# Video: Pack 1 • Paper 1 • Question 18

Pack 1 • Paper 1 • Question 18

02:04

### Video Transcript

In the diagram, 𝐴𝐵𝐶𝐷 is a square. The equation of the side 𝐴𝐵 is 𝑦 equals one-third 𝑥 plus five and the coordinates of point 𝐴 are three, six. Work out the equation of the side 𝐴𝐷.

As the equation 𝑦 equals one-third 𝑥 plus five is in the form 𝑦 equals 𝑚𝑥 plus 𝑐, its gradient must be one-third. As 𝐴𝐵𝐶𝐷 is a square, the lines 𝐴𝐵 and 𝐴𝐷 are perpendicular. If two lines are perpendicular, the product of their gradients is equal to minus one: 𝑚 one multiplied by 𝑚 two is equal to minus one.

In this case, one-third multiplied by 𝑚 two is equal to minus one. Multiplying both sides of this equation by three gives us 𝑚 two is equal to minus three. Therefore, the gradient of 𝐴𝐷 is minus three. As the gradient is equal to minus three, the equation of 𝐴𝐷 must be 𝑦 equals minus three 𝑥 plus 𝑐.

As this line passes through the point 𝐴 with coordinates three, six, we can substitute these values into the equation to calculate 𝑐. Substituting in the coordinates gives us six is equal to minus three multiplied by three plus 𝑐. Minus three multiplied by three is equal to minus nine. Adding nine to both sides of this equation gives us a value of 𝑐 equal to 15.

This means that the equation of 𝐴𝐷 is 𝑦 equals minus three 𝑥 plus 15.