The equation of a straight line passing through the point (-5,4) and which cuts off an intercept of sqrt2 units between the lines x + y + 1 = 0 and x + y - 1 = 0 is?

1 Answer
Sep 12, 2017

x-y+9=0.

Explanation:

Let the given pt. be A=A(-5,4), and, the given lines be

l_1 : x+y+1=0, and, l_2 : x+y-1=0.

Observe that, A in l_1.

If segment AM bot l_2, M in l_2, then, the dist. AM is given by,

AM=|-5+4-1|/sqrt(1^2+1^2)=2/sqrt2=sqrt2.

This means that if B is any pt. on l_2, then, AB > AM.

In other words, no line other than AM cuts off an intercept of

length sqrt2 between l_1, and, l_2, or, AM is the reqd. line.

To determine the eqn. of AM, we need to find the co-ords. of the

pt. M.

Since, AM bot l_2, &, the slope of l_2 is -1, the slope of

AM must be 1. Further, A(-5,4) in AM.

By the Slope-Pt. Form, the eqn. of the reqd. line, is,

y-4=1(x-(-5))=x+5, i.e., x-y+9=0.

Enjoy Maths.!