Drawn from origin are two mutually perpendicular straight lines forming an isosceles triangle together with straight line x + y=a. then the area of the triangle is in terms of ′a′?

- Hint: -The formula for calculating the distance which is the perpendicular distance between a point and a line is given as follows
\[d=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
(Where the equation of line is ax+by+c=0 and the point is \[\left( {{x}_{1}},{{y}_{1}} \right)\] )
Another fact that is important for solving this question is that the midpoint of a right isosceles triangle is equidistant from all the three vertices.

Complete step-by-step solution -

As mentioned in the question, we have to find the area of the triangle that is thus generated in terms of ‘a’.
Now, the perpendicular distance from the origin to the give line is as follows
\[\begin{align}
  & d=\dfrac{\left| 1\cdot 0+1\cdot 0+-a \right|}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \\
 & d=\dfrac{a}{\sqrt{2}} \\
\end{align}\]
Now, using the fact that is mentioned in the hint, we know that the midpoint of a right isosceles triangle is equidistant from all the three vertices, hence, the base of the formed triangle is 2 times the perpendicular distance from the origin to the given line.

Hence, area of the triangle can be calculated as follows
\[\begin{align}
  & =\dfrac{1}{2}\times base\times height \\
 & =\dfrac{1}{2}\times 2d\times d \\
 & ={{d}^{2}} \\
 & ={{\left( \dfrac{a}{\sqrt{2}} \right)}^{2}} \\
 & =\dfrac{{{a}^{2}}}{2} \\
\end{align}\]
Hence, this is the area of the formed triangle.


Note: -The students can make an error if they don’t know about the fact that is mentioned in the hint as follows
Another fact that is important for solving this question is that the midpoint of a right isosceles triangle is equidistant from all the three vertices.
Without knowing this formula, one could not get to the correct answer.