# C1: Parallel and Perpendicular Lines

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C1: Parallel and Perpendicular Lines
Learning Objective : to be able to calculate the gradients of two lines and identify the conditions for them to be parallel or perpendicular

Starter: Calculate the length of the line joining (5, 6) and (10, 18) and find the mid- point. The triangle ABC has vertices A(-5, 0), B(-1, 3) and C(2, 7). Determine the type of triangle it is.

Parallel lines If two lines have the same gradient they are parallel.
Show that the lines 3y + 6x = 2 and y = –2x + 7 are parallel. We can show this by rearranging the first equation so that it is in the form y = mx + c. 3y + 6x = 2 3y = –6x + 2 y = –6x + 2 3 y = –2x + 2/3 The gradient, m, is –2 for both lines and so they are parallel.

Exploring perpendicular lines
The gradient of each line in this activity is given by the coefficient of x in the line’s equation. Modify the red line by dragging its defining points and observe the relationship between its gradient and the gradient of the line perpendicular to it.

Perpendicular lines If the gradients of two lines have a product of –1 then they are perpendicular. In general, if the gradient of a line is m, then the gradient of the line perpendicular to it is . Find the gradient of the line perpendicular to the line joining the points A(–2, 2) and B(4, –1). The gradient of the line AB is -1/2 The exception to this generalization that the product of the gradient of two perpendicular lines is -1 is when the two perpendicular lines in question are horizontal and vertical. So the gradient of the line perpendicular to this is 2. Because -½ x 2 = -1

Task 1 : Work out the gradient of the line that is perpendicular to these gradients
3 -5 -2/5 0.1 -0.5

Task 2 : Work out whether these pairs of lines are parallel, perpendicular or neither
y = 4x + 2 y = -1/4 x – 7 y = 2/3 x – 1 y = 2/3 x – 11 y = 1/5 x + 9 y = 5x + 9 y = 5x – 3 5x – y + 4 = 0 4x – 5y + 1 = 0 8x – 10y -2 = 0 3x + 2y – 12 = 0 2x + 3y -6 = 0 5x – y + 2 = 0 2x + 10y – 4 = 0

Perpendicular lines If the gradients of two lines have a product of –1 then they are perpendicular. In general, if the gradient of a line is m, then the gradient of the line perpendicular to it is . Find the equation of the perpendicular bisector of the line joining the points A(–2, 2) and B(4, –1). The perpendicular bisector of the line AB has to pass through the mid-point of AB. The exception to this generalization that the product of the gradient of two perpendicular lines is -1 is when the two perpendicular lines in question are horizontal and vertical. Let’s call the mid-point of AB point M, so M is the point

Perpendicular lines The gradient of the line joining A(–2, 2) and B(4, –1) is mAB = The gradient of the perpendicular bisector of AB is therefore 2. Using this and the fact that is passes through the point we can use y – y1 = m(x – x1) to write Point out that in the general form ax + by + c =0, the value c has a different meaning to the value of c in the form y = mx + c. If an equation is given in the form ax + by + c = 0, it can be rearranged to find the gradient and the y-intercept. This would give y = -a/bx – c/b where the gradient is -a/b and the y-intercept is -c/b. So, the equation of the perpendicular bisector of the line joining the points A(–2, 2) and B(4, –1) is 2y – 4x + 3 = 0.

Task 3 Find an equation of the line that passes through the point (6, -2) and is perpendicular to the line y = 3x + 5. Find an equation of the line that passes through the point (-2, 7) and is parallel to the line y = 4x + 1. Write your answer in the form ax + by + c = 0. The line r passes through the points (1, 4) and (6, 8) and the line s passes through the points (5, -3) and (20, 9). Show that the lines r and s are parallel. The vertices of a quadrilateral ABCD have coordinates A(-1, 5), B(7, 1), C(5, -3) and D(-3, 1). Show that the quadrilateral is a rectangle.

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