MCQ 11 Mark
Statement-1 (A): The graph of the linear equation $4 x+3 y=24$ mects $x$-axis at (-6,0).
Statement-2 (R): Points on $x$-axis are of the form (a, 0), where a is a variable.
Statement-2 (R): Points on $x$-axis are of the form (a, 0), where a is a variable.
- AStatement-1 and Statement-2 are True; Statement-2 is a correct explanation for Statement-1.
- ✓Statement-1 and Statement-2 are True; Statement-2 is not a correct explanation for Statement-1.
- CStatement-1 is True, Statement-2 is False.
- DStatement-1 is False, Statement-2 is True.
Answer
View full question & answer→Correct option: B.
Statement-1 and Statement-2 are True; Statement-2 is not a correct explanation for Statement-1.
(b)
Let the line $3 x-2 y=12$ meet $y$-axis at $(0, a)$. Then, $x=0$ and $y=a$ is solution of the equation $3 x-2 y=12$.
$3 \times 0-2 \times a=12 \Rightarrow-2 a=12 \Rightarrow a=-6$
So, the required points is $(0,-6)$. Hence, statement -1 is true.
The graph of the linear equation $a x+b y=0$ is a line passing through the origin. So, the graph of the equation $2 y=3 x$ or, $3 x-2 y=0$ is a line passing through the origin. hence, statement-2 is true. Thus, both the statements are true. Hence, option (b) is correct.
Let the line $3 x-2 y=12$ meet $y$-axis at $(0, a)$. Then, $x=0$ and $y=a$ is solution of the equation $3 x-2 y=12$.
$3 \times 0-2 \times a=12 \Rightarrow-2 a=12 \Rightarrow a=-6$
So, the required points is $(0,-6)$. Hence, statement -1 is true.
The graph of the linear equation $a x+b y=0$ is a line passing through the origin. So, the graph of the equation $2 y=3 x$ or, $3 x-2 y=0$ is a line passing through the origin. hence, statement-2 is true. Thus, both the statements are true. Hence, option (b) is correct.