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

 Nov 13, 2021
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Find the distance between \(P_1(3,-4)\) and the line 4x+8y+7=0.

 

Hello Guest!

 

\(f_1(x)= -\frac{1}{2}x-\frac{7}{8}\\ P_1(3,-4)\)

a. We find the line that goes through \(P_1\) and is perpendicular to the line \(y=-\frac{1}{2}x-\frac{7}{8}\) .

b. We calculate the distance from \(P_1\) to the intersection of the lines.

\(m_1=-0.5\\ m_2=-\frac{1}{m_1}=2\)

\(f_2(x)=m_2(x-x_1)+y_1=2(x-3)-4\\ \color{blue}f_2(x)=2x-10 \)

\(-\frac{1}{2}x-\frac{7}{8}=2x-10\\ 2.5x=\frac{73}{8}\\ x_2=3.65\\ y_2=2\cdot 3.65-10\\ y_2=-2.7\)

\(P_2(3.65,-2.7)\)

\(d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\)

\(d=\sqrt{(-2.7-(-4))^2+(3.65-3)^2}\)

\(d=1.453\)

Te distance between \(P_1(3,-4)\) and the line 4x+8y+7=0 is 1.453.

laugh  !

 Nov 13, 2021
edited by asinus  Nov 13, 2021
edited by asinus  Nov 13, 2021

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