Question Number 38996 by Rio Mike last updated on 01/Jul/18
$${a}\:{line}\:{with}\:{equation} \\ $$$$\:{y}\:=\:\mathrm{2}{x}\:+\:\mathrm{5}.{the}\:{point}\:\left(\mathrm{1},\mathrm{7}\right) \\ $$$${lie}\:{on}\:{this}\:{line}.{find}\:{the}\: \\ $$$${distance}\:{between}\:{this}\:{line}\: \\ $$$${and}\:{the}\:{line}\:{joining}\: \\ $$$$\left(\mathrm{2},\mathrm{3}\right)\:{and}\:\left(\mathrm{6},\mathrm{7}\right). \\ $$
Commented by math khazana by abdo last updated on 02/Jul/18
$${let}\:{A}\left(\mathrm{2},\mathrm{3}\right)\:{and}\:{B}\left(\mathrm{6},\mathrm{7}\right) \\ $$$${M}\left({x},{y}\right)\:\in\left({AB}\right)\:\Leftrightarrow\:{det}\left(\:{A}\overset{\rightarrow} {{M}},{A}\overset{\rightarrow} {{B}}\right)=\mathrm{0}\:\Leftrightarrow \\ $$$$\begin{vmatrix}{{x}−\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{4}}\\{{y}−\mathrm{3}\:\:\:\:\:\:\:\:\:\:\mathrm{4}}\end{vmatrix}=\mathrm{0}\:\Leftrightarrow\:\mathrm{4}\left({x}−\mathrm{2}\right)−\mathrm{4}\left({y}−\mathrm{3}\right)=\mathrm{0}\:\Rightarrow \\ $$$${x}−\mathrm{2}\:−{y}\:+\mathrm{3}=\mathrm{0}\:\Rightarrow\:{x}−{y}\:+\mathrm{1}\:=\mathrm{0} \\ $$$${d}\left({D},{D}^{'} \right)\:=\:{d}\left({E},\:{D}\right)\:=\:\frac{\mid\mathrm{1}−\mathrm{7}+\mathrm{1}\mid}{\:\sqrt{\mathrm{1}^{\mathrm{2}\:} \:+\left(−\mathrm{1}\right)^{\mathrm{2}} }}\:=\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{2}}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18
$${the}\:{eq}^{{n}} \:{of}\:{line}\:{joining}\:\left(\mathrm{2},\mathrm{3}\right)\:{and}\left(\mathrm{6},\mathrm{7}\right)\:{is} \\ $$$${y}−\mathrm{3}=\left(\frac{\mathrm{7}−\mathrm{3}}{\mathrm{6}−\mathrm{2}}\right)\left({x}−\mathrm{2}\right) \\ $$$${y}−\mathrm{3}={x}−\mathrm{2} \\ $$$${x}−{y}+\mathrm{1}=\mathrm{0} \\ $$$${distance}\:=\mid\frac{\mathrm{1}−\mathrm{7}+\mathrm{1}}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }}\mid \\ $$$$=\frac{\mathrm{5}}{\:\sqrt{\mathrm{2}}} \\ $$