1.3 Equations of Lines in 2d

What, exactly, is $t$?

Both!

Since $t$ can be any real number, these equation describe the same line. They're both valid. For example, the point given by the first parametric equation with $t=7$, namely $\mathbf c + 7\mathbf d$, is exactly the same as the point given by the second parametric equation with $t=-7$, namely $\mathbf c -(-7)\mathbf d$.

What, exactly, is $\mathbf c$?

Generally, only the first.

In contrast to Question 1, the sign on $\mathbf c$ does generally matter. $\mathbf c$ is required to be a point on the line, but except in particular circumstances, there's no reason to believe that $-\mathbf c =-\mathbf c +t\mathbf d\big|_{t=0}$ is a point on the line. Indeed $-\mathbf c$ is on the line if and only if there is a $t$ with $\mathbf c + t\mathbf d = -\mathbf c$, i.e. $t\mathbf d = -2\mathbf c$. That is the case if and only if $\mathbf d$ is parallel to $\mathbf c$. So, only the first equation is correct in general.

Set $t=0$ in both equation to get two different points with integer coordinates; show that these two points are on both lines.

Since both lines pass through $(1,9)$ and $(9,13)$, the lines are identical.

Here is one answer of many.

Setting $t=0$ in the first equation shows that $(1,9)$ is on the first line. To see that $(1,9)$ is also on the second line, we substitute $x=1$, $y=9$ into the second equation to give

This equation is satisfied when $t=-8$. So $(1,9)$ is on both lines.

Setting $t=0$ in the second equation shows that $(9,13)$ is on the second line. To see that $(9,13)$ is also on the first line, we substitute $x=9$, $y=13$ into the first equation to give

This equation is satisfied when $t=1$. So $(9,13)$ is on both lines.

Since both lines pass through $(1,9)$ and $(9,13)$, the lines are identical.

A line is specified by two things: one point on the line, and a vector parallel to the direction of the line.

$\llt d_x,d_y\rgt$ can be any nonzero scalar multiple of $\llt 9,7\rgt$, and $\llt x_0,y_0\rgt$ can be any point on the line, i.e. any pair that satisfies $7x_0+24=9y_0$.

$\llt d_x,d_y\rgt$ is the direction of the line, so it can be any non-zero scalar multiple of $\llt 9,7\rgt$.

$\llt x_0,y_0\rgt$ can be any point on the line. Describing these is the same as describing the line itself. We're trying to find all doubles $\llt x_0,y_0\rgt$ that obey

for some real number $t$. That is,

Any of these steps could specify the possible values of $\llt x_0,y_0\rgt$. Say, they can be any pair satisfying $7x_0+24=9y_0$.

Remember that the parametric equation of a line with direction $\mathbf d$, passing through point $\mathbf c$, is $\llt x,y\rgt =\mathbf c + t\mathbf d$.

(a) $\llt x,y\rgt=\llt 1,2\rgt+t\llt 3,2\rgt$,
$x=1+3t,\ y=2+2t$,
$\frac{x-1}{3}=\frac{y-2}{2}$

(b) $\llt x,y\rgt=\llt 5,4\rgt+t\llt 2,-1\rgt$,
$x=5+2t,\ y=4-t$,
$\frac{x-5}{2}=\frac{y-4}{-1}$

(c) $\llt x,y\rgt=\llt -1,3\rgt+t\llt -1,2\rgt$,
$x=-1-t,\ y=3+2t$,
$\frac{x+1}{-1}=\frac{y-3}{2}$

(a) The vector parametric equation is $\llt x,y\rgt=\llt 1,2\rgt+t\llt 3,2\rgt$. The scalar parametric equations are $x=1+3t,\ y=2+2t$. The symmetric equation is $\frac{x-1}{3}=\frac{y-2}{2}$.

(b) The vector parametric equation is $\llt x,y\rgt=\llt 5,4\rgt+t\llt 2,-1\rgt$. The scalar parametric equations are $x=5+2t,\ y=4-t$. The symmetric equation is $\frac{x-5}{2}=\frac{y-4}{-1}$.

(c) The vector parametric equation is $\llt x,y\rgt=\llt -1,3\rgt+t\llt -1,2\rgt$. The scalar parametric equations are $x=-1-t,\ y=3+2t$. The symmetric equation is $\frac{x+1}{-1}=\frac{y-3}{2}$.

Review Equation 1.3.3 in the CLP-3 text.

(a) $\llt x,y\rgt=\llt 1,2\rgt+t\llt -2,3\rgt$,
$x=1-2t,\ y=2+3t$,
$\frac{x-1}{-2}=\frac{y-2}{3}$

(b) $\llt x,y\rgt=\llt 5,4\rgt+t\llt 1,2\rgt$,
$x=5+t,\ y=4+2t$,
$x-5=\frac{y-4}{2}$

(c) $\llt x,y\rgt=\llt -1,3\rgt+t\llt 2,1\rgt$,
$x=-1+2t,\ y=3+t$,
$\frac{x+1}{2}=y-3$

(a) The vector $\llt -2,3\rgt$ is perpendicular to $\llt 3,2\rgt$ (you can verify this by taking the dot product of the two vectors) and hence is a direction vector for the line. The vector parametric equation is $\llt x,y\rgt=\llt 1,2\rgt+t\llt -2,3\rgt$. The scalar parametric equations are $x=1-2t,\ y=2+3t$. The symmetric equation is $\frac{x-1}{-2}=\frac{y-2}{3}$.

(b) The vector $\llt 1,2\rgt$ is perpendicular to $\llt 2,-1\rgt$ and hence is a direction vector for the line. The vector parametric equation for the line is $\llt x,y\rgt=\llt 5,4\rgt+t\llt 1,2\rgt$. The scalar parametric equations are $x=5+t,\ y=4+2t$. The symmetric equation is $x-5=\frac{y-4}{2}$.

(c) The vector $\llt 2,1\rgt$ is perpendicular to $\llt -1,2\rgt$
and hence is a direction vector for the line. The vector parametric equation is $\llt x,y\rgt=\llt -1,3\rgt+t\llt 2,1\rgt$. The scalar parametric equations are the two component equations $x=-1+2t,\ y=3+t$. The symmetric equation is $\frac{x+1}{2}=y-3$.

Review Example 1.3.5 in the CLP-3 text.

$14/5$

$(0,1)$ is one point on the line $3x-4y=-4$. So $\llt-2-0,3-1\rgt =\llt-2,2\rgt$ is a vector whose tail is on the line and whose head is at $(-2,3)$. $\llt 3,-4\rgt$ is a vector perpendicular to the line, so $\frac{1}{5}\llt 3,-4\rgt$ is a unit vector perpendicular to the line. The distance from $(-2,3)$ to the line is the length of the projection of $\llt-2,2\rgt$ on $\frac{1}{5}\llt 3,-4\rgt$, which is the absolute value of $\frac{1}{5}\llt 3,-4\rgt\cdot\llt -2,2\rgt$. So the distance is $14/5$.

(a)

(b) $\frac{1}{3}(\va+\vb+\vc)$

(a) The midpoint of the side opposite $\va$ is $\half(\vb+\vc)$. The vector joining $\va$ to that midpoint is $\half\vb+\half\vc-\va$. The vector parametric equation of the line through $\va$ and $\half(\vb+\vc)$ is

Similarly, for the other two medians (but using $s$ and $u$ as parameters, rather than $t$)

(b) The three medians meet at a common point if there are values of $s,t$ and $u$ such that

Assuming that the triangle has not degenerated to a line segment, this is the case if and only if the coefficients of $\va,\ \vb$ and $\vc$ match

or

The medians meet at $\frac{1}{3}(\va+\vb+\vc)$.

The radius of the circle will serve as a normal vector to the line.

One way of writing the equation is $x+{\sqrt3}y=4+{\sqrt3}$.

A normal vector to the line is the vector with its tail at the centre of $C$, $(2,1)$, and its head at $\left(\frac{5}{2},1+\frac{\sqrt3}{2}\right)$. So, we set $\mathbf{n}=\llt \frac{5}{2},1+\frac{\sqrt3}{2} \rgt-\llt 2,1\rgt = \llt \frac12, \frac{\sqrt3}{2}\rgt$.

We know one point on the line is $\left(\frac{5}{2},1+\frac{\sqrt3}{2}\right)$, so following Equation 1.3.3 in the CLP-3 text: