Recall the following equations defining a line in the plane. These are equations in the coordinate variables \(x,y\), that is, the line \(\ell\) is the collection of points \(P(x,y)\) the coordinates of which satisfy the equation.

The *standard form* equation is \[
ax+by=c.
\] For this to describe a line, we need \(a,b\) not both zero. Every line can be described by the standard form.

In case a line has slope \(m\), there are two more equations we can use. Recall that if \(P(a_1,b_1)\) and \(Q(a_2,b_2)\) are two different points of the line, then its slope is \[ m=\frac{b_2-b_1}{a_2-a_1}. \] Note that this is defined precisely when \(a_2\ne a_1\), that is when the line is not vertical.

A line with slope \(m\) through point \(P(x_0,y_0)\) has the

*slope-point*form equation \[ y-y_0=m(x-x_0). \]If the line is not vertical, then it will intersect the \(y\)-axis at some point \(P(0,c)\). In this situation, we say that the line has \(y\)-intercept \(c\). Correspondingly, we get the

*slope-intercept*form equation \[ y=mx+c. \]A vertical line with \(x\)-intercept \(b\) has equation \[ x=b. \]

Note that all these three equations can be converted to standard form, and they cover every line in the plane.

*Exercise*. Let \(\ell\) be the line between the points \(P(1,2)\) and \(Q(-1,3)\). Write down its equation in slope-point form, and then in standard form.

Let \(P,Q,R\) be three points on a line \(\ell\). Then the vectors \(\vec{PQ},\vec{PR}\) need to be *parallel*, that is they have either the same direction or the opposite direction. We have seen that this happens precisely when there exists a scalar \(t\) such that \[
\vec{PR}=t\vec{PQ}.
\] Let's add \(\vec{OP}\) to both sides of the equation: \[
\vec{OR}=\vec{OP}+t\vec{PQ}.
\] Let \(\mathbf p=\vec{OP}\) and \(\mathbf v=\vec{PQ}\). What we have just shown is that the collection of the vectors \(\vec{OR}\) for the points \(R\) of the line \(\ell\) is exactly the collection of the vectors \[
\mathbf x(t)=\mathbf p+t\mathbf v
\] for scalars \(t\). This vector equation is called a *point-parallel* form equation of the line \(\ell\).

*Exercise*. Let \(\ell\) be the line with equation \(y=-\frac12x+3\). Write down an equation in point-parallel form.

Note that the line \(\ell\) has many point-parallel equations. You get a different one for any choice of the ordered pair of points \(P,Q\) on the line \(\ell\).

*Definition*. Any non-zero vector \(\mathbf v\) that is parallel to the line \(\ell\) is called a *direction vector* for the line \(\ell\).

These are called *parametric equations* for the line \(\ell\). The scalar \(t\) is referred to as the *parameter*. Note that just as for point-parallel form equations, there are many parametric equations for the same line.

Giving parametric equations for the line \(\ell\) is also called giving a *parametrization* for the line. The points \(R(t)=(p_1+tv_1,p_2+tv_2)\) trace out the line as the scalar \(t\) changes. If you think of \(t\) as a time parameter, then you can think of \(P\) as a starting point, and \(\mathbf v=\vec{PQ}\) as a velocity vector for this drawing process (allowing negative time values). So the fact that there are many parametrizations of the line can be rephrased as there are many ways to draw the same line with constant speed.

Let \(P,Q,R\) be three points on a line \(\ell\). Since the vectors \(\vec{PQ}\) and \(\vec{PR}\) are parallel, there exists a scalar \(t\) such that \[
\vec{PR}=t\vec{PQ}.
\] Adding \(\vec{OP}\) and rewriting the equation, we get \[
\vec{OR}=\vec{OP}+\vec{PR}=\vec{OP}+t\vec{PQ}=\vec{OP}+t(\vec{OQ}-\vec{OP})=(1-t)\vec{OP}+t\vec{OQ}.
\] We have gotten a *two-point* form equation for the line: \[
\mathbf x(t)=(1-t)\mathbf p+t\mathbf q.
\]

*Exercise*. Let \(\ell\) be the line between the points \(P(1,2)\) and \(Q(-1,3)\). Write down a two-point form equation for it.

Note that \(R\) is on the line segment \(\bar{PQ}\) precisely when \(\vec{PR}\) has the same direction as \(\vec{PQ}\), and its length is at most that of \(\vec{PQ}\). That is \(\vec{PR}=t\vec{PQ}\) for some \(0\le t\le 1\). This means that imposing \(0\le t\le1\) in the two-point form gives us a parametrization of \(\bar{PQ}\).

Let \(\mathbf n\) be a nonzero plane vector and \(P\) a point. Then there exists exactly one line \(\ell\) that goes through the point \(P\) and it is orthogonal to \(\mathbf n\). By definition, a point \(Q\) is on the line \(\ell\) precisely when the vector \(\vec{PQ}\) is orthogonal to \(\mathbf n\), that is we have \[
\mathbf n\cdot\vec{PQ}=0.
\] We can rewrite this as \[
\mathbf n\cdot(\vec{OQ}-\vec{OP})=0.
\] We have gotten a *point-normal* form equation for \(\ell\): \[
\mathbf n\cdot(\mathbf x-\mathbf p)=0.
\] Let \(\mathbf n=\left(\begin{smallmatrix}a\\b\end{smallmatrix}\right)\), \(P=P(x_0,y_0)\) and \(Q=Q(x,y)\). We get \[
a(x-x_0)+b(y-y_0)=0.
\] We can reorganize this to \[
ax+by=ax_0+by_0.
\] That is, writing out a point-normal form equation in components, we can get a standard form equation.

To find a point-normal form equation for a line, you can use the following.

*Fact*. We have \[
\left(\begin{smallmatrix}a & b \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix}-b\\a\end{smallmatrix}\right)=-ab+ab=0.
\]

*Exercise*. Let \(\ell\) be the line between the points \(P(1,2)\) and \(Q(-1,3)\). Find a point-normal form equation for it.

Note that a point-normal form equation does not give a parametrization for the line.

True to our motto of generalization, we employ the exact same methods in 3d space. The point-parallel form and two-point form equations give parametrizations of lines in space, and the point-normal form equations give equations of planes.

*Exercise*. Let \(\ell_1\) be the line between the points \(P(2,-1,4)\) and \(Q(0,3,2)\). (a) Find a point-parallel form equation, and a two-point form equation for it. (b) Let \(\ell_2\) be the line with parametric equations \(P(1-t,1+2t,4+t)\). Is \(\ell_1=\ell_2\)?

As for the point-normal form: let \(\mathbf n\) be a nonzero space vector, and \(P\) a point. The collection of points \(Q\) such that \(\vec{PQ}\) is orthogonal to \(\mathbf n\), that is the ones satisfying the point-normal form equation \[ \mathbf n\cdot\vec{PQ}=0 \] forms a plane \(\Pi\).

A plane \(\Pi\) is determined by three non-collinear (that is, not on the same line) points \(P,Q,R\). To find an equation for \(\Pi\), we need to find a normal vector \(\mathbf n\), that is a vector which is orthogonal to both \(\vec{PQ}\) and \(\vec{PR}\). The cross product supplies such a vector: we can choose \(\mathbf n=\vec{PQ}\times\vec{PR}\).

*Exercise*. Let \(\Pi\) be the plane determined by the points \(P(2,-1,4)\), \(Q(0,3,2)\), and \(R(1,0,3)\). Find a standard form equation for it.

*Definition*. Let \(P\) be a point and \(\ell\) be a line in \(\mathbf R^2\). The *distance* of \(P\) and \(\ell\) is the minimum of the distances \(|PQ|\) for the points \(Q\) of \(\ell\).

*Proposition*. Let \(\mathbf n\) be a normal vector for \(\ell\), and let \(Q\) be any point on \(\ell\). Then the distance from \(P\) to \(\ell\) is \[
\frac{|\mathbf n\cdot\vec{PQ}|}{\|\mathbf n\|}.
\]

*Exercise*. Let \(P=P(1,2)\). (a) Let \(\ell_1\) be the line with standard form equation \(2x+3y=6\). Find the distance from \(P\) to \(\ell_1\). (b) Let \(\ell_2\) be the line through the points \(Q(0,1)\) and \(R(3,2)\). Find the distance from \(P\) to \(\ell_2\).

*Definition*. Let \(P\) be a point and \(\Pi\) be a plane in \(\mathbf R^3\). The *distance* of \(P\) and \(\Pi\) is the minimum of the distances \(|PQ|\) for the points \(Q\) of \(\Pi\).

*Proposition*. Let \(\mathbf n\) be a normal vector for \(\Pi\), and let \(Q\) be any point on \(\Pi\). Then the distance from \(P\) to \(\Pi\) is \[
\frac{|\mathbf n\cdot\vec{PQ}|}{\|\mathbf n\|}.
\]

*Exercise*. Let \(\Pi\) be the plane determined by the points \(P(2,-1,4)\), \(Q(0,3,2)\), and \(R(1,0,3)\). Find its distance from the origin \(O(0,0,0)\).

Finally, we will consider finding intersections of a pair of lines, or a line and a plane. These are easiest to do when exactly one line is given by a parametrization, and the other line (in the \(\mathbf R^2\) case) or plane is given in point-normal form or standard form. In this case, you can substitute the parametric equations into the standard form and solve for the parameter.

*Exercise*. Let \(\ell_1\) be the line through the points \(P(1,2)\) and \(Q(0,-1)\). Let \(\ell_2\) be the line through \(O(0,0)\) with normal vector \(\mathbf n=\left(\begin{smallmatrix}2\\-1\end{smallmatrix}\right)\). Find their intersection.

*Exercise*. Let \(\ell\) be the line with parametrization \((2t,-3+t,1-t)\). Let \(\Pi\) be the plane with equation \(x+y-2z=4\). Find their intersection.

If both lines \(\ell_1\) and \(\ell_2\) are given by a parametrization, then you can either (i) equate the parametric equations coordinatewise, or (ii) in the \(\mathbf R^2\) case, find a standard form equation for one of the lines, and substitute.

**Warning**. Make sure to make the parameters for the two lines different. Following standard notation, you can make the parameter for \(\ell_1\) be \(s\), and the parameter for \(\ell_2\) be \(t\).

*Exercise*. Let \(\ell_1\) be the line with parametrization \((1-t,2-3t)\). Let \(\ell_2\) be the line with parametrization \((t,2t)\). Find their intersection.