{"id":4774,"date":"2021-08-25T15:34:42","date_gmt":"2021-08-25T15:34:42","guid":{"rendered":"https:\/\/mathemerize.com\/?p=4774"},"modified":"2022-01-16T16:57:53","modified_gmt":"2022-01-16T11:27:53","slug":"point-of-intersection-of-two-lines","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/point-of-intersection-of-two-lines\/","title":{"rendered":"Point of Intersection of Two Lines – Formula and Example"},"content":{"rendered":"
Here you will learn how to find point of intersection of two lines with examples.<\/p>\n
Let’s begin –<\/p>\n
Let the equations of two lines be<\/p>\n
\\(a_1x + b_1y + c_1\\) = 0<\/p>\n
and, \\(a_2x + b_2y + c_2\\) = 0<\/p>\n
Suppose these two lines intersect at a point P(\\(x_1, y_1\\)). Then, (\\(x_1, y_1\\)) satisfies each of the given equations.<\/p>\n
\\(\\therefore\\) \\(a_1x_1 + b_1y_1 + c_1\\) = 0 and \\(a_2x_1 + b_2y_1 + c_2\\) = 0<\/p>\n
Solving these two by cross-multiplication, we get<\/p>\n
\\(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}\\)<\/p>\n
\\(\\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}\\)<\/p>\n
Hence the coordinates of the point of the point of intersection of two lines are :<\/p>\n
( \\({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}\\))<\/p>\n
\n\\(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}\\))<\/p>\n<\/blockquote>\n
Note : <\/strong>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.<\/p>\n\n\n
Example : <\/span>Find the coordinates of the point of intersecton of the lines 2x – y + 3 = 0 and x + 2y – 4 = 0.<\/p>\n
Solution : <\/span>Solving simultaneously the equations 2x – y + 3 = 0 and x + 2y – 4 = 0, we obtain
\n\\(x\\over {4-6}\\) = \\(y\\over {3+8}\\) = \\(1\\over {4+1}\\)
\n\\(\\implies\\) \\(x\\over -2\\) = \\(y\\over 11\\) = \\(1\\over 5\\)
\n\\(\\implies\\) x = \\(-2\\over 5\\) , y = \\(11\\over 5\\)
\nHence, (-2\/5, 11\/5) is the required point of intersection
<\/p>\n\n\n\nExample : <\/span>Find the coordinates of the point of intersecton of the lines x – y + 4 = 0 and x + 2y – 1 = 0.<\/p>\n
Solution : <\/span>Solving simultaneously the equations x – y + 4 = 0 and x + 2y – 1 = 0, we obtain
\n\\(x\\over {1-8}\\) = \\(y\\over {4+1}\\) = \\(1\\over {2+1}\\)
\n\\(\\implies\\) \\(x\\over -7\\) = \\(y\\over 5\\) = \\(1\\over 3\\)
\n\\(\\implies\\) x = \\(-7\\over 3\\) , y = \\(5\\over 3\\)
\nHence, (-7\/3, 5\/3) is the required point of intersection
<\/p>\n\n\n
\nRelated Questions<\/h3>\n
Find the equation of line joining the point (3, 5) to the point of intersection of the lines 4x + y \u2013 1 = 0 and 7x \u2013 3y \u2013 35 = 0.<\/a><\/p>\n