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KEAM 2013 Math Paper

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Question : 56 of 120

Marks: +1, -0

The points on the line

x + y = 4

lying at a unit distance from the line

4x+3y-10 = 0

are

$(-7, 11), (3, 1)$
$(7, -11), (3, -1)$
$(-7,11), (-3, 7)$
$(7,-3), (11,-7)$
$(2,2), (3,1)$

Solution:

Let

(h, k)

be the point on the line

x+y=4

Then,

h+k=4

.....(i)

Also, given that

Perpendicular distance from

(h, k)

to the line

4x+3y-10=0

is 1

i.e.,

\;\; \left| \frac{4h+3k-10}{\sqrt{16+9}} \right| = 1

\Rightarrow \;\; 4h+3k-10 = \pm 5

\Rightarrow \;\; 4h+3k = 10 \pm 5

\therefore 4h+3k = 15 \dots (ii)

and

4h+3k=5

.....(iii)

Now, on solving Eqs. (i) and (ii), we get

4h+3(4-h)=15 \Rightarrow h=3

Then from Eq. (i),

k=1

Again, on solving Eqs. (i) and (iii), we get

4h+3(4-h)=5 \Rightarrow h=-7

Then from Eq. (i),

k=11

So, the required points are

(3,1)

and

(-7,11)

.

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