### Video Transcript

In the diagram, π΄π΅πΆπ· is a
square. The equation of the side π΄π΅ is π¦
equals one-third π₯ plus five and the coordinates of point π΄ are three, six. Work out the equation of the side
π΄π·.

As the equation π¦ equals one-third
π₯ plus five is in the form π¦ equals ππ₯ plus π, its gradient must be
one-third. As π΄π΅πΆπ· is a square, the lines
π΄π΅ and π΄π· are perpendicular. If two lines are perpendicular, the
product of their gradients is equal to minus one: π one multiplied by π two is
equal to minus one.

In this case, one-third multiplied
by π two is equal to minus one. Multiplying both sides of this
equation by three gives us π two is equal to minus three. Therefore, the gradient of π΄π· is
minus three. As the gradient is equal to minus
three, the equation of π΄π· must be π¦ equals minus three π₯ plus π.

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

This means that the equation of
π΄π· is π¦ equals minus three π₯ plus 15.