JEE Main 2023 — Straight Lines Question with Solution

Given:

$3x+4y=60$

$\Rightarrow \frac{x}{20}+\frac{y}{15}=1$

If $x=1,y=\frac{57}{4}=14.25$

So, points are

$\left(1,1\right)\left(1,2\right)-\left(1,14\right) \Rightarrow 14$ points.

If $x=2,y=\frac{27}{2}=13.5$

So, points are

$\left(2,2\right)\left(2,4\right)\dots \left(2,12\right) \Rightarrow 6$ points.

If $x=3,y=\frac{51}{4}=12.75$

So, points are

$\left(3,3\right)\left(3,6\right)-\left(3,12\right) \Rightarrow 4$ points.

If $x=4, y=12$

So, points are

$\left(4,4\right)\left(4,8\right) \Rightarrow 2$ points.

If $x=5,y=\frac{45}{4}=11.25$

So, points are

$\left(5,5\right),\left(5,10\right) \Rightarrow 2$ points.

If $x=6,y=\frac{21}{2}=10.5$

So, point is

$\left(6,6\right)\Rightarrow 1$ point

If $x=7,y=\frac{39}{4}=9.75$

So, point is

$\left(7,7\right)\Rightarrow 1$ point

If $x=8, y=9$

So, point is

$\left(8,8\right)\Rightarrow$ $1$ point

If $\mathrm{x}=9\Rightarrow \mathrm{y}=\frac{33}{4}=8.25\Rightarrow$ no point

Total points inside the triangle

$=31$ points