Here, we have x₁ = 3,    y = -1
x₂ = 8,    y₂ = 9
and    m₁ = k, m₂ = 1.
So, co-ordinates of P are


But P lies on x - y - 2 = 0    ...(i)
Substituting the values of x (abscissa) and y (ordinate) in (i), we get


Hence, the required ratio is 2 : 3.

We have,

           3AC = CB





Now, coordinates of C are

Here, we have x₁ = 1,    y₁ = 1;
x₂ ^{= 2,    y₂ = -3}
and    m₁ = 1, m₂ = 3
Hence,

Let the given points be A( 1, 3) and B(2, 7).

Let the line 3x + y - 9 = 0 divides the line segment AB in the ratio K : 1 at point P. Then the co-ordinates of P are given by

Here, we have
x₁ = 1,    y₁ = 3
x₂ = 2,    y₂ = 7
and    m₁ = K    m₂ = 1
So, co-ordinates of P are

But P lies on 3x + y - 9 = 0.    ...(i)
Substituting the values of x (abscissa) and y (ordinate) in (i), we get


So, the required ratio be (3 : 4).

Let AOB be a right angle triangle, with hypotenuse AB. We take OB along x-axis and OA along y-axis.

Let ‘P’ be the mid-point of the hypotenuse. Then, the co-ordinates of A, B and Pare respectively A(0, n), B(m, 0) and P  
Now,   




Hence, P is equidistant from the vertices of ∆ABC.