Prove that the points a , 0 , 0 , b and 1 , 1 are co-linear, if 1 a + 1 b = 1

Let, A x 1 , y 1 = a , 0 , B x 2 , y 2 = 0 , b and C x 3 , y 3 = 1 , 1 We have, 1 a + 1 b = 1 ⇒ b + a a b = 1 ⇒ b + a = a b ⇒ a b - a - b = 0 ....... 1 We shall prove that a b - a - b = 0 . The three points will be collinear if the area formed by the triangle with vertices A , B and C becomes 0 . So, a r △ A B C = 0 ⇒ 1 2 x 1 y 2 - y 3 + x 2 y 3 - y 1 + x 3 y 1 - y 2 = 0 ⇒ a b - 1 + 0 1 - 0 + 1 0 - b = 0 ⇒ a b - a + 0 - b = 0 ⇒ a b - a - b ⇒ a b - a - b = 0 [proved] Hence, the points A , B , C i.e. the given points, are collinear.

Find the coordinates of the point which divides internally the line joining the points 6 , - 9 and 4 , 6 in the ratio 3 : 4 .

Let the point P x , y divide internally the line joining the point ( 6 , - 9 ) and ( 4 , 6 ) in the ratio 3 : 4 , then, x = 3 ( 4 ) + 4 ( 6 ) 3 + 4 = 12 + 24 7 = 36 7 , y = 3 ( 6 ) + 4 ( - 9 ) 3 + 4 = 18 - 36 7 = - 18 7 Hence, the co-ordinates of the point of division are 36 7 , - 18 7 .

Find the co-ordinate of mid point of the line joining points 4 , - 1 and - 2 , - 3 .

Let, x 1 , x 1 = 4 , - 1 and x 2 , y 2 = - 2 , - 3 By, mid-point formula, coordinates of mid-point = x 1 + x 2 2 , y 1 + y 2 2 = 4 - 2 2 , - 1 - 3 2 = 1 , - 2

Find the point on the y - axis which is equidistant from the point 6 , 5 and - 4 , 3 .

Given points are A ( 6 , 5 ) and B ( - 4 , 3 ) . We need to find a point on y -axis which is equidistant from these points. Let us assume the point on y -axis be S 0 , y . We know that distance between the points x 1 , y 1 and x 2 , y 2 is x 1 - x 2 2 + y 1 - y 2 2 . From the problem, ⇒ SA = SB ⇒ SA 2 = SB 2 ⇒ ( 0 - 6 ) 2 + ( y - 5 ) 2 = ( 0 - ( - 4 ) ) 2 + ( y - 3 ) 2 ⇒ ( - 6 ) 2 + ( y - 5 ) 2 = ( 4 ) 2 + ( y - 3 ) 2 ⇒ 36 + y 2 - 10 y + 25 = 16 + y 2 - 6 y + 9 ⇒ 4 y = 36 ⇒ y = 36 4 ⇒ y = 9 ∴ The point on y -axis is ( 0 , 9 ) .