NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 3: Pair of Linear Equations in Two Variables - Exercise 3.2
Question 2

Question. 2

Do the following equations represent a pair of coincident lines? Justify your answer.

(i) \(3x + \dfrac{1}{7}y = 3\), \(7x + 3y = 7\)

(ii) \(-2x - 3y = 1\), \(6y + 4x = -2\)

(iii) \(\dfrac{x}{2} + \dfrac{y}{5} + \dfrac{5}{16} = 0\), \(4x + 8y + \dfrac{5}{4} = 0\)

Answer:

(i) No, (ii) Yes, (iii) No

Detailed Answer with Explanation:

Concept: Two equations represent coincident lines if and only if

\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\).

Here, each equation is written as \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\).

(i)

First equation: \(3x + \dfrac{1}{7}y - 3 = 0\)

Second equation: \(7x + 3y - 7 = 0\)

Now, compare the ratios:

\(\dfrac{a_1}{a_2} = \dfrac{3}{7}\)

\(\dfrac{b_1}{b_2} = \dfrac{1/7}{3} = \dfrac{1}{21}\)

Since \(\dfrac{3}{7} \ne \dfrac{1}{21}\), the lines are not coincident.

(ii)

First equation: \(-2x - 3y - 1 = 0\)

Second equation: \(4x + 6y + 2 = 0\)

If we multiply the first equation by \(-2\), we get:

\(4x + 6y + 2 = 0\)

which is exactly the second equation.

Thus, all three ratios are equal, so the lines are coincident.

(iii)

First equation: \(\dfrac{x}{2} + \dfrac{y}{5} + \dfrac{5}{16} = 0\)

Multiply throughout by 80 to clear fractions:

\(40x + 16y + 25 = 0\)

Second equation: \(4x + 8y + \dfrac{5}{4} = 0\)

Multiply throughout by 4:

\(16x + 32y + 5 = 0\)

Now compare ratios:

\(\dfrac{a_1}{a_2} = \dfrac{40}{16} = 2.5\)

\(\dfrac{b_1}{b_2} = \dfrac{16}{32} = 0.5\)

Since these are not equal, the lines are not coincident.

Final Answer:

(i) No, (ii) Yes, (iii) No

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 3: Pair of Linear Equations in Two Variables – Exercise 3.2 | Detailed Answers