Question

Find all integers a so that the area enclosed by the lines y = ax, y = 0, and x + 2y - 4 = 0 is a natural number. (Solve without using integrals)

Answer 1

The area enclosed by the lines,

• $\smally=ax\quad\quad\dots(1)$

• $\smally=0\quad\quad\dots(2)$

• $\smallx+2y-4=0\quad\quad\dots(3)$

is a triangle. First let us find the vertices of the triangle.

Solving (1) and (2) we get the vertex,

• $\smallA(0,\ 0)$

Solving (2) and (3) we get the vertex,

• $\smallB(4,\ 0)$

Solving (3) and (1) we get the vertex,

• $\smallC\left(\dfrac{4}{2a+1},\ \dfrac{4a}{2a+1}\right)$

Now the area of the triangle ABC is found out by determinant method.

The area of the triangle with vertices at $\small(x_1,\ y_1),\ (x_2,\ y_2)$ and $\small(x_3,\ y_3)$ is,

$\small\longrightarrow A=\Bigg|\dfrac{1}{2}\left|\begin{array}{cc}x_1-x_3&y_1-y_3\\x_2-x_3&y_2-y_3\end{array}\right|\Bigg|$

Then the area of triangle ABC is,

$\small\longrightarrow A=\left|\,\dfrac{1}{2}\left|\begin{array}{cc}4&0\\&\\\dfrac{4}{2a+1}&\dfrac{4a}{2a+1}\end{array}\right|\,\right|$

$\small\longrightarrow A=\left|\,\dfrac{1}{2}\cdot\dfrac{16a}{2a+1}\,\right|$

$\small\longrightarrow A=\left|\,\dfrac{8a}{2a+1}\,\right|$

$\small\longrightarrow \pm A=\dfrac{8a}{2a+1}$

Given that $\smallA$ is a natural number. Then $\small\pm A$ is a non - zero integer, so should $\small\dfrac{8a}{2a+1}$ be.

Then,

$\small\longrightarrow\dfrac{8a}{2a+1}\neq0$

$\small\Longrightarrow a\neq0$

Now,

$\small\longrightarrow \pm A=\dfrac{8a}{2a+1}$

$\small\longrightarrow \pm A=4\cdot\dfrac{2a}{2a+1}$

$\small\longrightarrow \pm A=4\cdot\dfrac{2a+1-1}{2a+1}$

$\small\longrightarrow \pm A=4\left(1-\dfrac{1}{2a+1}\right)$

Since $\smallA$ is a natural number, this equation implies that the term $\small2a+1$ exactly divides the number 1. Here $\smalla$ is an integer, so $\small2a+1$ should be an odd integer.

Then there are only two possibilities for $\small2a+1.$

$\small\longrightarrow 2a+1=1\quad\quad OR\quad\quad 2a+1=-1$

$\small\longrightarrow a=0\quad\quad OR\quad\quad a=-1$

Since $\smalla\neq0,$

$\small\text{ \longrightarrow\underline{\underline{a=-1}} }$

Hence this is the only possible value of $\smalla.$