Let S be the set of all triangles in the x y -plane, each having one vertex at the origin and the other two…

1. $36$
2. $32$
3. $9$
4. $18$

Solution

Since the vertex of given triangle is $O$ (origin) and other two vertices $A$ and $B$ lies on the coordinate axes, hence triangle must be a right-angled triangle.

Area $=\frac{1}{2}x.y ( x, y\in I)$

As per given condition

$\frac{1}{2}x.y=50\Rightarrow x. y=100$

Possible ordered pairs of $\left(x, y\right)$ when $∆$ formed in $1^{st}$ quadrant are $9$ i. e. $\left(1,100\right), \left(2,50\right), \left(4,25\right), \left(5,20\right), \left(10,10\right), \left(25, 4\right), \left(50, 2\right), \left(100,1\right)$

$∆$ can be formed in any of the quadrants, hence total number of cases $9\times 4=36.$

Asked in: JEE Main 2019 (09 Jan Shift 2)

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