 Hi and welcome to the session. Let us discuss the following question. The question says, if P is the length of perpendicular from the origin to the line, whose intercepts on the x's are a and b, then show that 1 by p squared is equal to 1 by a squared plus 1 by p squared. Now suppose this is the line a, b whose intercept, whose intercept on the x axis is a and intercept on the y axis is b. This is the perpendicular from origin to this line a, b and p is the length of this perpendicular. Let us now begin with the solution. Line a, b has intercept a on x axis and b on y axis. So equation of line a, b is x by a plus y by b is equal to 1 or we can say that b x plus a y minus a b is equal to 0. So equation of a b is b x plus a y minus a b is equal to 0. We know that distance of a point x 1 by 1 from the line a x plus b y plus c is equal to 0 is given by mod of a x 1 plus b y 1 plus c upon square root of a square plus b square right. Now here the line is b x plus a y minus a b is equal to 0 and the point x 1 by 1 is 0 0. So distance of a b from the origin is mod of b into 0 plus a into 0 minus a b upon square root of b square plus a square. Now in the question it is given that that this distance is equal to p. So put this equal to p. Now this implies a b upon square root of v square plus a square is equal to p squaring both sides of this equation we get a square v square upon v square plus a square is equal to p square. Now this implies 1 by p square is equal to p square plus a square by a square v square this implies 1 by p square is equal to p square by a square b square plus a square by a square b square this implies 1 by p square is equal to 1 by a square plus 1 by b square. Hence we have proved that 1 by p square is equal to 1 by a square plus 1 by v square. This completes the session by intake gear.