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?

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Answer 1 · 2017-09-12T18:26:55

# x-y+9=0.#

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.!