Question

calculate the distance between each line and the point. round your answer to one decimal place.
A) y = 4x - 2, (-3, 3)
B) y = -x + 5, (-1, -2)
How do I do these?? The answer for A) is about 4.1 units (but I got 5.8 ???) and for B) its about 5.7 units.

Answer 1

                                     Solving Equation A

Given the equation A

$y=4x-2$

and point (-3, 3)

The distance from the point (x₀, y₀) to the line Ax+By+C=0 is given by

$d=\frac{\left|Ax_0+By_0+C\right|}{\sqrt{A^2+B^2}}$

To apply this formula, we first need to express the line in standard form

$y=4x-2$

$4x-y-2=0$

After substituting: A=4, B=-1, C=2, x₀=-3, y₀=3

$d=\frac{\left|\left(4\right)\left(-3\right)+\left(-1\right)\left(3\right)+\left(-1\right)\right|}{\sqrt{4^2+\left(-1\right)^2}}$

$d=\frac{\left|-12-3-2\right|}{\sqrt{4^2+\left(-1\right)^2}}$

$d=\frac{17}{\sqrt{4^2+\left(-1\right)^2}}$

$d=\sqrt{17}$

$d=4.1$ units

Therefore, the distance between the line y = 4x - 2 and the point (-3, 3) will be:

• d=4.1 units

                                             Solving Equation B

Given the equation B

$y = -x + 5$

and point (-1, -2)

The distance from the point (x₀, y₀) to the line Ax+By+C=0 is given by

$d=\frac{\left|Ax_0+By_0+C\right|}{\sqrt{A^2+B^2}}$

To apply this formula, we first need to express the line in standard form

$y = -x + 5$

$x+y-5=0$

After substituting: A=4, B=1, C=-5, x₀=-1, y₀=-2

$d=\frac{\left|\left(1\right)\left(-1\right)+\left(1\right)\left(-2\right)+\left(-5\right)\right|}{\sqrt{1^2+\left(1\right)^2}}$

$d=\frac{\left|-1-2-5\right|}{\sqrt{1+1}}$

$d=\frac{\left|-8\right|}{\sqrt{2}}$

$d=\frac{8}{\sqrt{2}}$

$d=4\sqrt{2}$

$d=5.7$ units

Therefore, the distance between the line y = -x + 5 and the point (-1, -2) will be:

• d=5.7 units