# What is the distance between the point of contact of the line x + 4y + 8 = 0 with the lines x = 4 and y = 8?

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### 2 Answers

We have to first find the point of contact between the line x + 4y + 8 = 0 and x = 4.

For this we can substitute x = 4 in x + 4y + 8 = 0

=> 4 + 4y + 8 =0

=> 4y = -12

=> y = -12/4

=> y = -3

Therefore the point of contact is (4, -3)

To find the point of intersection of the line x + 4y + 8 = 0 and y = 8, we substitute y= 8 in x + 4y + 8 = 0.

=> x + 4*8 + 8 =0

=> x = -8 – 32

=> x = -40

Therefore the point of intersection is (-40, 8)

Now the distance between (4, -3) and (-40, 8) is

sqrt [(-40 – 4) ^2 + (8+3) ^2]

=> sqrt (44^2 + 11^2)

=> sqrt (1936 + 121)

=> sqrt (2057)

=> 45.35 approximately.

**The required distance is sqrt 2057.**

Let the line x+4y+8 = 0 and the line at x= 4 intersect at A(x1,y1).

Then we find the coordinates of A(x1,y1) which lie both on x+y+8 = and x= 4.So,

x1+4y1+8 = 0 and x1 = 4

Therefore 4+4y1 +8 = 0, 4y1 = 0-8-4 = -12. So y1 = -12/4 = -3

Therefore x1 = 4 and y1 = -3.

So A(x1,y1) = A(4,-3).

Similarly if B(x2,y2) is the point of intersection of x+4y+8, and y= 8, then the coordinates of B(x2,y2) should satisfy both x+4y+8 = 0 and y = 8:

x2+4y2+8 = 0 and y2 = 8.

So put y2 = 8 in x2+4y2+8 = 0 and we we get:

x2 +4*8+8 = 0.

x2 +40. Or x2 = -40.

So x2 = -40 and y2 = 8.

So B(x2,y2) = B(-40, 8).

Therefore the distance between A(x1,y1) and B(x2,y2) = sqrt{(x2-x1)^2+(y2-y1)^2} = sqrt{(-40-4)^2+(8-(-3))^2} = sqrt{44^2+11^2} = sqrt(2057).