Question

4. Prove that the feet of the perpendiculars from the origin on the lines
x + y = 4, x + 5y = 26 and 15x - 27y = 424 are collinear.​

Answer 1

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the lines perpendicular to the three mentioned lines and passing through origin are

x-y=0

5x-y=0

27x+15y=0

• solving x-y =0 & x+y = 4 we get the feet of the perpendicular from the origin as (2,2)

• solving 5x-y=0 & x+5y=26 ,we get the feet of the perpendicular from the origin as (1,5)

• solving 27x+15y=0 & 15 x -27y=424 , we get the feet of the perpendicular from the origin as (20/3,-32/3)

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the equation of the line through (2,2) and (1,5) is 3x+y=8

(20/3,-36/3) satisfies 3x+y=8

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So , the point also lies on the line & all the three points are collinear

Answer 2

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

Now,

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.