GCSE Secondary Maths Age 13-17 - Geometry & Measures: Parallel Lines - Explained

The video tutorial explains how to find an expression for b in terms of a, given two points P and Q and a line perpendicular to PQ. It covers calculating the gradient of the perpendicular line, determining the gradient of line PQ, and using rise over run to relate gradients to coordinates. The tutorial concludes with solving for b in terms of a and discussing the marking scheme for the problem.

What are the coordinates of point P?

• (b, a)

• (4, 3)

• (a, b)

• (3, 4)

What is the equation of the line perpendicular to PQ?

• 3x - 2y = 7

• 2x - 3y = 7

• 3x + 2y = 7

• 2x + 3y = 7

How do you find the gradient of a line perpendicular to another?

• Subtract the gradients to get 1

• Add the gradients to get 0

• Multiply the gradients to get -1

• Multiply the gradients to get 1

What is the gradient of the line 3x + 2y = 7?

• 3/2

• 2/3

• -3/2

• -2/3

How is the gradient of line PQ calculated using coordinates?

• Difference of x and y coordinates

• Sum of x and y coordinates

• Change in y over change in x

• Change in x over change in y

What is the expression for b in terms of a?

• b = 6 - 2a / 3

• b = 6 + 2a / 3

• b = 2 - 2a / 3

• b = 2 + 2a / 3

Which of the following is a key step in solving the problem?

• Finding the midpoint of PQ

• Finding the gradient of the perpendicular line

• Calculating the distance between P and Q

• Finding the equation of a parallel line