Analysing Triangle Vertices and Bisectors
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) (where m is the gradient) to obtain the required equation.
Co-ordinates of the mid-point of OA are ( ) = (1,3)
The gradient of the line joining the points O (0,0) and A(2,6) is a measure of the steepness of the line OA and it is the ratio of the change in the y co-ordinate to the change in the x co-ordinate
in going from O to A.
Thus, the gradient of OA = = = 3
If two lines are perpendicular, the product of their gradients is -1. This condition for perpendicular lines means that is one line has a gradient of m, a perpendicular line will have gradient
It follows that if the gradient of OA = 3, then the gradient of the perpendicular bisector of OA =
The perpendicular bisector has gradient and passes through the point (1,3).
So using y - y1 = m(x - x1) (where m is the gradient), I can find the equation of the line:
The equation of the perpendicular bisector of OA is 3y = 10 - x
Part 1
(c) The perpendicular bisectors of the lines OA and AB intersect at C.
Since the point of intersection has co-ordinates that satisfy both equations, it is possible to substitute one of the original equations into the other to show the co-ordinates of C.
Perpendicular bisector of AB is x = 7
Perpendicular bisector of OA is 3y = 10 - x
Substituting x = 7 into 3y = 10 - x gives 3y = 10 - 7
or 3y = 3
or y = 1
Hence showing that the co-ordinates of ...
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(c) The perpendicular bisectors of the lines OA and AB intersect at C.
Since the point of intersection has co-ordinates that satisfy both equations, it is possible to substitute one of the original equations into the other to show the co-ordinates of C.
Perpendicular bisector of AB is x = 7
Perpendicular bisector of OA is 3y = 10 - x
Substituting x = 7 into 3y = 10 - x gives 3y = 10 - 7
or 3y = 3
or y = 1
Hence showing that the co-ordinates of C are (7,1)
(d)
To find the equation of the circle with centre at point C and which passes through the points O(0,0), A(2,6) and B(12,6), the following equation can be used:
(x - a)² + (y - b)² = r²
where r is the radius of the circle and the point (a,b) is at its centre.
As the co-ordinates of the centre of the circle, point C, are already known from part (c) to be (7,1), all that is needed to find the equation of the circle is its radius.
The length of the line joining the point P(x1, y1) to Q(x2, y2) is given by
[(x2 - x1)² + (y2 - y1)²]
As the radius of the circle is the length of the line joining centre point C (7,1) to A (2,6), the above equation can be used to find the radius:
r = [(2 - 7)² + (1 - 6)²]
r² = (-5)² + (-5)² = 50
r = 50
The circle has centre (7,1) and a radius of 50; the equation of the circle can now be found:
Using (x - a)² + (y - b)² = r²
(x - 7)² + (y - 1)² = 50
x² - 14x + 49 + y² - 2y + 1 = 50
x² + y² - 14x - 2y + 50 = 50
x² + y² - 14x - 2y = 0
Part 1
(e) To find the y co-ordinate of the point other than O where the circle cuts the y-axis, I need to substitute the equation of the y-axis line (x = 0) into the equation for the circle (x² + y² - 14x - 2 = 0)
Substituting x = 0
into x² + y² - 14x - 2 = 0
gives y² - 2 = 0
which, when factorised, gives y(y - 2) = 0
resulting in y = 0 or y = 2
Since point O has the y co-ordinate of 0, y = 2 must be the y co-ordinate of the point other than O where the circle cuts the y-axis.
(f) To write down the area of triangle OAB, I firstly need to determine the lengths of the three sides. This can be done using the equation for the distance between two points, as mentioned in part (d):
The length of the line joining the point P(x1, y1) to Q(x2, y2) is given by
[(x2 - x1)² + (y2 - y1)²]
The length from point O(0,0) to A(2,6) = [(2 - 0)² + (6 - 0)²]
= (4 + 36)
= 40
The length from point O(0,0) to B(12,6) = [(12 - 0)² + (6 - 0)²]
= (144 + 36)
= 180
The length from point A(2,6) to B(12,6) = [(12 - 2)² + (6 - 6)²]
= 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:
a² = b² + c² - 2bc cos A
2bc cos A = b² + c² - a²
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 26.56505118º
= 30
Area of triangle OAB = 30
(g) The length of the perpendicular - labelled x on diagram - from point A to the line OB can be found by using the length of AB (10) and the angle ABO (26.6º (3s.f.)) - labelled ( on diagram.
Because the perpendicular (x) forms a right-angled triangle with hypotenuse AB and is the opposite side to the angle ABO, the trigonometric ratio
sin ( =
can be used to find the length of the perpendicular.
Hypotenuse = 10 sin 26.56505118° =
Opposite = x
( = 26.6° (3 s.f.) x = 10 x sin 26.56505118°
To show that the length of the perpendicular is 20, both sides of the equation need to squared:
x² = (10 x sin 26.56505118°)²
x² = 20
Therefore the length of the perpendicular from point A to the line OB is 20
(h) The perpendicular from point A to the line OB also forms a right-angled triangle with the side OA becoming the hypotenuse. By knowing the length of both the perpendicular and OA, the angle AOB - labelled ( on diagram - can be deduced by again using the same trigonometric ratio as in part (g):
sin ( =
sin ( = ( =
Thus the size of angle AOB = 45°
Part 2
(a) In order to find the point(s) at which a line and a curve intersect, the point(s) with co-ordinates which satisfy both equations have to be found.
This can be done by solving the equations of the line and the curve simultaneously.
The points of intersection of y = 2x² - 9x and y = x - 8 satisfy both equations.
By substituting y = x - 8 into the equation y = 2x² - 9x gives
x - 8 = 2x² - 9x
0 = 2x² - 10x + 8 ( (2x -2)(x - 4) = 0
giving x - 4 = 0 or 2x - 2 = 0
x = 4 x = 1
Substituting for x in y = x - 8 gives y = 4 - 8 or y = 1 - 8
y = -4 y = -7
( the points of intersection of y = 2x² - 9x and y = x - 8 are (1,-7) and (4,-4)
(b) To find for the curve y = 2x² - 9x, the rule for differentiating y = axn has to be
employed. This rule is = anxn - 1
If y = 2x² - 9x then = 4x - 9
From knowing for the curve, the gradient at any point on the curve can be
established and therefore the equation of the tangent at any point can be found.
Thus at point (1,-7), x = 1 giving = 4(1) - 9 =-5
( at the point (1,-7), the curve y = 2x² - 9x has a gradient of -5.
The equation of the tangent at the point (1,-7) can now be found using y - y1 = m(x - x1) (where m is the gradient determined by differentiation) as we know the gradient and one point on the line - as mentioned in Part 1(b).
Part 2
(b) Using y - y1 = m(x - x1) y - (-7) = -5(x - 1)
y + 7 = -5x + 5
y = -5x - 2
Thus the equation of the tangent to the curve at the point (1,-7) is y = -5x - 2.
This method can be used again to find the equation of the tangent to the curve at the point (4,-4), the other point of intersection with the line y = x - 8.
= 4x - 9
= 4(4) - 9 = 7
Thus the tangent at (4,-4) has gradient 7
Using y - y1 = m(x - x1) y - (-4) = 7(x - 4)
y + 4 = 7x - 28
y = 7x - 32
Thus the equation of the tangent to the curve at the point (4,-4) is y = 7x - 32.
(c) Finding the point of intersection of the two tangents found in part 2(b) can be done by solving the two equations of the tangents simultaneously.
Since the point of intersection satisfies both equations:
y = 7x - 32 ...[1]
y = -5x - 2 ...[2]
subtracting equation [2] from [1] gives 0 = 12x - 30
12x = 30
x = 2.5
substituting for x in equation [1] gives y = 17.5 - 32 = -14.5
Therefore the point of intersection of the two tangents is (2.5, -14.5)
(d) To determine whether or not the two tangents to the curve y = 2x² - 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)² + (y2 - y1)²]
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)² + (-14.5 + 7)²] = 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)² + (-14.5 + 4)²] = 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 = 2x² - 9x from their point of intersection (2.5,-14.5) are NOT equal in length.