The base of the triangle lies along the line x=a and is of length a. The area of the triangle is ${a^2}$, if the vertex lies on the line
(a)x=0
(b)x=-a
(c)x=3a
(d)x=-3a

Hint: Here we use the basic formula for the area of the triangle and then use the formula for distance between two parallel lines.
The area of the triangle as we know is \[\dfrac{1}{2}bh\] and the distance between the parallel lines is \[\dfrac{{{c}_{1}}-{{c}_{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]

Complete step-by-step solution -



Now, let us begin with the area.
Area = \[\dfrac{1}{2}bh\]
Since base length = a
{a^2} = \[\dfrac{1}{2}ah\]
Cancelling a on both the sides we get,
a = \[\dfrac{1}{2}h\]
Cross-multiplying we get,
2a=h
Now the height is perpendicular to the base. So the triangle should lie in between parallel lines which are a distance 2a apart.
One of the lines is x=a.
Now let us apply the formula of distance between two parallel lines
Distance = \[\dfrac{|{{c}_{1}}-{{c}_{2}}|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
Let c1 be a and the value of a is 1 and b is 0. Thus,
2a = \[\dfrac{|a-{{c}_{2}}|}{\sqrt{{{1}^{2}}+{{0}^{2}}}}\]
2a = \[\dfrac{|a-{{c}_{2}}|}{\sqrt{1}}\]
Removing the modulus sign we get \[\pm \] 2a.
\[\begin{align}
  & \pm 2a=a-{{c}_{2}} \\
 & {{c}_{2}}=a\pm 2a \\
\end{align}\]
Thus we get two values for c2.
c2=a+2a and c2=a-2a
Which is
c2=3a and c2=-a
Thus the equation of the line parallel to the line x=a is
x=3a or x=-a.
Now as we mentioned earlier the vertex should lie on this/these lines so that the triangle area becomes a2.
Thus here two options are right, option(b) and option(c)

Note: Students can forget to put the modulus sign in the distance between the parallel lines formula. This might result in just one answer and might result in losing one point. So remember this is a “multiple answer correct” type of question. So there are two answers to this question.