Question Number 47821 by mondodotto@gmail.com last updated on 15/Nov/18

$$\boldsymbol{\mathrm{L}}_{\mathrm{1}} \boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{L}}_{\mathrm{2}} \:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{lines}}\:\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{intersects}} \\ $$$$\boldsymbol{\mathrm{each}}\:\boldsymbol{\mathrm{other}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{right}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{point}}\:\left(\mathrm{1},\mathrm{3}\right),\boldsymbol{\mathrm{L}}_{\mathrm{1}} \:\boldsymbol{\mathrm{cuts}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{axis}}\:\boldsymbol{\mathrm{at}} \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{point}}\:\left(\mathrm{0},\mathrm{2}\right)\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{equations}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{L}}_{\mathrm{1}} \boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{L}}_{\mathrm{2}} \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Nov/18
![eqn L_1 y−3=((2−3)/(0−1))(x−1) y−3=(x−1) y−x−2=0 slope L_1 =1 slope L_2 =−1 [m_1 m_2 =−1] eqn L_2 y−3=(−1)(x−1) y−3=−x+1 y+x−4=0](https://www.tinkutara.com/question/Q47826.png)
$${eqn}\:{L}_{\mathrm{1}} \:\:{y}−\mathrm{3}=\frac{\mathrm{2}−\mathrm{3}}{\mathrm{0}−\mathrm{1}}\left({x}−\mathrm{1}\right) \\ $$$${y}−\mathrm{3}=\left({x}−\mathrm{1}\right)\:\:\:{y}−{x}−\mathrm{2}=\mathrm{0} \\ $$$${slope}\:{L}_{\mathrm{1}} =\mathrm{1}\:\:\:{slope}\:{L}_{\mathrm{2}} =−\mathrm{1}\:\:\left[{m}_{\mathrm{1}} {m}_{\mathrm{2}} =−\mathrm{1}\right] \\ $$$${eqn}\:{L}_{\mathrm{2}} \:{y}−\mathrm{3}=\left(−\mathrm{1}\right)\left({x}−\mathrm{1}\right) \\ $$$${y}−\mathrm{3}=−{x}+\mathrm{1} \\ $$$${y}+{x}−\mathrm{4}=\mathrm{0} \\ $$$$\:\: \\ $$