# Analysing Triangle Vertices and Bisectors

Extracts from this document...

Introduction

Part 1 The diagram shows a triangle with vertices O(0,0), A(2,6) and B(12,6). The perpendicular bisectors of OA and AB meet a C. (a) In order to write down the perpendicular bisector of the line joining the points A(2,6) and B(12,6), I need to find the line's mid-point. The mid-point of the line joining P(x1, y1) to Q(x2, y2) has the co-ordinates ( ) So the co-ordinates of the midpoint of AB are ( ) = (7,6) As the two points A(2,6) and B(12,6) have the same y-value, the gradient of the line joining the points is 0. This means that the line's perpendicular bisector also has a gradient of 0. Thus the equation of the bisector is x = 7 (b) To find the equation of the perpendicular bisector of the line joining the points O (0,0) and A(2,6), I again need to find the co-ordinates of the mid-point of OA. The gradient, and hence that of the perpendicular bisector, can also be found. Thus, knowing the gradient of the perpendicular bisector and one point on it, I can use y - y1 = m(x - x1) ...read more.

Middle

to B(12,6) = [(12 - 0)� + (6 - 0)�] = (144 + 36) = 180 The length from point A(2,6) to B(12,6) = [(12 - 2)� + (6 - 6)�] = 100 = 10 OA = 40 OB = 180 AB = 10 By knowing these three sides, I can determine an angle between two of them by rearranging the cosine rule into terms of cos A: a� = b� + c� - 2bc cos A 2bc cos A = b� + c� - a� cos A = By labelling OA, AB and OB with a, b and c respectively, I can deduce the angle ABO, labelled A on the diagram. Part 1 (f) Using cos A = cos A = = If cos A = then A = By knowing two sides of the triangle and the angle between them, I can find out the area of the triangle using the equation Area = 1/2 bc sin A Area = 1/2 bc sin A = 1/2 x 10 x 180 x sin 26.56505118� = 30 Area of triangle OAB = 30 (g) ...read more.

Conclusion

(d) To determine whether or not the two tangents to the curve y = 2x� - 9x from their point of intersection are equal in length, I again need to use the equation that gives the distance between two points - as mentioned in part 1(d) and part 1(f): The length of the line joining the point P(x1, y1) to Q(x2, y2) is given by [(x2 - x1)� + (y2 - y1)�] Firstly I shall find the length between the point (1,-7) on the curve and the point of intersection (2.5, -14.5) - as found in part 2(c): [(2.5 - 1)� + (-14.5 + 7)�] = 58.5 = 7.65(3 s.f.) I shall compare this to the length between the point (4, -4) on the curve and the point (2.5,-14.5) [(2.5 - 4)� + (-14.5 + 4)�] = 112.5 = 10.6(3 s.f.) As the length between the point (4,-4) and the point (2.5,-14.5) > the length between the point (1,-7) and the point (2.5,-14.5), the two tangents to the curve y = 2x� - 9x from their point of intersection (2.5,-14.5) are NOT equal in length. ...read more.

This student written piece of work is one of many that can be found in our GCSE Gradient Function section.

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month