Give the point-normal form for the following 2D line with the following description:
– Through P(-2,5) and perpendicular to the line x + 2y = 5
I know that the point-normal form is n * (x – p) = 0
when n is the normal vector, x = (x,y,z), and p is a point.
So, the point was given, therefore I get the following
n * (x-(-2,5)) = 0
but how do I find the normal vector?
Here's a picture of the exercise (#19)
Best Answer
As you said, $p$ is a point on the line (in this case, $(-2, 5)$ works) and $n$ is the normal vector, but since this is a line in $\mathbb R_2$, $x$ is a 2D point that can be represented as $(x_1, y_1)$, not $(x_1, y_1, z_1)$.
Now, we are given that this line is perpendicular to $x+2y=5$. We can rewrite this as $x-0+2y-5=0$, which is the same as $1(x-0)+2(y-\frac 5 2)=0$. From this, we can notice that $(1, 2)$ is the normal vector of this line and $(0, \frac 5 2)$ is a point on this line because this line can be written as $(1, 2) \cdot (\mathbf x-(0, \frac 5 2))=0$. Thus, the normal vector for $x+2y=5$ is $(1, 2)$.
However, our line is perpendicular to this line, which means its normal vector is perpendicular to the normal vector of $(1, 2)$. To find the vector perpendicular to $(1, 2)$, we flip the coordinates of the vector and make the first coordinate negative to get $(-2, 1)$. If you take the dot product of $(1, 2)$ and $(-2, 1)$, you'll find that it comes out to $0$, showing that these two normal vectors are perpendicular. Thus, $(-2, 1)$ is the real normal vector of our line.
Now that we have the normal vector and a point on the line, we can write the line in point-normal form: $(-2, 1) \cdot (\mathbf x-(-2, 5))=0$.