# Ex.3.2 Q2 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

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## Question

On comparing the ratios \begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}, \end{align} find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i)

\begin{align}\quad 5x-4y + 8 &= 0\\7x + 6y-9 &= 0\end{align}

(ii)

\begin{align}\quad 9x + 3y + 12& = 0\\18x + 6y + 24 &= 0\end{align}

(iii)

\begin{align}\quad 6x-3y + 10 &= 0\\2x-y + 9 &= 0\end{align}

Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.2 | Question 2

## Text Solution

(i) What is Known?

\begin{align}5x-4y + 8 &= 0\\7x + 6y-9 &= 0\end{align}

What is Unknown?

Whether the lines are

(i) Intersecting

(ii) Parallel

(iii) Coincident

Reasoning:

For any pair of linear equation

$$a_1 x + b_1 y + c_1 = 0\\ a_2 x + b_2 y + c_2 = 0$$

a) \begin{align}\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\end{align} (Intersecting Lines)

b)  \begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\end{align} (Coincident Lines)

c)  \begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\end{align} (Parallel Lines)

Steps:

\begin{align}{a_1} &= 5 \;\,\qquad{b_1} = - 4 \qquad {c_1} = 8\\{a_2} &= 7 \qquad \;\,{b_2} = 6 \;\;\;\qquad {c_2} = - 9\end{align}

\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{5}{7} \qquad \;\;\; \dots(1)\\ \frac{{{b_1}}}{{{b_2}}} = \frac{{ - 4}}{6} &= \frac{{ - 2}}{3} \qquad \dots (2)\end{align}

From (i) and (ii)

$\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}$

Therefore, they are intersecting lines at a point

(ii) What is Known?

\begin{align}9x + 3y + 12& = 0\\18x + 6y + 24 &= 0\end{align}

What is Unknown?

Whether the lines are

(i) Intersecting

(ii) Parallel

(iii) Coincident

Reasoning:

For any pair of linear equation

\begin{align}{a_1}x + {b_1}y + {c_1} &= 0\\{a_2}x + {b_2}y + {c_2} &= 0\end{align}

a) \begin{align}\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\end{align}(Intersecting Lines)

b) \begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\end{align}(Coincident Lines)

c) \begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\end{align}(Parallel Lines)

Steps:

\begin{align}{a_1} &= 9, \quad {b_1} = 3 \quad {c_1} = 12\\{a_2}& = 18\quad {b_2} = 6\quad {c_2} = 24\end{align}

\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{9}{{18}} = \frac{1}{2} \quad \dots(1)\\ \frac{{{b_1}}}{{{b_2}}} &= \frac{3}{6} = \frac{1}{2} \quad \;\;\dots(2)\\ \frac{{{c_1}}}{{{c_2}}} &= \frac{{12}}{{24}} = \frac{1}{2} \quad \dots(3) \end{align}

From (1), (2) and (3)

$\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}} = \frac{1}{2}$

Therefore, they are coincident lines

(iii) what is Known?

\begin{align}6x-3y + 10 &= 0\\2x-y + 9 &= 0\end{align}

What is Unknown?

Whether the lines

(i) Intersecting

(ii) Parallel

(iii) Coincident

Reasoning:

For any pair of linear equation

\begin{align}{a_1}x + {b_1}y + {c_1} = 0\\{a_2}x + {b_2}y + {c_2} = 0\end{align}

a) \begin{align}\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\end{align} (Intersecting Lines)

b) \begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\end{align} (Coincident Lines)

c) \begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\end{align} (Parallel Lines)

Steps:

\begin{align}{a_1} &= 6,\quad {b_1} = - 3\quad{c_1} = 10\\{a_2} &= 2 \quad \; {b_2} = - 1 \quad \,{c_2} = 9\end{align}

\begin{align}\frac{{{a_1}}}{{{a_2}}}& = \frac{6}{2} = 3 \qquad \quad \dots(1)\\\frac{{{b_1}}}{{{b_2}}} &= \frac{{ - 3}}{{ - 1}} = 3 \qquad \dots(2)\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{10}}{9} \qquad \qquad \;\dots(3)\end{align}

From $$(1), (2)$$ and $$(3)$$

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

Therefore, they are parallel lines.

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