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⊥p:ax+by+c=0 a=4 b=3 q:4x+3y+c=0   4⋅ (−1)+3⋅ 5+c=0  c=−11  c=−11   x=∣a⋅ m+a⋅ n+c∣a2+b2=∣4⋅ 2+4⋅ 1+(−11)∣42+32=15=0.2p: \ \\ 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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