Question:
Line A has a gradient of 4 and passes through point (5,6).
Line B passes through points C (0,5) and D (2,0).
Find the coordinates of the point where the two lines intersection.
Solution:
First of all find the equation of line A:
Using y= mx + c,
Applying the gradient, line A has equation, y = 4x + c
To find c, substitute in the coordinates of point P,
6 = (4x5) + c
6 = 20 + c
c = 6 - 20 = -14
Therefore the equation of line A is y = 4x - 14
Now find the equation of line B:
Using ( y2 - y1 ) / ( x2 - x1 ) = gradient of a line
Substitute in coordinates of points C and D,
( yC - yB ) / ( xC - xB ) = ( 5 - 0 ) / ( 0 - 2 ) = 5/-2 or -5/2
Using y = mx + c
Applying the gradient found, line B has the equation, y = -5/2 x + c
To find c, substitute in the coordinates of point C,
5 = ( -5/2 x 0 ) + c
c = 5
Therefore the equation of line B is y = -5/2 x + 5
This can be rearranged,
(multiply everything by 2) --> 2y = -5x + 10
(rearrange) ---> 5x + 2y = 10
You can check your answer by using the coordinates of point D,
( 5 x 2 ) + ( 2 x 0 ) = 10 ---> Yes
Finally find the coordinates where the lines intersect:
A y = 4x - 14
B 5x + 2y = 10
A x2 2y = 8x - 28
Rearrange 8x - 2y = 28
Using simultaneous equations, add A x2 and B, to eliminate y,
5x + 8x + 2y - 2y = 10 + 28
13x = 38
x = 38/13
Substitute in x to A to find y,
y = ( 4 x 38/13 ) - 14
y = 152/13 - 182/13
y = -30/13
Put these coordinates into the equation for line B to check it works,
( 5 x 38/13 ) + (2 x -30/13 ) = 10
190/13 - 60/13 = 130/13 = 10 ----> Yes
Answer:
The lines cross at coordinate ( 38/13, -30/13 )