Solving Equation A
Given the equation A
![y=4x-2](https://tex.z-dn.net/?f=y%3D4x-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}}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B%5Cleft%7CAx_0%2BBy_0%2BC%5Cright%7C%7D%7B%5Csqrt%7BA%5E2%2BB%5E2%7D%7D)
To apply this formula, we first need to express the line in standard form
![y=4x-2](https://tex.z-dn.net/?f=y%3D4x-2)
![4x-y-2=0](https://tex.z-dn.net/?f=4x-y-2%3D0)
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}}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B%5Cleft%7C%5Cleft%284%5Cright%29%5Cleft%28-3%5Cright%29%2B%5Cleft%28-1%5Cright%29%5Cleft%283%5Cright%29%2B%5Cleft%28-1%5Cright%29%5Cright%7C%7D%7B%5Csqrt%7B4%5E2%2B%5Cleft%28-1%5Cright%29%5E2%7D%7D)
![d=\frac{\left|-12-3-2\right|}{\sqrt{4^2+\left(-1\right)^2}}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B%5Cleft%7C-12-3-2%5Cright%7C%7D%7B%5Csqrt%7B4%5E2%2B%5Cleft%28-1%5Cright%29%5E2%7D%7D)
![d=\frac{17}{\sqrt{4^2+\left(-1\right)^2}}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B17%7D%7B%5Csqrt%7B4%5E2%2B%5Cleft%28-1%5Cright%29%5E2%7D%7D)
![d=\sqrt{17}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B17%7D)
units
Therefore, the distance between the line y = 4x - 2 and the point (-3, 3) will be:
Solving Equation B
Given the equation B
![y = -x + 5](https://tex.z-dn.net/?f=y%20%3D%20-x%20%2B%205)
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}}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B%5Cleft%7CAx_0%2BBy_0%2BC%5Cright%7C%7D%7B%5Csqrt%7BA%5E2%2BB%5E2%7D%7D)
To apply this formula, we first need to express the line in standard form
![y = -x + 5](https://tex.z-dn.net/?f=y%20%3D%20-x%20%2B%205)
![x+y-5=0](https://tex.z-dn.net/?f=x%2By-5%3D0)
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}}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B%5Cleft%7C%5Cleft%281%5Cright%29%5Cleft%28-1%5Cright%29%2B%5Cleft%281%5Cright%29%5Cleft%28-2%5Cright%29%2B%5Cleft%28-5%5Cright%29%5Cright%7C%7D%7B%5Csqrt%7B1%5E2%2B%5Cleft%281%5Cright%29%5E2%7D%7D)
![d=\frac{\left|-1-2-5\right|}{\sqrt{1+1}}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B%5Cleft%7C-1-2-5%5Cright%7C%7D%7B%5Csqrt%7B1%2B1%7D%7D)
![d=\frac{\left|-8\right|}{\sqrt{2}}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B%5Cleft%7C-8%5Cright%7C%7D%7B%5Csqrt%7B2%7D%7D)
![d=\frac{8}{\sqrt{2}}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B8%7D%7B%5Csqrt%7B2%7D%7D)
![d=4\sqrt{2}](https://tex.z-dn.net/?f=d%3D4%5Csqrt%7B2%7D)
units
Therefore, the distance between the line y = -x + 5 and the point (-1, -2) will be: