Question

The points on the line $x+y=4$ lying at a unit distance from the line $4x+3y-10=0$ are

• A. $(-7,11),(3,1)$
• B. $(7,-11),(3,-1)$
• C. $(-7,11),(-3,7)$
• D. $(7,-3),(11,-7)$
• E. $(2,2),(3,1)$

Answer 1

Answer: (A)

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

Let $(h,k)$ be the point on the line $x+y=4$
Then, $h+k=14$ .... (i)
Also, given that
Perpendicular distance from $(h,k)$ to the line
$4x+3y-10=0$ is $1$.
i.e., $\left|\frac{4h+3k-10}{\sqrt{16+9}}\right|=1$
$\Rightarrow 4h+3k-10=\pm 5$
$\Rightarrow 4h+3k=10\pm 5$
$\Rightarrow 4h+3k=15$ ...(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 $(-7,11)$.