# Number of its at unit distance from the lines 3x-4y+11=0 and 5x-12+7=0

Let the points to be found be (x

_{1},y

_{1}) and (x

_{2},y

_{2}).

By the formula for finding perpendicular distance of a point from a line,

3x

_{1}-4y

_{1}+11= 1

3x

_{1}-4y

_{1}+10 = 0

**(I)**

and

5x

_{1}-12y

_{1}+7 = 1

5x

_{1}-12y

_{1}+6 = 0

**(II)**

Solving

**I**and

**II**,

9x

_{1}-12y

_{1}+30 = 0

**-**15x

_{1}-12y

_{1}+6 = 0 ==> -6x

_{1}+24 = 0

6x

_{1}= 24

x

_{1}= 4

=>y1 = 11/2.

Since Distance is taken with a modulus sign, x

_{2}= -x

_{1}, and y

_{2}= -y

_{1}.

Therefore, the required points are:

(x1 , y1) = (4 , 5.5)

(x2 , y2) = (-4 , -5.5)

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