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

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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 \(4x + 6y-16 = 0\)