Question

Show that the points (5,5) , (h,0) , (0,k) are collinear if 1/h + 1/K = 1/5

Answer 1

Answer:

Step-by-step explanation:

For any 3 given points to be collinear, all the given points should lie

on the same straight line.

If the given 3 points A(5,5) , B(h,0) and C(0, k) to be collinear,

we need to have A, B and C lie on same straight line.

If 3 points lie on the same line then the slope of line joining any 2 points

would be the same.

Hence, Slope of AB = Slope of BC

Slope of AB = 5/5-h

Slope of BC = -k/h

Thus,

5/5-h = -k/h

=> 5h = -5k +hk

=>5h + 5k = hk

Dividing by 5hk on both sides we get

1/k + 1/h =1/5.

Answer 2

Solution:

If the three points A(5,5), B(h,0) and C(0,k) are collinear than they can not form a triangle,so Area of triangle = 0

$| \frac{1}{2}(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) | = 0 \\ \\ | \frac{1}{2}(5(0 - k) + h(k - 5) + 0(5 - 0) | = 0 \\ \\ - 5k + hk - 5h = 0 \\ \\ - 5k - 5h = - hk \\ \\ \frac{5k}{hk} + \frac{5h}{hk} = 1 \\ \\ \frac{5}{h} + \frac{5}{k} = 1 \\ \\ 5( \frac{1}{k} + \frac{1}{h} ) = 1 \\ \\ \frac{1}{k} + \frac{1}{h} = \frac{1}{5}$


Hence proved.