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Chapter 6: Problem 52

Find the distance between the point and the line. \((0,0) \quad 2 x-y=4\)

Short Answer

Expert verified
The distance between the point (0,0) and the line \(2x - y = 4\) is \(\frac{4\sqrt{5}}{5}\).

Step by step solution

01

Identify A, B, and C in the Line Equation

Rewrite the equation \(2x - y = 4\) in the form \(Ax + By + C = 0\). This gives \(2x - y - 4 = 0\). So, A = 2, B = -1, and C = -4.
02

Identify the Given Point

The given point is (0,0). So, x1 = 0 and y1 = 0.
03

Substitute A, B, C ,x1, and y1 into Distance Formula

Substitute A = 2, B = -1, C = -4, x1 = 0, and y1 = 0 into the formula \(d = \frac{|Ax1 + By1 + C|}{\sqrt{A^2 + B^2}}\). This gives \(d = \frac{|2*0 -1*0 - 4|}{\sqrt{2^2 + (-1)^2}}\) which simplifies to \(d = \frac{|- 4|}{\sqrt{5}}\)
04

Calculation

The calculation gives \(d = \frac{4}{\sqrt{5}} = \frac{4\sqrt{5}}{5}\)

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