Direction vector

The line p is given by the point P [- 0,5; 1] and the direction vector s = (1,5; - 3) determines:
A) value of parameter t for points X [- 1,5; 3], Y [1; - 2] lines p
B) whether the points R [0,5; - 1], S [1,5; 3] lies on the line p
C) parametric equations of the line m || p if the line m passes through the origin of the coordinate system
D) parametric equations of the line n _ | _ p, if the line n passes through the point N [- 2; 4]

Result

t₁ =  -0.6667
t₂ =  1
R=

Yes

No

t₂ =S=

Yes

No

t₂ =mx = 1.5t
my = -3t
nx = -2+3t
ny = 4-1.5t

Step-by-step explanation:

$$\begin{gathered} P_x=-0.5;P_y=1 \\ {s}_x=1.5;s_y=-3 \\ X[-1,5;3] \\ -1.5=P_x+t_1\cdot s_x \\ {t}_1=(-1.5-P_x)/s_x=(-1.5-(-0.5))/1.5=-0.6667 \end{gathered}$$

$$\begin{gathered} Y[1;-2] \\ 1=P_x+t_1\cdot s_x \\ {t}_2=(1-P_x)/s_x=(1-(-0.5))/1.5=1 \end{gathered}$$

$$\begin{gathered} R[0,5;-1] \\ {p}_1=(0.5-P_x)/s_x=(0.5-(-0.5))/1.5=\frac{2}{3}\doteq 0.6667 \\ {p}_2=(-1-P_y)/s_y=(-1-1)/(-3)=\frac{2}{3}\doteq 0.6667 \\ {p}_1=p_2 \\ R=1 \end{gathered}$$

$$\begin{gathered} S[1,5;3] \\ {p}_3=(1.5-P_x)/s_x=(1.5-(-0.5))/1.5=\frac{4}{3}=1\frac{1}{3}\doteq 1.3333 \\ {p}_4=(3-P_y)/s_y=(3-1)/(-3)=-\frac{2}{3}\doteq -0.6667 \\ {p}_3\ne p_4 \\ S=0 \end{gathered}$$

$mx=m_x=1.5t$

$my=m_y=-3t$

$$\begin{gathered} (1.5;-3)\perp (3;1.5) \\ nx=n_x=-2+3t \end{gathered}$$

$ny=n_y=4-1.5t$

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