Classify the following pairs of lines as coincident, parallel or intersecting:
(i) 2x + y − 1 = 0 and 3x + 2y + 5 = 0
(ii) x − y = 0 and 3x − 3y + 5 = 0
(iii) 3x + 2y − 4 = 0 and 6x + 4y − 8 = 0.

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Solution

Let $a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0$ be the two lines.

(a) The lines intersect if $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ is true.

(b) The lines are parallel if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ is true.

(c) The lines are coincident if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$ is true.

(i) 2x + y − 1 = 0 and 3x + 2y + 5 = 0

Here, $\frac{2}{3} \neq \frac{1}{2}$
Therefore, the lines 2x + y − 1 = 0 and 3x + 2y + 5 = 0 intersect.

(ii) x − y = 0 and 3x − 3y + 5 = 0

Here, $\frac{1}{3} = \frac{- 1}{- 3} \neq \frac{0}{5}$
Therefore, the lines x − y = 0 and 3x − 3y + 5 = 0 are parallel.

(iii) 3x + 2y − 4 = 0 and 6x + 4y − 8 = 0

Here, $\frac{3}{6} = \frac{2}{4} = \frac{- 4}{- 8}$
Therefore, the lines 3x + 2y − 4 = 0 and 6x + 4y − 8 = 0 are coincident.

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