Point of Intersection of Two Lines

We will learn how to find the co-ordinates of the point of intersection of two lines.

Let the equations of two intersecting straight lines be

a\(_{1}\) x + b\(_{1}\)y + c\(_{1}\)  = 0 ………….. (i) and   

a\(_{2}\) x + b\(_{2}\) y + c\(_{2}\) = 0 …….…... (ii)

Suppose the above equations of two intersecting lines intersect at P(x\(_{1}\), y\(_{1}\)). Then (x\(_{1}\), y\(_{1}\)) will satisfy both the equations (i) and (ii).

Therefore, a\(_{1}\)x\(_{1}\) + b\(_{1}\)y\(_{1}\)  + c\(_{1}\) = 0 and

a\(_{2}\)x\(_{1}\) + b\(_{2}\)y\(_{1}\) + c\(_{2}\) = 0               

Solving the above two equations by using the method of cross-multiplication, we get,

\(\frac{x_{1}}{b_{1}c_{2} - b_{2}c_{1}} = \frac{y_{1}}{c_{1}a_{2} - c_{2}a_{1}} = \frac{1}{a_{1}b_{2} - a_{2}b_{1}}\)

Therefore, x\(_{1}\)  = \(\frac{b_{1}c_{2} - b_{2}c_{1}}{a_{1}b_{2} - a_{2}b_{1}}\) and

y\(_{1}\)  = \(\frac{c_{1}a_{2} - c_{2}a_{1}}{a_{1}b_{2} - a_{2}b_{1}}\),  a\(_{1}\)b\(_{2}\) - a\(_{2}\)b\(_{1}\) ≠ 0

Therefore, the required co-ordinates of the point of intersection of the lines (i) and (ii) are

(\(\frac{b_{1}c_{2} - b_{2}c_{1}}{a_{1}b_{2} - a_{2}b_{1}}\), (\(\frac{c_{1}a_{2} - c_{2}a_{1}}{a_{1}b_{2} - a_{2}b_{1}}\)), a\(_{1}\)b\(_{2}\) - a\(_{2}\)b\(_{1}\) ≠ 0


Notes: 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 so obtained determine the coordinates of the point of intersection.

If a\(_{1}\)b\(_{2}\) - a\(_{2}\)b\(_{1}\) = 0 then a\(_{1}\)b\(_{2}\) = a\(_{2}\)b\(_{1}\)

\(\frac{a_{1}}{b_{1}}\) = \(\frac{a_{2}}{b_{2}}\)

- \(\frac{a_{1}}{b_{1}}\) = - \(\frac{a_{2}}{b_{2}}\)  i.e., the slope of line (i) = the slope of  line  (ii)

Therefore, in this case the straight lines (i) and (ii) are parallel and hence they do not intersect at any real point.


Solved example to find the co-ordinates of the point of intersection of two given intersecting straight lines:

Find the coordinates of the point of intersection of the lines 2x - y + 3 = 0 and x + 2y - 4 = 0.

Solution:

We know that the co-ordinates of the point of intersection of the lines a\(_{1}\) x+ b\(_{1}\)y+ c\(_{1}\)  = 0 and a\(_{2}\) x + b\(_{2}\) y + c\(_{2}\) = 0 are

(\(\frac{b_{1}c_{2} - b_{2}c_{1}}{a_{1}b_{2} - a_{2}b_{1}}\), (\(\frac{c_{1}a_{2} - c_{2}a_{1}}{a_{1}b_{2} - a_{2}b_{1}}\)), a\(_{1}\)b\(_{2}\) - a\(_{2}\)b\(_{1}\) ≠ 0

Given equations are

2x - y + 3 = 0 …………………….. (i)

x + 2y - 4 = 0 …………………….. (ii)

Here a\(_{1}\) = 2, b\(_{1}\) = -1, c\(_{1}\) = 3, a\(_{2}\) = 1, b\(_{2}\) = 2 and c\(_{2}\) = -4.

(\(\frac{(-1)\cdot (-4) - (2)\cdot (3)}{(2)\cdot (2) - (1)\cdot (-1)}\), \(\frac{(3)\cdot (1) - (-4)\cdot (2)}{(2)\cdot (2) - (1)\cdot (-1)}\))

(\(\frac{4 - 6}{4 + 1}\), \(\frac{3 + 8}{4 + 1}\))

(\(\frac{11}{5}, \frac{-2}{5}\))

Therefore, the co-ordinates of the point of intersection of the lines 2x - y + 3 = 0 and x + 2y - 4 = 0 are (\(\frac{11}{5}, \frac{-2}{5}\)).





11 and 12 Grade Math

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