Question

Match the linear equations in Column-I with their solution is Column-II.

COLUMN I	COLUMN II
(A) $2\mathrm{x}+3\mathrm{y}=5\sqrt{2}$	(p) $(\sqrt{2},0)$
(B) $\sqrt{2}\mathrm{x}+4\mathrm{y}=6$	(q) $(\sqrt{2},\sqrt{2})$
(C) $3\mathrm{x}+\frac{\mathrm{y}}{2}=\frac{1}{\sqrt{2}}$	(r) $(\sqrt{2},1)$
(D) $\sqrt{2}\mathrm{x}+5\mathrm{y}=2$	(s) $(0,\sqrt{2})$

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

Answer 1

The correct option is C

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

Explanation for (A)

Checking the solution for option A

For option (A), the equation is $2\mathrm{x}+3\mathrm{y}=5\sqrt{2}$

Substituting (p) → $(\sqrt{2},0)$ in the equation,

$$\begin{gathered} \Rightarrow LHS=2\left(\sqrt{2}\right)+3\left(0\right)=2\sqrt{2} \\ \Rightarrow RHS=5\sqrt{2} \\ \Rightarrow LHS\ne RHS \end{gathered}$$

Substituting (q) →$(\sqrt{2},\sqrt{2})$ in the equation,

$$\begin{gathered} \Rightarrow \mathrm{RHS}=5\sqrt{2} \\ \Rightarrow LHS=2\left(\sqrt{2}\right)+3\left(\sqrt{2}\right)=5\sqrt{2} \\ \Rightarrow LHS=RHS \end{gathered}$$

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

Explanation for (B)

Checking the solution for option B

The equation is $\sqrt{2}\mathrm{x}+4\mathrm{y}=6$

Substituting (p) → $(\sqrt{2},0)$ in the equation

$$\begin{gathered} \Rightarrow RHS=6 \\ \Rightarrow LHS=\sqrt{2}\left(\sqrt{2}\right)=2 \\ \Rightarrow LHS\ne RHS \end{gathered}$$

Hence, option (p) is incorrect.

Substituting (r) → $(\sqrt{2},1)$ in the equation

$$\begin{gathered} \Rightarrow RHS=6 \\ \Rightarrow LHS=\sqrt{2}\left(\sqrt{2}\right)+4\left(1\right)=6 \\ \Rightarrow LHS=RHS \end{gathered}$$

Hence, the correct match for option (B) →(r)

Explanation for (C)

Checking the solution for option C

The equation is $3\mathrm{x}+\frac{\mathrm{y}}{2}=\frac{1}{\sqrt{2}}$

Substituting (p) → $(\sqrt{2},0)$ in the equation

$$\begin{gathered} \Rightarrow LHS=3\left(\sqrt{2}\right) \\ \Rightarrow RHS=\frac{1}{\sqrt{2}} \\ \Rightarrow LHS\ne RHS \end{gathered}$$

Hence, option (p) is incorrect.

Substituting (s) → $(0,\sqrt{2})$ in the equation

$$\begin{gathered} \Rightarrow RHS=\frac{1}{\sqrt{2}} \\ \Rightarrow LHS=3\left(0\right)+\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \\ \Rightarrow LHS=RHS \end{gathered}$$

Therefore, the correct match for option C → (s)

Explanation for (D)

Checking the solution for option D

The equation is $\sqrt{2}\mathrm{x}+5\mathrm{y}=2$

Substituting (p) → $(\sqrt{2},0)$ in the equation

$$\begin{gathered} \Rightarrow RHS=2 \\ \Rightarrow LHS=\sqrt{2}\left(\sqrt{2}\right)+5\left(0\right)=2 \\ \Rightarrow LHS=RHS \end{gathered}$$

Hence, the correct match for option D→ (p)

Thus we conclude the correct match 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 C.