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Finding Points of Intersection

There are times when we need to find the point(s) common to two graphs.  In other words, both graphs pass through the same point, and we may need to find out that point of intersection.  We could do this to optimize a distribution of resources, or to find a point of equilibrium or balance.  We can look at both graphs and see that they intersect, and sometimes, we may even be able to approximate the point or points of intersection just by looking at the two graphs together.  Very often, though, we can use algebra to find the exact intersection point(s) of two graphs.

First, we need the equations of the graphs.  Let’s look at an example to see how we might use algebra to find their intersection.

Example 1

Find the intersection of the two lines  and .

Solution 1

As we look at the graph above, we see that the two lines intersect in a single point.  We are looking for the coordinates of that point (x, y) that satisfies both equations.  Recall that “y” and “f(x)” are interchangeable.  Because y = 3x+2 and also 2x1, we can find the x-coordinate of the point of intersection by setting 3x+2 equal to 2x1, and solving for x.  This is equivalent to substituting the expression 3x+2 for y in the equation y = 2x – 1.  After all, the same y will be used in both equations since the point of intersection satisfies both.  So we have:

.  Now, we solve for x:


We’ve found that the lines intersect when .  To find the
 y-coordinate of the intersection, we use the value 3 for x and find y.  If we have correctly found the right x-coordinate, we should get the same y-coordinate, regardless of which equation we use.  Let’s check:

y = 3x + 2, x = 3 gives
y  =  3(-3) + 2
y  =  9 + 2
y  =  7

y  =  2x - 1, x = -3 gives
y  =  2(3) – 1
y  =  6 – 1
y  =  7

Thus, the point (3, 7) is the point of intersection of the two lines. 

We are not limited to finding points of intersection of two straight lines.  We can also do it when one or both graphs are, for instance, parabolas.  And while the algebra is a little more complicated, the process is exactly the same!

Example 2

Given the graphs of equations , and , find their point(s) (if any) of intersection.

Solution 2

Since y = x + 1 and also y = x2 – 1, we can substitute –x + 1 for y in the second equation, to get –x + 1 = x2 – 1.

Let’s solve that equation:

x + 1 = x2 – 1
0 = x2 – 1 + x – 1
0 = x2 + x – 2
0 = (x + 2)(x – 1)    leads to x = 2 or x = 1.


Thus, we have two values for x:  2 and 1.  We check these values in our two functions to make sure they are points of intersection.


So when x = 2, y = 3 for both equations, and when x = 1, y = 0 for both equations.  Our two points of intersection are, therefore,
(2, 3) and (1,0).  (If you’re wondering how we ended up with two points of intersection, remember we were working with a line and a parabola.)


Example 3 

Given the equations  and , find their points of intersection.

Solution 3

We will solve the equation x2 – 8x + 20 = x2 + 4x + 2


Now, we have only one x-value to check:  x = 3.


We have f(3) = g(3) = 5.  In other words, when we use the value 3 for x, the value for y is 5 for both equations.

So, we have one point of intersection, (3, 5).