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Co-ordinate Geometry III Equations of Lines II By Mr Porter 0 246-2 X-axis Y-axis -2 2 4 Parallel and Perpendicular Lines. m 1 = m 2 m 1 x m 2 = -1.

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Presentation on theme: "Co-ordinate Geometry III Equations of Lines II By Mr Porter 0 246-2 X-axis Y-axis -2 2 4 Parallel and Perpendicular Lines. m 1 = m 2 m 1 x m 2 = -1."— Presentation transcript:

1 Co-ordinate Geometry III Equations of Lines II By Mr Porter X-axis Y-axis Parallel and Perpendicular Lines. m 1 = m 2 m 1 x m 2 = -1

2 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 2y =-3x + 6 b) 4x – 2y + 7 = 0 4x + 7 = 2y y = mx + b

3 Parallel Lines Definition: Two lines y = m 1 x + b 1 and y = m 2 x + b 2 are parallel, if their gradients are equal: m 1 = m 2. Example 1 Find the equation of the line parallel to y = 3x – 2, passing through the point (2,-5). 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 m = 3. Gradient, m i, of ALL lines parallel to the given line are equal. m 1 = m 2 Therefore, m = 3. The point-gradient form of a lines is: y - y 1 = m(x - x 1 ) y – -5 = 3(x – 2) y + 5 = 3x – 6 y = 3x – 1, is the required line. The point-gradient form of a lines is: y - y 1 = m(x - x 1 ) 3y – 9 = -2x – 2 2x +3y – 7 = 0, is the required line. Example 2 Find the equation of the line parallel to, passing through the point (-1,3). Gradient, m i, of ALL lines parallel to the given line are equal. m 1 = m 2 From,, in standard form, y = mx + b. The gradient of the given line is. Therefore, y – 3 = (x – -1)

4 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). Step 1: Rearrange the given equation to standard form: y = mx + b. 3x + 2y – 6 = 0 2y = -3x + 6 i.e y = mx + b The point-gradient form of a lines is: y - y 1 = m(x - x 1 ) 3x + 2y + 1 = 0, is the required line. 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. 5x – 2y + 4 = 0 5x + 4 = 2y i.e y = mx + b The point-gradient form of a lines is: y - y 1 = m(x - x 1 ) 5x – 2y + 26 = 0, is the required line.

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

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

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

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


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