Let sides of an isosceles triangle are $7 x - y - 4 = 0$ and $x + y + 1 = 0$ if $(1, 2)$ is on the base then the perpendicular distance from origin to the base is

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$\frac{3}{\sqrt{10}}$

$\frac{5}{\sqrt{10}}$

answer is D.

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Detailed Solution

Let eq of AB is $7 x - y - 4 = 0 \Rightarrow m_{1} = 7$
Let eq of AC is $x + y + 1 = 0 \Rightarrow m_{2} = - 1$
Slope of BC is m
$T a n θ = | \frac{m_{1} - m}{1 + m m_{1}} = | \frac{m - m_{2}}{1 + m m_{2}} | | \Rightarrow = 1 / 3, - 3$
$\begin{aligned} (y - 2) = \frac{1}{3} (x - 1) \\ 3 y - 6 = x - 1 \\ 3 y - x - 5 = 0 . \end{aligned}$
The perpendicular distance from origin to the base is $\frac{5}{\sqrt{3^{2} + 1^{2}}} = \frac{5}{\sqrt{10}}$