Select the correct options:
In a triangle ABC, the bisector of the angles B and C lie along the line x = y and y = 0. If A is (1, 2) then \[\sqrt{10}d\left( A,B,C \right)\] where \[d\left( A,B,C \right)\] represents distance of point A from side BC
A. 1
B. 4
C. 3
D. 5
E. 0

Hint: Draw \[\Delta ABC\] and bisect \[\angle B\] and \[\angle C\] by line x = y and y = 0. Now find the image of A w.r.t line y =x which can be taken as D and find the image of a w.r.t to line y = 0, which is E on BC. Find equation of line BC. Draw \[AF\bot BC\]. Using a perpendicular distance formula, get the value of \[\sqrt{10}d\left( A,B,C \right)\].

Complete Step-by-Step solution:
Let us draw a triangle ABC, as in figure. Angle B is bisected by the line y = x and angle C is bisected by the line y = 0. Given point, A = (1, 2).

Now we need to find the image of point A, with respect to the line y = x.
Thus from the figure we can see that the image of point A w.r.t to line y = x is (2, 1).

Thus let us mark it as E = (1, -2).

Thus equation of line BC can be formed by the equation,
\[\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}\Rightarrow \dfrac{y-\left( -2 \right)}{1-\left( -2 \right)}=\dfrac{x-1}{2-1}\]
\[\left\{ \begin{align}
  & \because \left( {{x}_{1}},{{y}_{1}} \right)=\left( 1,-2 \right) \\
 & \left( {{x}_{2}},{{y}_{2}} \right)=\left( 2,1 \right) \\
\end{align} \right\}\]
\[\therefore \dfrac{y+2}{1+2}=\dfrac{x-1}{2-1}\], use cross multiplication property.
\[\begin{align}
  & 1\left( y+2 \right)=3\left( x-1 \right) \\
 & \therefore y+2=3x-3 \\
 & 3x-y=5 \\
\end{align}\]
Now we need the distance of AF, which is perpendicular to the base BC.
Now let us use the perpendicular distance formula.
Here \[3x-y=5\] is of the form \[ax+by+c\] and \[\left( x,y \right)=\left( 1,2 \right)\].
\[\begin{align}
  & d=\dfrac{\left| ax+by+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \\
 & \therefore d=\dfrac{\left| \left( 3\times 1 \right)+\left( 2\times -1 \right)-5 \right|}{\sqrt{{{3}^{2}}+{{\left( -1 \right)}^{2}}}}=\dfrac{\left| 3-2-5 \right|}{\sqrt{9+1}}=\dfrac{\left| -4 \right|}{\sqrt{10}}=\dfrac{4}{\sqrt{10}} \\
\end{align}\]
\[\therefore d\left( A,B,C \right)=\dfrac{4}{\sqrt{10}}\], cross multiply \[\sqrt{10}\].
Hence, we got, \[\sqrt{10}d\left( A,B,C \right)=4\].
Hence we got the value of \[\sqrt{10}d\left( A,B,C \right)=4\].
\[\therefore \] Option (b) is the correct answer.

Note: It is important to take the image of point A w.r.t the line y = x and y = 0. Then only you get the equation of line of BC. We have been given the hint that d (A, B, C) is the distance from A to side BC, which means that we have to use the perpendicular distance formula.