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Find the distance between (3,4) and the line 4x+8y+7=0.

 Nov 23, 2020
 #1
avatar+11086 
+2

Find the distance between \(P_1\)(3,4) and the line 4x+8y+7=0.
 

Hello Guest!

 

\(f_1(x)=y=-\frac{1}{2}x-\frac{7}{8}\)

\(m_1=-\frac{1}{2}\\ {\color{blue}m_2=-\frac{1}{m_1}}=-\frac{1}{-\frac{1}{2}}\\ \color{blue}m_2=2\)

\(\color{blue}f_2=y=m_2(x-x_p)+y_p\\ y=2(x-3)+4\\ y=2x-2\)

\(-\frac{1}{2}x-\frac{7}{8}=2x-2\\ \frac{5}{2}x=\frac{9}{8}\\ \color{blue}x_2=0.45\\ \color{blue}P_2\\ \color{blue}y_2=-1.1 \)

\(\color{blue}D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\\ D=\sqrt{(-1.1-4)^2+(0.45-3)^2}\\\)

\(D=5.702\)

The distance between \(P_1\)(3,4) and the line 4x+8y+7=0 is D = 5.702.

laugh  !


 

 Nov 24, 2020
 #2
avatar+116126 
+1

Thx, Asinus....here's another way

 

Distance  between  (3,4)  and  line  4x + 8y  + 7  = 0

 

l 4(3)  + 8(4)  + 7 l                   51               51                51 sqrt (5)

________________ =         ______  =  ________  =  _________  ≈   5.70 units

sqrt ( 4^2  + 8^2)                 sqrt (80)      4sqrt (5)              20

 

 

cool cool cool

 Nov 24, 2020

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