NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.1 Question 6
NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables
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.
Given:
The first linear equation is:
$$ 2x + 3y - 8 = 0 $$To Find:
Another linear equation in two variables such that the pair of equations represents:
- (i) Intersecting lines
- (ii) Parallel lines
- (iii) Coincident lines
Formula:
For two linear equations in the form:
$$ \begin{aligned} a_1x + b_1y + c_1 &= 0 \\ a_2x + b_2y + c_2 &= 0 \end{aligned} $$The nature of the pair is determined as follows:
- Intersecting lines: \(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\)
- Parallel lines: \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\)
- Coincident lines: \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\)
Solution:
Step 1: Write the given equation and compare with the standard form.
$$ 2x + 3y - 8 = 0 $$So, \(a_1 = 2\), \(b_1 = 3\), \(c_1 = -8\).
Step 2: Case (i): Intersecting lines
For intersecting lines, \(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\).
Let us choose another equation:
$$ x + y + 1 = 0 $$Here, \(a_2 = 1\), \(b_2 = 1\), \(c_2 = 1\).
Check the ratios:
$$ \dfrac{a_1}{a_2} = \dfrac{2}{1} = 2 \\ \dfrac{b_1}{b_2} = \dfrac{3}{1} = 3 $$Since \(2 \neq 3\), the lines are intersecting.
Step 3: Case (ii): Parallel lines
For parallel lines, \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\).
Let us choose another equation:
$$ 4x + 6y + 5 = 0 $$Here, \(a_2 = 4\), \(b_2 = 6\), \(c_2 = 5\).
Check the ratios:
$$ \dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2} \\ \dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2} \\ \dfrac{c_1}{c_2} = \dfrac{-8}{5} $$Since \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\) but \(\dfrac{c_1}{c_2} \neq \dfrac{a_1}{a_2}\), the lines are parallel.
Step 4: Case (iii): Coincident lines
For coincident lines, \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\).
Multiply the given equation by 2:
$$ 2 \times (2x + 3y - 8) = 0 \implies 4x + 6y - 16 = 0 $$Here, \(a_2 = 4\), \(b_2 = 6\), \(c_2 = -16\).
Check the ratios:
$$ \dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2} \\ \dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2} \\ \dfrac{c_1}{c_2} = \dfrac{-8}{-16} = \dfrac{1}{2} $$All ratios are equal, so the lines are coincident.
Result:
The required equations are:
- (i) For intersecting lines: \(x + y + 1 = 0\)
- (ii) For parallel lines: \(4x + 6y + 5 = 0\)
- (iii) For coincident lines: \(4x + 6y - 16 = 0\)
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