+14915 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.
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