Question

Match the two columns:

Column I	COLUMN II
(A) $4x+3y=12$ cuts $x$ and $y$ axes	(p) $4x-3y=7$
(B) $4-3x=0$ can be written as	(q) at $\left(3,0\right)$ and $\left(0,4\right)$
(C) $\left(3,4\right)$ lies on $3y=ax+7$, then $a$ =	(r) $-3x+0y+4=0$
(D) $\left(1,-1\right)$ is a solution of the equation	(s) $\frac{5}{3}$

Choose the correct option

• A. $\left(A\right)\to q;\left(B\right)\to r;\left(C\right)\to s;\left(D\right)\to p$
• B. $\left(A\right)\to p;\left(B\right)\to r;\left(C\right)\to s;\left(D\right)\to q$
• C. $\left(A\right)\to q;\left(B\right)\to r;\left(C\right)\to p;\left(D\right)\to s$
• D. $\left(A\right)\to r;\left(B\right)\to p;\left(C\right)\to q;\left(D\right)\to s$

Answer 1

The correct option is A

$\left(A\right)\to q;\left(B\right)\to r;\left(C\right)\to s;\left(D\right)\to p$

Explanation for (A)

Checking $4\mathrm{x}+3\mathrm{y}=12$ cuts the x and y axes at which point

For $x$-axis, the $y$-coordinate will be $0$

Substituting $\mathrm{y}=0$ in the equation we get,

$$\begin{gathered} 4\mathrm{x}+3\mathrm{y}=12 \\ \Rightarrow 4\mathrm{x}=12 \\ \Rightarrow \mathrm{x}=3 \end{gathered}$$

The point at which it will cut the $x$-axis is $(3,0)$

For $y$-axis, the $x$-coordinate will be $0$

Substituting $\mathrm{x}=0$ in the equation we get,

$$\begin{gathered} 4\mathrm{x}+3\mathrm{y}=12 \\ \Rightarrow 3\mathrm{y}=12 \\ \Rightarrow \mathrm{y}=4 \end{gathered}$$

The point at which it will cut the $y$-axis is $(0,4)$

Hence, the correct match for equation (A)→q

Explanation for (B)

Rewriting this equation $4-3\mathrm{x}=0$

Here, the coefficient of $\mathrm{y}$ variable is zero.

$$\begin{gathered} \Rightarrow 4-3\mathrm{x}=0 \\ \Rightarrow 4-3\mathrm{x}+0\mathrm{y}=0 \\ \Rightarrow -3\mathrm{x}+0\mathrm{y}+4=0 \end{gathered}$$

Hence, the correct match for equation (B) →r.

Explanation for (C)

Finding the value of $\mathrm{a}$

$3\mathrm{y}=\mathrm{ax}+7$ passes through the point $(3,4)$

Hence, $(3,4)$ will satisfy the equation.

Substituting $x=3$ and $y=4$ in the equation.

$$\begin{gathered} \Rightarrow 3\mathrm{y}=\mathrm{ax}+7 \\ \Rightarrow 3\left(4\right)=\mathrm{a}\left(3\right)+7 \\ \Rightarrow 12=3\mathrm{a}+7 \\ \Rightarrow 5=3\mathrm{a} \\ \Rightarrow \mathrm{a}=\frac{5}{3} \end{gathered}$$

Hence, the correct match for equation (C) →s

Explanation for (D)

Checking if $(1,-1)$ is a solution of the equation $4x-3y=7$

Substituting $\mathrm{x}=1$ and $y=-1$ in the equation

$$\begin{gathered} \Rightarrow RHS=7 \\ \Rightarrow LHS=4\left(1\right)+3\left(1\right)=7 \\ \Rightarrow LHS=RHS \end{gathered}$$

Hence, the correct match for $D$ is (D) →p

Thus, the obtained matching options are $\left(A\right)\to q;\left(B\right)\to r;\left(C\right)\to s;\left(D\right)\to p$

Hence, the correct answer is option A.