# 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. ..

Result

x =  0.2

#### Solution:

$p: \ \\ x = -1+3t \ \\ y = 5-4t \ \\ \ \\ A[2,1] = A[m,n] \ \\ m = 2 \ \\ n = 1 \ \\ \ \\ q \perp p: ax+by+c = 0 \ \\ a = 4 \ \\ b = 3 \ \\ q: 4x+3y+c = 0 \ \\ \ \\ \ \\ 4 \cdot \ (-1)+3 \cdot \ 5+c = 0 \ \\ \ \\ c = -11 \ \\ \ \\ c = -11 \ \\ \ \\ \ \\ x = \dfrac{ |a \cdot \ m+a \cdot \ n+c| }{ \sqrt{ a^2+b^2 } } = \dfrac{ |4 \cdot \ 2+4 \cdot \ 1+(-11)| }{ \sqrt{ 4^2+3^2 } } = \dfrac{ 1 }{ 5 } = 0.2$

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