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If two distinct points lying at a unit distance from $4 \mathrm{x}+3 \mathrm{y}-10=0$ and passing through $\mathrm{x}+\mathrm{y}=4$ are separated by a distance $\mathrm{d}$, then the value of $\mathrm{d}$ is
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The correct answer is:
$10 \sqrt{2}$
The point lie on $x+y=4$
Coordinates of point $(\alpha, 4-\alpha)$
Distnace from $4 x+3 y-10=0$ is 1
$\left|\frac{4 \alpha+3(4-\alpha)-10}{\sqrt{4^2+3^2}}\right|=1$
$\left|\frac{4 \alpha+12-3 \alpha-10}{5}\right|=1$
$\begin{aligned} & |\alpha+2|=5 \\ & \alpha+2= \pm 5 \\ & \alpha=3,-7\end{aligned}$
at $\alpha=3,4-\alpha=1$
at $\alpha=-7, \quad 4-\alpha=11$
The 2 points are $(3,1)$ and $(-7,11)$
$\therefore d=\sqrt{(-7-3)^2+(11-1)^2}$
$=\sqrt{100+100=10 \sqrt{2}}$
Coordinates of point $(\alpha, 4-\alpha)$
Distnace from $4 x+3 y-10=0$ is 1
$\left|\frac{4 \alpha+3(4-\alpha)-10}{\sqrt{4^2+3^2}}\right|=1$
$\left|\frac{4 \alpha+12-3 \alpha-10}{5}\right|=1$
$\begin{aligned} & |\alpha+2|=5 \\ & \alpha+2= \pm 5 \\ & \alpha=3,-7\end{aligned}$
at $\alpha=3,4-\alpha=1$
at $\alpha=-7, \quad 4-\alpha=11$
The 2 points are $(3,1)$ and $(-7,11)$
$\therefore d=\sqrt{(-7-3)^2+(11-1)^2}$
$=\sqrt{100+100=10 \sqrt{2}}$
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