# Co-ordinate Geometry III

## Presentation on theme: "Co-ordinate Geometry III"— Presentation transcript:

Co-ordinate Geometry III
2 4 6 -2 X-axis Y-axis Parallel and Perpendicular Lines. m1x m2 = -1 m1 = m2 Equations of Lines II By Mr Porter

Assumed Knowledge: Gradient, m, using 2 points.
Standard form of a line: y = mx + b General form of a line: Ax + By + C = 0 Ability to rearrange algebraic expression / equation: Change of Subject Examples Rearrange the general line Ax + By + C = 0, to standard form, y = mx + b. Where m is the gradient and b is the y-intercept, for each of the following. a) 3x + 2y – 6 = 0 b) 4x – 2y + 7 = 0 2y =-3x + 6 4x + 7 = 2y y = mx + b y = mx + b

Parallel Lines Definition: Two lines y = m1x + b1 and y = m2 x + b2 are parallel, if their gradients are equal: m1 = m2. Example 2 Find the equation of the line parallel to , passing through the point (-1,3). Example 1 Find the equation of the line parallel to y = 3x – 2, passing through the point (2,-5). From, , in standard form, y = mx + b. To find the equation of a line, you need to have a gradient, m, and a point on the line. In this case, we have the point, need to find the gradient. From, y =3x – 2, in standard form, y = mx + b. The gradient of the given line is The gradient of the given line is m = 3. Gradient, mi, of ALL lines parallel to the given line are equal. m1 = m2 Gradient, mi, of ALL lines parallel to the given line are equal. m1 = m2 Therefore, m = 3. Therefore, The point-gradient form of a lines is: y - y1 = m(x - x1) The point-gradient form of a lines is: y - y1 = m(x - x1) y – 3 = (x – -1) y – -5 = 3(x – 2) 3y – 9 = -2x – 2 y + 5 = 3x – 6 y = 3x – 1, is the required line. 2x +3y – 7 = 0, is the required line.

Parallel lines to a given general line: Ax + By + C = 0.
If the given line is in general form, it has to be rearrange to standard form to obtain the gradient, m. Example 1 Find the equation of the line parallel to 3x + 2y – 6 = 0, through the point (3,-5). Example 2 Find the equation of the line parallel to 5x – 2y + 4 = 0, through the point (-2,8). Step 1: Rearrange the given equation to standard form: y = mx + b. Step 1: Rearrange the given equation to standard form: y = mx + b. 5x – 2y + 4 = 0 3x + 2y – 6 = 0 5x + 4 = 2y 2y = -3x + 6 i.e y = mx + b i.e y = mx + b The point-gradient form of a lines is: y - y1 = m(x - x1) The point-gradient form of a lines is: y - y1 = m(x - x1) 3x + 2y + 1 = 0, is the required line. 5x – 2y + 26 = 0, is the required line.

Exercise 1: Find the equation of the line parallel to the given line passing through the given point.

Perpendicular Lines. Definition: Two lines y = m1x + b1 and y = m2 x + b2 are perpendicular, if the  product of the gradients is equal to -1: m1 x m2 = -1. Alternative: The gradients are negative reciprocals. Example 1 Find the equation of the line perpendicular to y = 3x – 2, passing through the point (2,-5). Example 2 Find the equation of the line perpendicular to , passing through the point (-1,3). From, y =3x – 2, in standard form, y = mx + b. From, , in standard form, y = mx + b. The gradient of the given line is m1 = 3. The gradient of the given line is Perpendicular gradient, m2, are such that 3 x m2 = -1. Therefore, Perpendicular gradient, m2, are such that x m2 = -1. Therefore, Negative reciprocal! Negative reciprocal! The point-gradient form of a lines is: y - y1 = m2(x - x1) The point-gradient form of a lines is: y - y1 = m2(x - x1) x + 3y + 13 = 0, is the required line. 3x – 2y + 9 = 0, is the required line.

Perpendicular gradient, m2, are such that x m2 = -1. Therefore,
Example 3 Find the equation of the line perpendicular to 3x + 2y – 6 = 0, through the point (3,-5). Example 4 Find the equation of the line perpendicular to 5x – 2y + 4 = 0, through the point (-2,8). Step 1: Rearrange the given equation to standard form: y = mx + b. Step 1: Rearrange the given equation to standard form: y = mx + b. 3x + 2y – 6 = 0 5x – 2y + 4 = 0 2y = -3x + 6 5x + 4 = 2y i.e y = mx + b i.e y = mx + b Perpendicular gradient, m2, are such that x m2 = -1. Therefore, Perpendicular gradient, m2, are such that x m2 = -1. Therefore, The point-gradient form of a lines is: y - y1 = m(x - x1) The point-gradient form of a lines is: y - y1 = m(x - x1) 2x – 3y – 21 = 0, is the required line. 2x + 5y – 36 = 0, is the required line.

Exercise 2: Find the equation of the line perpendicular to the given line passing through the given point.

Similar presentations