How to use vectors and dot products to show that $y=5-x$ and $y=x$ are orthogonal I have two lines in $\Bbb R^2$. One is described by the equation
$$y=5-x$$
while the other line is described by
$$y=x$$
From the fact that one line's slope is the negative of the other, it is directly obvious that they are perpendicular lines. However, I am hoping to show that they are perpendicular by writing both equations as vectors and then showing the dot product is zero. 
If I let the first line be the vector $v$, then the line is described by
$$v=(c,5-c):c\in\Bbb R$$
while the second is
$$u=(d,d):d\in\Bbb R$$
However, when I try to take the dot product, I get
$$\langle v,u\rangle=cd+d(5-c)=5d\ne0$$
Does anyone know what I am misunderstanding here? Thanks.
 A: One way to show that two lines $Ax + By = C $ and $A'x + B'y = C'$ are orthogonal is to show that their normals are orthogonal. That is, since $\mathbf{n_1} = (A,B)$ and $\mathbf{n_2} = (A',B')$ then, the two lines are orthogonal means that 
$$ \mathbf{n_1} \cdot \mathbf{ n_2} = 0 $$
For your specific problem, we know $\mathbf{n_1} = (1,1)$, the normal corresponding to the line $y=5-x$ and $\mathbf{n_2} = (-1,1)$, the normal corresponding to the other line. We have 
$$ \mathbf{n_1} \cdot \mathbf{n_2} = 1(-1)+1\cdot 1 = 0 $$
Therefore, the lines are orthogonal.
A: $v = (c,5-c)$ is a vector from the origin to a point on the line $y = 5-x$. Such a vector is not parallel to that line (nor perpendicular except for one particular choice of $c$).
What you want is to take vectors parallel to those lines and take the dot product of those (using normal lines also works). One way to do this is to take the difference of two points on a given line.
Pick $c$ and $d$ as $0$ and $1$:
$$v = (1,5-1) - (0, 5-0) = (1,4) - (0,5) = (1,-1)$$
$$u = (1,1) - (0,0) = (1,1)$$
Then
$$\langle u, v\rangle = \langle (1,-1), (1,1) \rangle = 1 \cdot 1 + 1 \cdot(-1) = 0$$
A: If we have a line described by equation $ax+by+c=0$ than this line is orthogonal to vector $(a, b)$. It's easy to understand: take some point on this line, move along the line to some other point, the displacement is $(dx, dy)$, but the equation still must hold, so $a*dx + b*dy=0$.
Well, now we have that the first of your lines is orthogonal to $(1, 1)$, second one is orthogonal to $(1, -1)$. Dot product of these two vectors is 0, so the are orthogonal, so the lines are orthogonal to each other.
A: Line $y = x $ has direction vector $(1,1)$ and line $y = 5 - x$ has direction vector $(-1,1)$. Then $<(1,1),(-1,1)> = 0$ so they are orthogonal.
The vector equation of a line is given by:
 $$r = p + tv, t \in \mathbb{R}$$
With $p$ a point of the line and $v$ the direction vector. To check orthogonality you have to check orthogonality between the direction vectors.
A: $$l_1:(c,5-c),c\in\Bbb R$$
$$l_2:(d,d),d\in\Bbb R$$
The representations of the lines are correct. What went into the dot product, however, were not the correct direction vectors (coefficients of $c$ or $d$ in each component of the equations for $l_1$ and $l_2$), but rather the vectors themselves.
The direction vector of $l_1$ is $(1,-1)$; that of $l_2$ is $(1,1)$. The dot product of these yields zero as expected:
$$(1,-1)\cdot(1,1)=0$$
