The points on the line $x+y=4$ which lie at a unit distance from the line $4x+3y=10,$ are

The correct option is A

$(3,1),(–7,11)$

The explanation for the correct answer.

Solve for the points on the line $x+y=4$

Let $P(h,k)$ lies on $x+y=4$

$\Rightarrow h+k=4...\left(i\right)$

We know the distance of the point $(x_1,y_1)$ from the line $ax+by+c=0$ is

$d=\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}$

$$\begin{gathered} \Rightarrow 1=\frac{\left|4h+3k-10\right|}{\sqrt{4^2+3^2}} \\ \Rightarrow 1=\frac{\left|4h+3k-10\right|}{5} \\ \Rightarrow 4h+3k–10=\pm 5 \\ \Rightarrow 4h+3k=15\dots \left(ii\right) \\ \Rightarrow 4h+3k=5\dots \left(iii\right) \end{gathered}$$

The distance of the point$P(h,k)$ from the line $4x+3y=10$is $1$unit.

Solving $\left(i\right)$ and $\left(ii\right)$, we get $(h,k)=(3,1)$

$h+k=4...\left(i\right)$

$4h+3k=15...\left(ii\right)$

$$\begin{gathered} \Rightarrow 4h+3(4-h)=15 \\ \Rightarrow h=15-12=3 \end{gathered}$$

Similarly,solving $\left(i\right)$ and $\left(iii\right)$, we get $(h,k)=(-7,11)$

Hence, option(A) is the correct answer.