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Find the distance between  (3,4) and the line 4x+3y+7=2x-y+5.

 Nov 11, 2021
 #1
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12/sqrt(5)

 Nov 12, 2021
 #2
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Find the distance between  \(P_1 \)(3,4) and the line 4x+3y+7=2x-y+5.

 

Hello Guest!

 

\(4x+3y+7=2x-y+5\\ 4y=-2x-2\\ \color{blue}y=-0.5x+0.5\)

\(m_1=-0.5\\ m_2=2\\ y=m_2(x-x_1)+y_1\\ y=2(x-3)+4\\ \color{blue}y=2x-2\)

\(-0.5x+0.5=2x-2\\ 2.5x=2.5\)

\(x_2=1\\ y_2=0\)

\(d=\sqrt{4^2+2^2}=\sqrt{4\cdot 5}\)

\(d=2\cdot \sqrt{5}\)

\({\color{blue}The\ distance}\ between\ P_1 (3,4)\ and\ the\ line\ 4x+3y+7=2x-y+5\ \color{blue}is\ 2\cdot \sqrt{5}.\)

laugh  !

 Nov 12, 2021

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