Q3(i):

Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: (i) $(0, 2)$

Step 1: Initial Setup & Given Variables

We are given the linear equation in two variables:

$x - 2y = 4$

We need to determine whether the coordinate point $(0, 2)$ is a valid solution to this equation. In the Cartesian coordinate system, an ordered pair $(x, y)$ represents a solution to an equation if and only if substituting the $x$ and $y$ values into the equation results in a true mathematical statement [Per the Fundamental Principle of Algebraic Substitution].

• Given $x$-coordinate (abscissa): $x = 0$
• Given $y$-coordinate (ordinate): $y = 2$

Step 2: Substitution into the Left-Hand Side (LHS)

We isolate the Left-Hand Side (LHS) of the given equation and substitute the given coordinates:

$\text{LHS} = x - 2y$

Substituting $x = 0$ and $y = 2$:

$\text{LHS} = (0) - 2(2)$

$\text{LHS} = 0 - 4$

$\text{LHS} = -4$

Step 3: Comparison with the Right-Hand Side (RHS)

The Right-Hand Side (RHS) of the original equation is a constant:

$\text{RHS} = 4$

Comparing the evaluated LHS with the RHS:

$-4 \neq 4$

$\text{LHS} \neq \text{RHS}$

Because the substitution yields an inequality, the ordered pair $(0, 2)$ does not satisfy the equation.

Step 4: Geometric Interpretation & Verification

Geometrically, a linear equation in two variables represents a straight line on a Cartesian plane. Every point that lies exactly on this line is a solution to the equation. Since $(0, 2)$ does not satisfy the algebraic equation, the point $(0, 2)$ will not lie on the line $x - 2y = 4$.

To graph the line $x - 2y = 4$, we find its intercepts:

• x-intercept (set $y = 0$): $x - 2(0) = 4 \implies x = 4$. Point: $(4, 0)$
• y-intercept (set $x = 0$): $0 - 2y = 4 \implies y = -2$. Point: $(0, -2)$

The visual representation below proves that the point $(0, 2)$ is spatially disconnected from the solution set (the line).

Final Solution: Since substituting $x = 0$ and $y = 2$ results in $\text{LHS} \neq \text{RHS}$ ($-4 \neq 4$), the point $(0, 2)$ is NOT a solution to the equation $x - 2y = 4$.