Find the point of intersection of the following pairs of lines:
(i) 2x − y + 3 = 0 and x + y − 5 = 0
(ii) bx + ay = ab and ax + by = ab.
(iii) $y = m_{1} x + \frac{a}{m_{1}} a n d y = m_{2} x + \frac{a}{m_{2}} .$

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Solution

(i)
The equations of the lines are as follows:

2x − y + 3 = 0 ... (1)

x + y − 5 = 0 ... (2)

Solving (1) and (2) using cross-multiplication method:

$\frac{x}{5 - 3} = \frac{y}{3 + 10} = \frac{1}{2 + 1} \Rightarrow \frac{x}{2} = \frac{y}{13} = \frac{1}{3} \Rightarrow x = \frac{2}{3} and y = \frac{13}{3}$

Hence, the point of intersection is $(\frac{2}{3}, \frac{13}{3})$ .

(ii)
The equations of the lines are as follows:

bx + ay = ab

$\Rightarrow$ bx + ay − ab = 0 ... (1)

ax + by = ab

$\Rightarrow$ ax + by − ab = 0 ... (2)

Solving (1) and (2) using cross-multiplication method:

$\frac{x}{- a^{2} b + a b^{2}} = \frac{y}{- a^{2} b + a b^{2}} = \frac{1}{b^{2} - a^{2}} \Rightarrow \frac{x}{a b (b - a)} = \frac{y}{a b (b - a)} = \frac{1}{(a + b) (b - a)} \Rightarrow x = \frac{a b}{a + b} and y = \frac{a b}{a + b}$

Hence, the point of intersection is $(\frac{a b}{a + b}, \frac{a b}{a + b})$ .
(iii)
The equations of the lines are $y = m_{1} x + \frac{a}{m_{1}} a n d y = m_{2} x + \frac{a}{m_{2}} .$
Thus, we have:
$m_{1} x - y + \frac{a}{m_{1}} = 0$ ... (1)

$m_{2} x - y + \frac{a}{m_{2}} = 0$ ... (2)

Solving (1) and (2) using cross-multiplication method:

$\frac{x}{- \frac{a}{m_{2}} + \frac{a}{m_{1}}} = \frac{y}{\frac{a m_{2}}{m_{1}} - \frac{a m_{1}}{m_{2}}} = \frac{1}{- m_{1} + m_{2}} \Rightarrow x = \frac{\frac{- a}{m_{2}} + \frac{a}{m_{1}}}{- m_{1} + m_{2}}, y = \frac{\frac{a m_{2}}{m_{1}} - \frac{a m_{1}}{m_{2}}}{- m_{1} + m_{2}} \Rightarrow x = \frac{a}{m_{1} m_{2}} and y = \frac{a (m_{1} + m_{2})}{m_{1} m_{2}}$

Hence, the point of intersection is $(\frac{a}{m_{1} m_{2}}, \frac{a (m_{1} + m_{2})}{m_{1} m_{2}}) or (\frac{a}{m_{1} m_{2}}, a (\frac{1}{m^{1}} + \frac{1}{m_{2}}))$ .

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Find the point of intersection of the following pairs of lines :

(i) 2 x - y + 3 = 0 and x + y - 5 = 0 (ii) bx + ay = ab and ax + by = ab. (iii) $y = m_{1} x + \frac{a}{m_{1}} a n d y = m_{2} x + \frac{a}{m_{2}} .$

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