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(A) Let the point on the line $x+y=4$ be $(\alpha, 4-\alpha)$.Given that the perpendicular distance from this point to the line $4x+3y-10=0$ is $1$.Us

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$10\sqrt{2}$

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(A) Let the point on the line $x+y=4$ be $(\alpha, 4-\alpha)$.Given that the perpendicular distance from this point to the line $4x+3y-10=0$ is $1$.Using the distance formula: $\left|\frac{4\alpha+3(4-\alpha)-10}{\sqrt{4^2+3^2}}\right|=1$.$\left|\frac{4\alpha+12-3\alpha-10}{5}\right|=1$.$|\alpha+2|=5$.This gives $\alpha+2=5$ or $\alpha+2=-5$.So,$\alpha=3$ or $\alpha=-7$.For $\alpha=3$,the point is $(3, 1)$.For $\alpha=-7$,the point is $(-7, 11)$.The distance $d$ between $(3, 1)$ and $(-7, 11)$ is given by $\sqrt{(-7-3)^2+(11-1)^2}$.$d=\sqrt{(-10)^2+(10)^2} = \sqrt{100+100} = \sqrt{200} = 10\sqrt{2}$.

If two distinct points lying on the line $x+y=4$ are at a unit distance from the line $4x+3y-10=0$,and the distance between these two points is $d$,then the value of $d$ is:

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