How to find the point of intersection of two lines formula

Point of intersection is the point where two lines or two curves meet each other. The point of intersection of two lines of two curves is a point. If two planes meet each other then the point of intersection is a line. More precisely it is defined as the common point of both the lines or curves that satisfy both the curves which can be derived by solving the equation of the curves.

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Sample Problems
Things to try
Limitations
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If we consider two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 the point of intersection of these two lines is given by:

Point of Intersection (x, y) = ((b1×c2 − b2×c1)/(a1×b2 − a2×b1), (c1×a2 − c2×a1)/(a1×b2 − a2×b1))

Derivation of point of intersection of two lines:

Given equations:

a1x + b1y + c1 = 0 -> eq-1

a2x + b2y + c2 = 0 -> eq-2

Solving the equations using cross multiplication method:

x y 1

b1 c1 a1 b1

b2 c2 a2 b2

On cross-multiplying the constants we obtain:

x/(b1*c2 – b2* c1) = y/(c1*a2-c2*a1) = 1/(a1*b2-a2*b1)

Solving for x:

=> x/(b1*c2 – b2* c1) = 1/(a1*b2-a2*b1)

=> x = (b1*c2 – b2* c1)/(a1*b2-a2*b1)

Solving for y:

=> y/(c1*a2-c2*a1) = 1/(a1*b2-a2*b1)

=> y=(c1*a2−c2*a1)/(a1*b2−a2*b1)

Hence point of intersection:

(x,y) = ((b1×c2 − b2×c1)/(a1×b2 − a2×b1), (c1×a2 − c2×a1)/(a1×b2 − a2×b1))

If two lines are parallel they never intersect each other:

Condition for two lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 to be parallel

a1/b1 = a2/b2.

Sample Problems

Question 1: Find the point of intersection of line 3x + 4y + 5 = 0, 2x + 5y +7 = 0.

Solution:

The point of intersection of two lines is given by :

(x, y) = ((b1*c2−b2*c1)/(a1*b2−a2*b1), (c1*a2−c2*a1)/(a1*b2−a2*b1))

a1 = 3, b1 = 4, c1 = 5

a2 = 2, b2 = 5, c2 = 7

(x,y) = ((28-25)/(15-8), (10-21)/(15-8))

(x,y) = (3/7,-11/7)

Question 2: Find the point of intersection of line 9x + 3y + 3 = 0, 4x + 5y + 6 = 0.

Solution:

The point of intersection of two lines is given by :

(x,y) = ((b1*c2−b2*c1)/(a1*b2−a2*b1), (c1*a2−c2*a1)/(a1*b2−a2*b1))

a1 = 9, b1 = 3, c1 = 3

a2 = 4, b2 = 5, c2 = 6

(x, y) = ((18-15)/(45-15), (54-12)/(45-15))

(x, y) = (1/10, 7/5)

Question 3: Check if the two lines are parallel or not 2x + 4y + 6 = 0, 4x + 8y + 6 = 0

Solution:

To check whether the lines are parallel or not we need to check a1/b1 = a2/b2

a1 = 2, b1 = 4

a2 = 4, b2 = 8

2/4 = 4/8

1/2 = 1/2

Since the condition is satisfied the lines are parallel and can’t intersect each other.

Question 4: Check if the two lines are parallel or not 3x + 4y + 8 = 0, 4x + 8y + 6 = 0

Solution:

To check whether the lines are parallel or not we need to check a1/b1 = a2/b2

a1 = 3, b1 = 4

a2 = 4, b2 = 8

3/4 is not equal to 4/8

Since the condition is not satisfied the lines are not parallel.

Question 5: Check whether the point (3, 5) is point of intersection of lines 2x + 3y – 21 = 0, x + 2y – 13 = 0.

Solution:

A point to be a point of intersection it should satisfy both the lines.

Substituting (x,y) = (3,5) in both the lines

Check for equation 1: 2*3 + 3*5 – 21 =0 —-> satisfied

Check for equation 2: 3 + 2* 5 -13 =0 —-> satisfied

Since both the equations are satisfied it is a point of intersection of both the lines.

Question 6: Check whether the point (2, 5) is point of intersection of lines x + 3y – 17 = 0, x + y – 13 = 0

Solution:

A point to be a point of intersection it should satisfy both the lines.

Substituting (x,y) = (2,5) in both the lines

Check for equation 1: 2+ 3*5 – 17 =0 —-> satisfied

Check for equation 2: 7 -13 = -6 —>not satisfied

Since both the equations are not satisfied it is not a point of intersection of both the lines.

Question 7: Find the point of intersection of lines x = -2 and 3x + y + 4 = 0

Solution:

On substituting x = -2 in 3x + y + 4 = 0

-6 + y + 4 = 0;

y = 2;

So the point of intersection is (x,y) = (-2,2)

Article Tags :

Point of intersection means the point at which two lines intersect. These two lines are represented by the equation a1x + b1y + c1= 0 and a2x + b2y + c2 = 0, respectively. Given figure illustrate the point of intersection of two lines.

We can find the point of intersection of three or more lines also. By solving the two equations, we can find the solution for the point of intersection of two lines.

The formula of the point of Intersection of two lines is:

(x, y) = [

\(\begin{array}{l}\frac{b_{1}c_{2}-b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\end{array} \)

\(\begin{array}{l}\frac{a_{2}c_{1}-a_{1}c_{2}}{a_{1}b_{2}-a_{2}b_{1}}\end{array} \)

]

Question: Find out the point of intersection of two lines x + 2y + 1 = 0 and 2x + 3y + 5 = 0.

Solution:

Given straight line equations are:

x + 2y + 1 = 0 and 2x + 3y + 5 = 0

Here,
a1 = 1, b1 = 2, c1 = 1

a2 = 2, b2 = 3, c2 = 5

Intersection point can be calculated using this formula,

x =

\(\begin{array}{l}\frac{b_{1}c_{2}-b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\end{array} \)

; y =

\(\begin{array}{l}\frac{a_{2}c_{1}-a_{1}c_{2}}{a_{1}b_{2}-a_{2}b_{1}}\end{array} \)

(x,y) = (

\(\begin{array}{l}\frac{2\times5-3\times1}{1\times3-2\times2}\end{array} \)

\(\begin{array}{l}\frac{2\times1-1\times5}{1\times3-2\times2}\end{array} \)

)

(x,y) = (

\(\begin{array}{l}\frac{10-3}{3-4}\end{array} \)

\(\begin{array}{l}\frac{2-5}{3-4}\end{array} \)

)

(x,y) = (-7, 3)

The point of intersection of two non-parallel lines can be found from the
equations of the two lines.

Try this Drag any of the 4 points below to move the lines. Note where they intersect.

To find the intersection of two straight lines:

First we need the equations of the two lines. If you do not have the equations, see Equation of a line - slope/intercept form and Equation of a line - point/slope form (If one of the lines is vertical, see the section below).
Then, since at the point of intersection, the two equations will have the same values of x and y, we set the two equations equal to each other. This gives an equation that we can solve for x
We substitute that x value in one of the line equations (it doesn't matter which) and solve it for y.

This gives us the x and y coordinates of the intersection.

Example

So for example, if we have two lines that have the following equations (in slope-intercept form):

y = 3x-3

y = 2.3x+4

At the point of intersection they will both have the same y-coordinate value, so we set the equations equal to each other:

3x-3 = 2.3x+4

This gives us an equation in one unknown (x) which we can solve: Re-arrange to get x terms on left

3x - 2.3x = 4+3

Combining like terms

0.7x = 7

Giving

x = 10

To find y, simply set x equal to 10 in the equation of either line and solve for y: Equation for a line (Either line will do)

y = 3x - 3

Set x equal to 10

y = 30 - 3

Giving

y = 27

We now have both x and y, so the intersection point is (10, 27) Recall that lines can be described by the slope/intercept form and point/slope form of the equation. Finding the intersection works the same way for both. Just set the equations equal as above. For example, if you had two equations in point-slope form:

y = 3(x-3) + 9

y = 2.1(x+2) - 4

simply set them equal:

3(x-3) + 9 = 2.1(x+2) - 4

and proceed as above, solving for x, then substituting that value into either equation to find y.

The two equations need not even be in the same form. Just set them equal to each other and proceed in the usual way.

When one of the lines is vertical, it has no defined slope, so its equation will look something like

x=12

. See Vertical lines (Coordinate Geometry). We find the intersection slightly differently. Suppose we have the lines whose equations are

y = 3x-3	A line sloping up and to the right
x = 12	A vertical line

On the vertical line, all points on it have an x-coordinate of 12 (the definition of a vertical line), so we simply set x equal to 12 in the first equation and solve it for y.

Equation for a line:

y = 3x - 3

Set x equal to 12 Using the equation of the second (vertical) line

y = 36 - 3

Giving

y = 33

So the intersection point is at (12,33).

If both lines are vertical, they are parallel and have no intersection (see below).

When two lines are parallel, they do not intersect anywhere. If you try to find the intersection, the equations will be an absurdity. For example the lines

y=3x+4

and

y=3x+8

are parallel because their slopes (3) are equal. See Parallel Lines (Coordinate Geometry). If you try the above process you would write

3x+4 = 3x+8

. An obvious impossibility.

Fig 1. Segments do not intersect

In the case of two non-parallel lines, the intersection will always be on the lines somewhere. But in the case of line segments or rays which have a limited length, they might not actually intersect.

In Fig 1 we see two line segments that do not overlap and so have no point of intersection. However, if you apply the method above to them, you will find the point where they would have intersected if extended enough.

Things to try

In the above diagram, press 'reset'.
Drag any of the points A,B,C,D around and note the location of the intersection of the lines.
Drag a point to get two parallel lines and note that they have no intersection.
Click 'hide details' and 'show coordinates'. Move the points to any new location where the intersection is still visible. Calculate the slopes of the lines and the point of intersection. Click 'show details' to verify your result.

Limitations

In the interest of clarity in the applet above, the coordinates are rounded off to integers and the lengths rounded to one decimal place. This can cause calculatioons to be slightly off.

For more see Teaching Notes