# Find the equation of the line perpendicular to y=x and passing at a distance 3sqrt2 from (4,1).

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### 2 Answers

The equation is perpendicular to the line y = x. The line y = x has a slope 1. As perpendicular lines have slopes which are negative reciprocals the slope of the required line is -1. The equation of the line is x + y + k = 0

The required line passes through a point which is at a distance 3*sqrt 2 from (4,1).

We use the relation for determining the distance d of a point (x1, y1) from the line ax+by +c = 0, which is:

d = |ax1+by1+c|/ sqrt (a^2+b^2)

=> 3*sqrt 2 = | 4 + 1 + k|/sqrt 2

=> 5 + k = 3*2 = 6

=> k = 1

**Therefore the equation of the required line is x + y + 1 = 0.**

The equation of the line perpendicular to y = x is y = -x+ constant or x+y+k = 0....(1)

If ax+by+c = 0 is any line , then its distance d from a point (x1,y1) is given by:

d = (ax1+by1+c)|/(a^2+b^2).

Therefore if the the point (4,1) and the line x+y+k are at distance of 3sqrt2, then:

3*2^(1/2) = (4+1+k)/(1^2+1^2)^(1/2) .

3*2^(1/2) = (5+k)/2^(1/2).

3* 2(1/2)*2^(1/2) = 5+k.

3*2 = 5+k.

k = 6-5 = 1.

So we substitute k = in (1) and get : x+y+1 = 0

**Therefore the required equation of the line is x+y+1 = 0.**