Higher June 2024 Paper 2 Q22
22 \(A\) and \(B\) are points on a circle, centre \(O\).
\(MAP\) and \(NBP\) are tangents to the circle.
Prove that \(AP = BP\) (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| Proof | C1 | for angle \(OAP\) = angle \(OBP = 90\) angle between radius and tangent is 90° |
| C1 | for \(OA = OB\) both radii | |
| C1 | \(OP\) is common | |
| C1 | (dep on C3) for a complete proof with all reasons given, eg triangles \(OAP\) and \(OBP\) are congruent RHS so \(AP = BP\) or \(AP = \sqrt{OP^2 - OA^2}\) and \(BP = \sqrt{OP^2 - OB^2}\) and \(OA = OB\) so \(AP = BP\) |
Additional guidance
May be seen as 2 separate diagrams
All reasons given must be clearly linked to the appropriate statement
Underlined words need to be shown
Alternative
| Answer | Mark | Mark scheme |
|---|---|---|
| Proof | C1 | for \(\cos AOP = \dfrac{AO}{OP} = \dfrac{r}{OP}\) or \(\cos BOP = \dfrac{BO}{OP} = \dfrac{r}{OP}\) angle between radius and tangent is 90° |
| C1 | for \(\cos AOP = \dfrac{AO}{OP} = \dfrac{r}{OP}\) and \(\cos BOP = \dfrac{BO}{OP} = \dfrac{r}{OP}\) and \(AOP = BOP\) both radii | |
| C1 | \(OP\) is common | |
| C1 | (dep on C3) for a complete proof with all reasons given, eg triangles \(OAP\) and \(OBP\) are congruent SAS so \(AP = BP\) |
May be seen as 2 separate diagrams
Underlined words need to be shown