# Ex.3.2 Q6 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

Go back to  'Ex.3.2'

## Question

Given the linear equation $$2x + 3y-8 = 0,$$ write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) intersecting lines

(ii) parallel lines

(iii) coincident lines

Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.2 | Question 6

## Text Solution

What is Known?

One linear equation $$2x + 3y-8 = 0$$

What is Unknown?

Another linear equation such that given is satisfied.

Reasoning:

Same as Exercise 3.2 (2)

(i) Intersecting lines

Condition: \begin{align} \frac{{{a_1}}}{{{a_2}}} \ne \,\frac{{{b_1}}}{{{b_2}}}\, \end{align}

\begin{align}2x + 3y-8 &= 0\\{a_1}& = 2\\{b_1} &= 3\end{align}

So, considering $${a_2} = 3$$ and $${b_2} = 2$$ will satisfy the condition for intersecting lines $${c_2}$$ can be any value.

\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{2}{3} \quad \frac{{{b_1}}}{{{b_2}}} = \frac{3}{2}\\ \frac{2}{3} &\ne \frac{3}{2}\,\end{align}

$$\therefore$$ Another linear equation is $$3x + 2y-6 = 0$$

(ii) Parallel Lines

Condition: \begin{align} \frac{{{a_1}}}{{{a_2}}} = \,\frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}} \end{align}

\begin{align}2x + 3y-8 &= 0\\{a_1} &= 2\\{b_1} &= 3\\{c_1} &= - 8\end{align}

So, considering $${a_2} = 4,\;\;{b_2} = 6,\;\;{c_2} = 9$$ will satisfy the condition for parallel lines.

\begin{align}\frac{{{a_1}}}{{{a_2}}}&= \frac{2}{4} = \frac{1}{2}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{3}{6} = \frac{1}{2}\\\frac{{{c_1}}}{{{c_2}}}& = \frac{{ - 8}}{9}\end{align}

From above:

$\frac{{{a_1}}}{{{a_2}}} = \,\frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}$

Therefore, another linear equation is $$= 4x + 6y + 9 = 0$$

(iii) Coincident lines:

Condition: \begin{align} \frac{{{a_1}}}{{{a_2}}} = \,\frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}} \end{align}

\begin{align}2x + 3y-8 &= 0\\{a_1} &= 2\\{b_1} &= 3\\{c_1}& = - 8\end{align}

So, considering $${a_2} = 4,\;\;{b_2} = 6,\;\;{c_2} = - 16$$ will satisfy the condition for parallel lines.

\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{2}{4} = \frac{1}{2}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{3}{6} = \frac{1}{2}\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 8}}{{ - 16}} = \frac{1}{2}\end{align}

From above:

$\frac{{{a_1}}}{{{a_2}}} = \,\frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}$

Therefore, linear equation is $$4x + 6y-16 = 0$$

Learn math from the experts and clarify doubts instantly

• Instant doubt clearing (live one on one)
• Learn from India’s best math teachers
• Completely personalized curriculum