# Parametric form

Calculate the distance of point A [2,1] from the line p:
X = -1 + 3 t
Y = 5-4 t
Line p has a parametric form of the line equation. ..

Correct result:

x =  0.2

#### Solution:

$p: \ \\ x=-1+3t \ \\ y=5-4t \ \\ \ \\ A[2,1]=A[m,n] \ \\ m=2 \ \\ n=1 \ \\ \ \\ q \perp p: ax+by+c=0 \ \\ a=4 \ \\ b=3 \ \\ q: 4x+3y+c=0 \ \\ \ \\ \ \\ 4 \cdot \ (-1)+3 \cdot \ 5+c=0 \ \\ \ \\ c=-11 \ \\ \ \\ c=-11 \ \\ x=\dfrac{ |a \cdot \ m+a \cdot \ n+c| }{ \sqrt{ a^2+b^2 } }=\dfrac{ |4 \cdot \ 2+4 \cdot \ 1+(-11)| }{ \sqrt{ 4^2+3^2 } }=\dfrac{ 1 }{ 5 }=0.2$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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