three whole square plus one square that is AC is equal to square root of 49 plus 1 that is equal to square root 50 which means we get AC is equal to 5 root 2 units. Thus we have AB is equal to 5 units, BC is equal to 5 units and AC is equal to 5 root 2 units. Now as AB is equal to BC they both are of measure 5 units therefore triangle ABC is an isosceles triangle. Now we need to show that it is an isosceles right angle triangle. So for that we will first find out AB square which is equal to 5 square that is we have AB square is equal to 20 5 units then BC square is equal to again 5 square therefore BC square is equal to 25 units then AC square is equal to 5 root 2 square therefore AC square is equal to 50 units. So as you can see we get that AB square plus BC square is equal to AC square so AB is of measure 5 units BC is also of measure 5 units and AC is of measure 5 root 2 units. Now as you can see we are getting AC square is equal to AB square plus BC square so this means triangle ABC is right angled B thus triangle ABC is a right angled isosceles triangle. So the points minus 3 0 1 minus 3 and 4 1 are the vertices of an isosceles right angled triangle. So this completes the session hope you have understood the solution for this question.