Point of Intersection of Two Lines – Formula and Example

Here you will learn how to find point of intersection of two lines with examples.

Let’s begin –

How to find Point of Intersection of Two Lines

Let the equations of two lines be

\(a_1x + b_1y + c_1\) = 0

and,  \(a_2x + b_2y + c_2\) = 0

Suppose these two lines intersect at a point P(\(x_1, y_1\)). Then, (\(x_1, y_1\)) satisfies each of the given equations.

\(\therefore\)  \(a_1x_1 + b_1y_1 + c_1\) = 0   and   \(a_2x_1 + b_2y_1 + c_2\) = 0

Solving these two by cross-multiplication, we get

\(x_1\over {b_1c_2 – b_2c_1}\) = \(y_1\over {c_1a_2 – c_2a_1}\) = \(1\over {a_1b_2 – a_2b_1}\)

\(\implies\)  \(x_1\) = \({b_1c_2 – b_2c_1}\over {a_1b_2 – a_2b_1}\),  \(y_1\) = \({c_1a_2 – c_2a_1}\over {a_1b_2 – a_2b_1}\)

Hence the coordinates of the point of the point of intersection of two lines are :

( \({b_1c_2 – b_2c_1}\over {a_1b_2 – a_2b_1}\), \({c_1a_2 – c_2a_1}\over {a_1b_2 – a_2b_1}\))

Formula to find Point of Intersection :

\(x_1\) = ( \({b_1c_2 – b_2c_1}\over {a_1b_2 – a_2b_1}\), \(y_1\) = \({c_1a_2 – c_2a_1}\over {a_1b_2 – a_2b_1}\))

Note : To find the coordinates of the point of intersection of two non-parallel lines, we solve the given equations simultaneously and the values of x and y are so obtained determine the coordinates of the point of intersection.

Example : Find the coordinates of the point of intersecton of the lines 2x – y + 3 = 0 and x + 2y – 4 = 0.

Solution : Solving simultaneously the equations 2x – y + 3 = 0 and x + 2y – 4 = 0, we obtain

\(x\over {4-6}\) = \(y\over {3+8}\) = \(1\over {4+1}\)

\(\implies\) \(x\over -2\) = \(y\over 11\) = \(1\over 5\)

\(\implies\) x = \(-2\over 5\) , y = \(11\over 5\)

Hence, (-2/5, 11/5) is the required point of intersection

Example : Find the coordinates of the point of intersecton of the lines x – y + 4 = 0 and x + 2y – 1 = 0.

Solution : Solving simultaneously the equations x – y + 4 = 0 and x + 2y – 1 = 0, we obtain

\(x\over {1-8}\) = \(y\over {4+1}\) = \(1\over {2+1}\)

\(\implies\) \(x\over -7\) = \(y\over 5\) = \(1\over 3\)

\(\implies\) x = \(-7\over 3\) , y = \(5\over 3\)

Hence, (-7/3, 5/3) is the required point of intersection


Related Questions

Find the equation of line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0.

Find the coordinates of the point of intersecton of the lines 2x – y + 3 = 0 and x + y – 5 = 0.

Find the equation of line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

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