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# Coordinates

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

Nov 13, 2021

#2
+12710
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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.

!

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