MCQ
The graph of the line $x - y = 0$ passes through the point:
- A$\Big(\frac{-1}{2},\frac{1}{2}\Big)$
- B$\Big(\frac{3}{2},\frac{-3}{2}\Big)$
- C$(0,-1)$
- ✓$(1, 1)$
✓
Answer
Correct option: D.
$(1, 1)$
The given linear equation is $x = y = 0.$
We have to check which of the point satisfy the given equation.
consider option (a):
Substituting $\text{x}=-\frac{1}{2}$ and $\text{y}=\frac{1}{2}$ in the $LHS$ if the given linear equation
$\therefore\ \text{x}-\text{y}=-\frac{1}{2}-\frac{1}{2}=-1\neq\text{RHS}$
$\therefore\ \text{x}=-\frac{1}{2}$ and $\text{y}=\frac{1}{2}$ does not satisfy the given linear equation.
Consider option (b):
Substituting $\text{a}=\frac{3}{2}$ and $\text{y}=-\frac{3}{2}$ in the $LHS$ if the given linear equation on
$\therefore\ \text{x}-\text{y}=\frac{3}{2}+\frac{3}{2}=3\neq\text{RHS}$
$\therefore\ \text{x}=-\frac{3}{2}$ and $\text{y}=-\frac{3}{2}$ does not satisfy the given linear eqation on.
Consider option (d):
Substitution $x = 1 $and $y = 1$ in the $LHS$ if the given linear equation
$\therefore\ $$x - y = 1 - 1 = 0 = RHS$
$\therefore\ $$x = 1$ and $y = 1$ satisfies the given linear equation.
We have to check which of the point satisfy the given equation.
consider option (a):
Substituting $\text{x}=-\frac{1}{2}$ and $\text{y}=\frac{1}{2}$ in the $LHS$ if the given linear equation
$\therefore\ \text{x}-\text{y}=-\frac{1}{2}-\frac{1}{2}=-1\neq\text{RHS}$
$\therefore\ \text{x}=-\frac{1}{2}$ and $\text{y}=\frac{1}{2}$ does not satisfy the given linear equation.
Consider option (b):
Substituting $\text{a}=\frac{3}{2}$ and $\text{y}=-\frac{3}{2}$ in the $LHS$ if the given linear equation on
$\therefore\ \text{x}-\text{y}=\frac{3}{2}+\frac{3}{2}=3\neq\text{RHS}$
$\therefore\ \text{x}=-\frac{3}{2}$ and $\text{y}=-\frac{3}{2}$ does not satisfy the given linear eqation on.
Consider option (d):
Substitution $x = 1 $and $y = 1$ in the $LHS$ if the given linear equation
$\therefore\ $$x - y = 1 - 1 = 0 = RHS$
$\therefore\ $$x = 1$ and $y = 1$ satisfies the given linear equation.
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