Question

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) $y=m_{1}x+\frac{a}{m_1}andy=m_{2}x+\frac{a}{m_2}.$

Answer 1

$\left(i\right)2x-y+3=0\Rightarrow y=2x+3$

$\text{Putting this values in the second equation, we get}$

$x+y-5=0$

$x+(2x+3)-5=0$

$x+2x+3-5=0$

$3x-2=0$

$x=\frac{2}{3}$

$\text{Putting this value in the first equation, we get}$

$\Rightarrow y=2x+3=\frac{2\times 2}{3}+3=\frac{4}{3}+3=\frac{13}{3}$

$\therefore \text{Point of intersection is}(\frac{2}{3},\frac{13}{3})$

$\left(ii\right)bx+ay=ab\Rightarrow x=\frac{ab-ay}{b}$

$\text{Putting this value in the second equation, we get}$

$ax+by=ab$

$a\left(\frac{ab-ay}{b}\right)+by=ab$

$a^{2}b-a^{2}y+b^{2}y=a{b}^2$

$y(b^2-a^2)=ab(b-a)$

$y=\frac{ab(b-a)}{b^2-a^2}=\frac{ab}{b+a}$

$\text{Putting this value in the firt equation, we get}$

$\Rightarrow x=\frac{ab-\frac{a\left(ab\right)}{a+b}}{b}=\frac{ab}{a+b}$

$\therefore \text{Point is}(\frac{ab}{a+b},\frac{ab}{a+b})$

$\left(iii\right)y=m_{1}x+\frac{a}{m_1}andy=m_{2}x+\frac{a}{m_2}$

$\text{Putting value of y from one equation to another}$

$m_{1}x+\frac{a}{m_1}=m_{2}x+\frac{a}{m_2}$

$x(m_1-m_2)=\frac{a}{m_2}-\frac{a}{m_1}=a\left(\frac{m_1-m_2}{m_1{m}_2}\right)$

$\Rightarrow x=\frac{a}{m_1{m}_2}$

$\Rightarrow y=m_{1}x+\frac{a}{m_1}$

$=m_1\left(\frac{a}{m_1{m}_2}\right)+\frac{a}{m_1}$

$=\frac{a}{m_2}+\frac{a}{m_1}$

$=a\left(\frac{m_1+m_2}{m_1{m}_2}\right)$

$\text{Thus, the point of intersection is}$

$(\frac{a}{m_1{m}_2},a(\frac{1}{m_1}+\frac{1}{m_2}))$