Question

prove the feet of the perpendicular from the origin o the lines x+y=4 x+5y =26 and 15x-27y=424 are collinear

Answer 1

Answer:

Given Lines

L1 : x + y =4    ........ (1)

or y = -x + 4

Slope of line L1 = -1

Slope of the line perpendicular to L1 will be 1 (∵ m₁m₂ = -1, for perpendicular lines)

Therefore equation of the line with slope 1 and passing through the origin is

y = x

or x - y = 0   ....... (2)

Solving equation (1) and (2) we get

x = 2, y = 2

Therefore foot of the perpendicular A = (2, 2)

Again

Line L2 : x + 5y =26       ........ (3)

or, y = -x/5 + 26/5

Slope of line L2 = -1/5

Slope of the line perpendicular to L2 will be 5

Therefore equation of the line with slope 5 and passing through the origin is

y = 5x

or y - 5x = 0         ........... (4)

Solving equation (3) and (4) we get

x = 1, y = 5

Therefore foot of the perpendicular B = (1, 5)

Line L3 : 15x -27y = 424        ........ (5)

or y = 15x/27 - 424

Slope of line L3 = 15/27

Slope of the line perpendicular to L3 will be -27/15

Therefore equation of the line with slope -27/15 and passing through the origin is

y = -27x/15

or 15y + 27x = 0         ........... (4)

Solving equation (3) and (4) we get

x = 20/3, y = -12

Therefore foot of the perpendicular C = (20/3, -12)

Now the distance

$AB = \sqrt{(2-1)^2+(2-5)^2} =\sqrt{1+9} =\sqrt{10}$

$BC = \sqrt{(1-20/3)^2+(5-(-12))^2} =\sqrt{17^2/9+17^2} =\frac{17}{3}\sqrt{10}$

$CA = \sqrt{(2-20/3)^2+(2-(-12))^2} =\sqrt{14^2/9+14^2} =\frac{14}{3}\sqrt{10}$

Now,

$CA+AB=\frac{14}{3}\sqrt{10}+\sqrt{10} =\frac{17}{3}\sqrt{10}=BC$

i.e. the sum of the distance from point C to point A and from point A to point B is equal to the distance from point C to point B

Therefore, A, B, C , which are the foot of the perpendiculars, are collinear.