June 2016 Paper 2 Q9
9 \(PQ\) is parallel to \(BC\).
Not drawn accurately
Prove that \(\quad x = 3y\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 \(PBC\) in terms of \(x\) and in terms of \(y\) | ||
| \(180 - x\) or \(360 - (y + 100 + 2y + 80)\) or \(360 - (3y + 180)\) | M1 | May be on diagram \(PBC\) (allow \(B\)) |
| \(180 - x\) and \(360 - (y + 100 + 2y + 80)\) or \(360 - (3y + 180)\) | M1 | |
| \(180 - x\) and \(180 - 3y\) and \(x = 3y\) | A1 | Must have seen correct working for M1 M1 |
| Both reasons given | A1 | Must have M1 M1 (Co-)interior angles or allied angles (add up to 180°) and angles at a point (add up to 360°) |
| Alternative method 2 \(PBC\) in terms of \(x\) + reflex \(PBC\) in terms of \(y\) = 360 | ||
| \(180 - x\) or \(y + 100 + 2y + 80\) or \(3y + 180\) | M1 | May be on diagram \(PBC\) (allow \(B\)) or reflex \(PBC\) |
| \(180 - x + y + 100 + 2y + 80 = 360\) or \(180 - x + 3y + 180 = 360\) | M1 | oe unsimplified correct equation |
| Simplifies to \(x = 3y\) | A1 | Must have seen correct working for M1 M1 |
| Both reasons given | A1 | Must have M1 M1 (Co-)interior angles or allied angles (add up to 180°) and angles at a point (add up to 360°) |
| Alternative method 3 \(x = 180 - PBC\) in terms of \(y\) | ||
| \(360 - (y + 100 + 2y + 80)\) or \(360 - (3y + 180)\) | M1 | May be on diagram \(PBC\) (allow \(B\)) |
| \(x = 180 - (360 - (y + 100 + 2y + 80))\) \(x = 180 - (360 - (3y + 180))\) | M1 | oe unsimplified correct equation |
| Simplifies to \(x = 3y\) | A1 | Must have seen correct working for M1 M1 |
| Both reasons given | A1 | Must have M1 M1 (Co-)interior angles or allied angles (add up to 180°) and angles at a point (add up to 360°) |
| Alternative method 4 \(x + PBC = 180\) and reflex \(PBC\) in terms of \(y + PBC = 360\) | ||
| \(x + PBC = 180\) or \(y + 100 + 2y + 80 + PBC = 360\) or \(3y + 180 + PBC = 360\) | M1 | \(PBC\) (allow \(B\)) |
| \(x + PBC = 180\) and \(y + 100 + 2y + 80 + PBC = 360\) or \(3y + 180 + PBC = 360\) | M1 | |
| \(x + PBC = 180\) and \(3y + PBC = 180\) and \(x = 3y\) | A1 | Must have seen correct working for M1 M1 |
| Both reasons given | A1 | Must have M1 M1 (Co-)interior angles or allied angles (add up to 180°) and angles at a point (add up to 360°) |
| Alternative method 5 (Produces \(CB\) to \(X\)) \(PBX\) = reflex \(PBC\) in terms of \(y - 180\) | ||
| \(y + 100 + 2y + 80\) or \(3y + 180\) | M1 | May be on diagram reflex \(PBC\) |
| \(y + 100 + 2y + 80 - 180\) or \(3y + 180 - 180\) | M1 | \(PBX\) |
| Simplifies to \(3y\) and states \(x = 3y\) | A1 | Must have seen correct working for M1 M1 |
| Both reasons given | A1 | Must have M1 M1 Angles on a (straight) line (add up to 180) and alternate angles (are equal) |
| Alternative method 6 (Produces \(PB\) to \(Y\)) \(CBY\) = reflex \(PBC\) in terms of \(y - 180\) | ||
| \(y + 100 + 2y + 80\) or \(3y + 180\) | M1 | May be on diagram reflex \(PBC\) (allow reflex \(B\)) |
| \(y + 100 + 2y + 80 - 180\) or \(3y + 180 - 180\) | M1 | \(CBY\) |
| Simplifies to \(3y\) and states \(x = 3y\) | A1 | Must have seen correct working for M1 M1 |
| Both reasons given | A1 | Must have M1 M1 Angles on a (straight) line (add up to 180) and corresponding angles (are equal) |
| Alternative method 7 (Produces \(CB\) to \(X\)) \(PBX = 180 - PBC\) in terms of \(y\) | ||
| \(360 - (y + 100 + 2y + 80)\) or \(360 - (3y + 180)\) | M1 | May be on diagram \(PBC\) (allow \(B\)) |
| \(180 - (360 - (y + 100 + 2y + 80))\) or \(180 - (360 - (3y + 180))\) | M1 | \(PBX\) |
| Simplifies to \(3y\) and states \(x = 3y\) | A1 | Must have seen correct working for M1 M1 |
| All reasons given | A1 | Must have M1 M1 Angles at a point and angles on a (straight) line (add up to 180) and alternate angles (are equal) |
| Alternative method 8 (Produces \(PB\) to \(Y\)) \(CBY = 180 - PBC\) in terms of \(y\) | ||
| \(360 - (y + 100 + 2y + 80)\) or \(360 - (3y + 180)\) | M1 | May be on diagram \(PBC\) (allow \(B\)) |
| \(180 - (360 - (y + 100 + 2y + 80))\) or \(180 - (360 - (3y + 180))\) | M1 | \(CBY\) |
| Simplifies to \(3y\) and states \(x = 3y\) | A1 | Must have seen correct working for M1 M1 |
| All reasons given | A1 | Must have M1 M1 Angles at a point and angles on a (straight) line (add up to 180) and corresponding angles (are equal) |
Additional guidance
Recovery of brackets is not allowed as it is a proof
Acceptable reasons must include the word ‘angles’
Angles at a point can be angles round a point
These reasons are not allowed:
Alternating angles, alternative angles, angles in a circle, straight line,
at a point, round a point, parallel lines
Supplementary angles are not allowed unless accompanied by an acceptable reason eg Allied angles are supplementary is allowed
Other variations on these methods will be seen. Escalate if necessary.
Starting with \(x = 3y\) or substituting values for \(x\) and \(y\) is zero unless M marks seen in working