Equation of a line that goes through $A(-3,-7,-5)$ and $B(2,3,0)$ and find $C(x, -1, z)$ on the same line Problem: Find the equation of a line that passes through $A(-3,-7,-5)$ and $B(2,3,0)$ and find $C(x, -1, z)$ on the same line.
I have completely forgotten how to solve this and I've been reading about how to solve this and it's super confusing. How do I find the equation of this line? I've found the parametric, vector and symmetric forms for the equation, but I don't have a clue how to actually solve it. 
 A: vector between A and B is
$$\overrightarrow{AB}=5i+10j+5k$$
so the parametric equation of line is
$$\frac{x-x_0}{v_x}=t$$
$$\frac{x+3}{5}=t$$
$$\frac{y-y_0}{v_y}=t$$
$$\frac{y+7}{10}=t$$
$$\frac{z-z_0}{v_z}=t$$
$$\frac{z+5}{5}=t$$
now use the point C(x,-1,z)
at $y=-1$
$$\frac{-1+7}{10}=t$$
$$t=0.6$$
then you can use the value of $t$ to find the value of $x$ and $z$
$$\frac{x+3}{5}=0.6$$
$$x=0$$
$$\frac{z+5}{5}=0.6$$
$$z=-2$$
A: By definition, if $P(x,y,z)$ belongs to the line $AB$, then $\exists \lambda \in \mathbb{R}$ such that $\vec{AP}=\lambda \vec{AB}$.
In your case that means the following set of equations for $C(x,-1,z)$:
$x+3=5\lambda$, 
$y+7=6=10 \lambda$ and 
$z+5=5\lambda$
We have $\lambda=\dfrac{3}{5}$ hence
$C=(0,-1,-2)$...
A: The equation of a line that passes through $(x_1,y_1,z_1),(x_2,y_2,z_2)$ can be expressed as
$$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}.$$
Then, set $y=-1$ to find $x,z$.
A: In general, the equation of the line passing trough the points $\color{blue}{(x_{1}, y_{1}, z_{1})}$ & $\color{blue}{(x_{2}, y_{2}, z_{2})}$ is given as
$$\color{blue}{\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}=k \space \text{(any arbitrary constant)}}$$
As per your question, the equation of the line passing through the points $\color{blue}{A(-3, -7, -5)}$ & $\color{blue}{B(2, 3, 0)}$ s given as  $$\frac{x-(-3)}{2-(-3)}=\frac{y-(-7)}{3-(-7)}=\frac{z-(-5)}{0-(-5)} $$
$$\frac{x+3}{5}=\frac{y+7}{10}=\frac{z+5}{5} $$ $$\implies \color{red}{\frac{x+3}{1}=\frac{y+7}{2}=\frac{z+5}{1}=k }$$
Now, the consider a general point on the above line as $\color{blue}{((k-3), (2k-7), (k-5) )}$ & compare its y-coordinate with that of the given point $\color{blue}{C(x, -1, z)}$ we get
$$2k-7=-1 $$$$\implies k=\frac{6}{2}=3$$
Hence, the point $\color{blue}{C(x, -1, z)}$ on the above line is obtained by substituting $k=3$ in the general point as $\color{blue}{((3-3), (2\times 3-7), (3-5) )}\equiv \color{red}{(0, -1, -2)}$
