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Question Number 38996 by Rio Mike last updated on 01/Jul/18

a line with equation   y = 2x + 5.the point (1,7)  lie on this line.find the   distance between this line   and the line joining   (2,3) and (6,7).

$${a}\:{line}\:{with}\:{equation} \\ $$$$\:{y}\:=\:\mathrm{2}{x}\:+\:\mathrm{5}.{the}\:{point}\:\left(\mathrm{1},\mathrm{7}\right) \\ $$$${lie}\:{on}\:{this}\:{line}.{find}\:{the}\: \\ $$$${distance}\:{between}\:{this}\:{line}\: \\ $$$${and}\:{the}\:{line}\:{joining}\: \\ $$$$\left(\mathrm{2},\mathrm{3}\right)\:{and}\:\left(\mathrm{6},\mathrm{7}\right). \\ $$

Commented by math khazana by abdo last updated on 02/Jul/18

let A(2,3) and B(6,7)  M(x,y) ∈(AB) ⇔ det( AM^→ ,AB^→ )=0 ⇔   determinant (((x−2        4)),((y−3          4)))=0 ⇔ 4(x−2)−4(y−3)=0 ⇒  x−2 −y +3=0 ⇒ x−y +1 =0  d(D,D^′ ) = d(E, D) = ((∣1−7+1∣)/(√(1^(2 )  +(−1)^2 ))) = (5/(√2)) .

$${let}\:{A}\left(\mathrm{2},\mathrm{3}\right)\:{and}\:{B}\left(\mathrm{6},\mathrm{7}\right) \\ $$$${M}\left({x},{y}\right)\:\in\left({AB}\right)\:\Leftrightarrow\:{det}\left(\:{A}\overset{\rightarrow} {{M}},{A}\overset{\rightarrow} {{B}}\right)=\mathrm{0}\:\Leftrightarrow \\ $$$$\begin{vmatrix}{{x}−\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{4}}\\{{y}−\mathrm{3}\:\:\:\:\:\:\:\:\:\:\mathrm{4}}\end{vmatrix}=\mathrm{0}\:\Leftrightarrow\:\mathrm{4}\left({x}−\mathrm{2}\right)−\mathrm{4}\left({y}−\mathrm{3}\right)=\mathrm{0}\:\Rightarrow \\ $$$${x}−\mathrm{2}\:−{y}\:+\mathrm{3}=\mathrm{0}\:\Rightarrow\:{x}−{y}\:+\mathrm{1}\:=\mathrm{0} \\ $$$${d}\left({D},{D}^{'} \right)\:=\:{d}\left({E},\:{D}\right)\:=\:\frac{\mid\mathrm{1}−\mathrm{7}+\mathrm{1}\mid}{\sqrt{\mathrm{1}^{\mathrm{2}\:} \:+\left(−\mathrm{1}\right)^{\mathrm{2}} }}\:=\:\frac{\mathrm{5}}{\sqrt{\mathrm{2}}}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

the eq^n  of line joining (2,3) and(6,7) is  y−3=(((7−3)/(6−2)))(x−2)  y−3=x−2  x−y+1=0  distance =∣((1−7+1)/(√(1^2 +(−1)^2 )))∣  =(5/(√2))

$${the}\:{eq}^{{n}} \:{of}\:{line}\:{joining}\:\left(\mathrm{2},\mathrm{3}\right)\:{and}\left(\mathrm{6},\mathrm{7}\right)\:{is} \\ $$$${y}−\mathrm{3}=\left(\frac{\mathrm{7}−\mathrm{3}}{\mathrm{6}−\mathrm{2}}\right)\left({x}−\mathrm{2}\right) \\ $$$${y}−\mathrm{3}={x}−\mathrm{2} \\ $$$${x}−{y}+\mathrm{1}=\mathrm{0} \\ $$$${distance}\:=\mid\frac{\mathrm{1}−\mathrm{7}+\mathrm{1}}{\sqrt{\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }}\mid \\ $$$$=\frac{\mathrm{5}}{\sqrt{\mathrm{2}}} \\ $$

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