**Problem: **Find the vertices of the two right angles triangles, each having area 18 and such that the point (2, 4) lies on the hypotenuse, and the other two sides are formed by the x and y axes.

**Discussion: **Suppose the vertices are (a, 0) and (0, b). Clearly \(\frac{1}{2} ab = 18 \) or ab = 36.

Also the equation of the line through (0,a), (b,0) is \(\displaystyle{ \frac{x}{a} + \frac{y}{b} = 1 } \). Since we know that (2, 4) is on that line, there fore \(\displaystyle { \frac{2}{a} + \frac{4}{b} = 1 } \).

In this equation, lets replace \(a \) by \(\frac{36}{b} \). Hence we get \(\displaystyle { \frac{2}{\frac{36}{b}} + \frac{4}{b} = 1 } \) or \(\displaystyle { \frac{b}{18} + \frac{4}{b} = 1 } \).

Therefore we get a quadratic in b.

\(\displaystyle { b^2 -18b + 72 = 0 } \). We can simply middle term factorize this to find \(\displaystyle {(b-12)(b-6) = 0 } \). Thus b = 12 or 6 implying a = 3 or 6.

The other two vertices are: (0,12) and (3, 0) OR (0,6) and (6,0).

## Chatuspathi:

**What is this topic:**Coordinate Geometry**What are some of the associated concept:**Intercept form of straight line equation.**Where can learn these topics:**Cheenta**Book Suggestions:**Coordinate Geometry Volume I by S.L. Loney