Question Number 133885 by bemath last updated on 25/Feb/21
$$\:\mathrm{Consider}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{two} \\ $$$$\mathrm{intersecting}\:\mathrm{straight}\:\mathrm{lines} \\ $$$$\begin{cases}{{ax}+{by}+{c}=\mathrm{0}}\\{{a}_{\mathrm{1}} {x}+{b}_{\mathrm{1}} {y}+{c}_{\mathrm{1}} =\mathrm{0}}\end{cases} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\mathrm{passing}\:\mathrm{through}\:\mathrm{a}\:\mathrm{given}\:\mathrm{point} \\ $$$$\left(\mathrm{x}_{\mathrm{0}} ,\mathrm{y}_{\mathrm{0}} \right)\:\mathrm{and}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{point} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{straight}\:\mathrm{lines}. \\ $$
Commented by mathocean1 last updated on 25/Feb/21
$${Hello} \\ $$$${is}\:{it}\:{a}_{\mathrm{1}} {x}+{b}_{\mathrm{1}} {y}+{c}_{\mathrm{1}} =\mathrm{0}\:\:{the}\:{second}\:{eq}? \\ $$
Commented by bemath last updated on 25/Feb/21
$$\mathrm{yes} \\ $$
Answered by EDWIN88 last updated on 25/Feb/21
$$\mathrm{The}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{specified}\:\mathrm{by}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\lambda\left({ax}+{by}+{c}\right)+\mu\left({a}_{\mathrm{1}} {x}+{b}_{\mathrm{1}} {y}+{c}_{\mathrm{1}} \right)=\mathrm{0} \\ $$$${passes}\:{through}\:{the}\:{point}\:{of}\:{intersection} \\ $$$${of}\:{the}\:{given}\:{straight}\:{lines}.\:\mathrm{We}\:\mathrm{required}\:\mathrm{that} \\ $$$$\mathrm{it}\:\mathrm{should}\:\mathrm{pass}\:\mathrm{also}\:\mathrm{through}\:\left(\mathrm{x}_{\mathrm{0}} ,\mathrm{y}_{\mathrm{0}} \right). \\ $$$$\mathrm{Then}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{desired}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\mathrm{will}\:\mathrm{be}\:\left({ax}+{by}+{c}\right)\left({a}_{\mathrm{1}} {x}_{\mathrm{0}} +{b}_{\mathrm{1}} {y}_{\mathrm{0}} +{c}\right)−\left({a}_{\mathrm{1}} {x}+{b}_{\mathrm{1}} {y}+{c}_{\mathrm{1}} \right)\left({ax}_{\mathrm{0}} +{by}_{\mathrm{0}} +{c}\right)=\:\mathrm{0}\: \\ $$