The Equation of a Straight Line Can Be Written As Y = Mx + C, Where M Is the Gradient And

The equation of a straight line can be written as The gradient of a line passing through the The equation of the straight line with gradient m y = mx + c, where m is the gradient and c is the y2- y 1 that passes through the point (x1, y 1 ) is points (x1, y 1) and ( x 2 , y 2 ) is . intercept with the vertical axis. x- x 2 1 y- y1 = m( x - x 1 ) . Lines are parallel if they have the same If the gradient of a line is m, then the gradient of gradient. 1 a perpendicular line is - . Two lines are perpendicular if the product of Coordinate Geometry m their gradients is -1. The distance between the points with coordinates The midpoint of the line joining the points The equation of a circle centre (a, b) with radius 2 2 2 2 2 骣x1+ x 2 y 1 + y 2 r is (x- a ) + ( y - b ) = r . (x1, y 1) and ( x 2 , y 2) is ( x 2- x 1) +( y 2 - y 1 ) . (x1, y 1) and ( x 2 , y 2 ) is 琪 , . 桫 2 2 Example: Example: Example: Find the equation of the perpendicular bisector Find the point of intersection of the lines: Find the centre and the radius of the circle with of the line joining the points (3, 2) and (5, -6). 2x + y = 3  equation x2+2 x + y 2 - 6 x + 6 = 0 . and y = 3x – 1.  Solution: Solution: The midpoint of the line joining (3, 2) and Solution: We begin by writing x2 + 2 x in completed 骣3+ 5 2 + ( - 6) To find the point of intersection we need to (5, -6) is 琪 , ,i . e .( 4,- 2) . square form: solve the equations  and  simultaneously. 2 2 2 桫 2 2 x2 + 2 x = (x+ 1) - 1 = ( x + 1) - 1 . We can substitute  into equation : The gradient of the line joining these two points We then write y2 - 6 x in completed square -6 - 2 - 8 2x + (3x – 1) = 3 is: = = -4 . i.e. 5x – 1 = 3 form: 5- 3 2 2 2 2 2 y- 6 x = (y- 3) - 3 = ( y - 3) - 9 . The equation of the perpendicular bisector must i.e. x = 4/5 Substituting this into equation : So we can rewrite the equation of the circle as -1= 1 therefore be -4 4 . y = 3(4/5) – 1 = 7/5. (x+ 1)2 - 1 + ( y - 3) 2 - 9 + 6 = 0 We need the equation of the line through (4, -2) Therefore the lines intersect at the point i.e. (x+ 1)2 + ( y - 3) 2 = 4 . with gradient ¼ . This is (4/5, 7/5). 1 This is a circle centre (-1, 3), radius 2. y-( - 2) =4 ( x - 4) 1 y+2 =4 x - 1 1 y=4 x - 3