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Determining the point at which two linear lines in a plane intersect is a simple process.


Line Intersection

Given two planar lines with equations in the slope-intercept form \(y=mx+b\), we can find the point where these lines intersect by finding the point where both equations are equal. We could do this by setting the equations equal to each other, then rearrange to solve for \(x\) by working the order of operations backwards, but I prefer using polynomial subtraction; I find it to be quicker and easier than doing it longhand.

For example, given lines with equations \(y=\frac{1}{2}x+1\) and \(y=-3x+8\):

\(\begin{array}{cccccc} & -3x & + & 8 & = & 0\\ - & \frac{1}{2}x & + & 1 & = & 0\\\hline & -\frac{7}{2}x & + & 7 & = & 0\\ \end{array}\)

Solving for \(x\), we get:

\(-\frac{7}{2}x + 7 = 0\)  →  \(-\frac{7}{2}x = -7\)  →  \(x = 2\)

Now that we know the value of \(x\) we can calculate \(y\) by plugging it into either of our line equations:

\(y=\frac{1}{2}x+1\)  →  \(y=\frac{1}{2}(2)+1\)  →  \(y=1+1\)  →  \(y=2\)

(or)

\(y=-3x+8\)  →  \(y=-3(2)+8\)  →  \(y=-6+8\)  →  \(y=2\)

So the lines intersect at \((2,2)\). Easy, right?


Note:

This method of determining line intersection cannot be used if either of the lines is vertical (has an infinite slope). This is because a vertical line cannot be represented in slope-intercept form. Instead, since a vertcal line has \(y=\{ℝ\}\), it has an equation of \(x=n\), where \(n\) is the x-coordinate of the line. Simply plug the \(x\) value into the second line's equation and solve for \(y\).



Analytic Geometry: Lines Analytic Geometry: Vectors