Intersection

Section 1.4 Intersection

Subsection 1.4.1

Figure 1.4.1. Intersections of Lines

If two lines in the plane are not parallel, there will be exactly one point where they meet. This is called the intersection point. Figure 1.4.1 shows the lines \(y=x+1\) and \(y=3x-x\) and their intersection point at \((2,3)\text{.}\) At this intersection point, the coordinates \((x,y)\) satisfy both equations. Algebraically, we can find this point by solving the system composed of both equation. I’ll do the calculation. Since the \(y\) must be the same in both equations to find the interesction, I can make both of the \(x\) expressions equal as well. Then I can solve for the \(x\) coordinates. Finally, I can use that \(x\) coordinate to solve for the \(y\) coordinate using either line equation.

\begin{align*} y \amp = x+1\\ y \amp = 3x-3\\ x+1 \amp = 3x-3\\ -2x \amp = -4\\ x \amp = 2\\ y \amp = x+1 = 2+1 = 3 \end{align*}

The intersection point is \((2,3)\text{.}\)

This example show the general idea. A point is on the intersection of two loci if its coordinates satisfy both equations. Algebraically, I find this point by solving the system of equations using the equations for both loci. There are many techniques for solving systems: isolating one variable and replacing it, adding equations together, solving for one variable in both equations, etc. Feel free to rely on whatever techniques you may have learned for solving systems in your previous mathematical experience.

Subsection 1.4.2 Intersection of Lines and Conics

With any two loci, there is the possibility of intersection; I would simply have to find some point \((x,y)\) that satisfied both equations. The algebraic approach to finding intersection is also the same as above for line. Finding an intersection point is accomplished by solving the system of equations. However, once the equations are no longer linear, solving that system can get much more difficult and complication. There many be no solutions or many solutions. It may be impossible to find the solution by exact value, meaning that approximation is the only technique. I’m not going to go do deeply into the theory of intersections in this course, but I’ll I do want to cover intersection involving both lines and conics. I’ll do this by example.

Example 1.4.2.

Figure 1.4.3. Intersection of a Line and an Ellipse

In Figure 1.4.3, the ellipse \(\frac{x^2}{4} + \frac{y^2}{1} = 1\) and the line \(y=-x\) have two intersection points. I can solve for those point by using the equation of the line to replace a variable. If I replace \(y\) with \(-x\) in the equation of the ellipse, that equation simplifies as follows.

\begin{equation*} \frac{5x^2}{4} = 1 \implies x = \pm \frac{2}{\sqrt{5}} \end{equation*}

Now I know the \(x\) values of the intersection point. The line equation \(y = -x\) reminds me that the \(y\) values are just the negative of the \(x\) values. With that reminder, I can write down the coordinates of the two intersection points: \(p_1 = (\frac{-2}{\sqrt{5}},\frac{2}{\sqrt{5}})\) and \(p_2 = (\frac{2}{\sqrt{5}},\frac{-2}{\sqrt{5}})\text{.}\)

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