geometry 2 tutorial solutions q1
TRANSCRIPT
Internal
Geometry 2 Tutorial โ Solutions
Q1.
a) Find the slope using slope formula
๐ =(๐ฆ2โ๐ฆ1)
(๐ฅ2โ๐ฅ1) =
(โ6โ2)
(6โ2) =
โ8
4 = โ2
Slope = -2, A = (2,2)
Equation of AB: y - 2 = -2 * (x - 2)
y - 2 = -2x +4
2x + y = 6
b) Line intersects the y-axis means that x=0 at this point
Substitute x=0 into Equation of AB: 2x + y = 6
y = 6
D = (0,6)
Internal
c) Formula in Tables Page 19
C = (-2, -3) and AB = 2x + y = 6 or 2x + y โ 6 = 0
๐ซ๐๐๐๐๐๐๐ =|๐๐๐+๐๐๐+๐|
โ(๐๐+๐๐) =
|๐(โ๐)+๐(โ๐)โ๐|
โ(๐๐+๐๐)=
|โ๐๐|
โ๐ =
๐๐
โ๐
d) Area of Triangle is ยฝ x base x perpendicular height
Perpendicular height is 13
โ5 from last question
Distance between A(2,2) and D(0,6) is the base
โ(0 โ 2)2 + (6 โ 2)2
โ4 + 16 = โ20
Area = ยฝ * โ20* 13
โ5 = 13 units squared
Note: To use formula for area of triangle on page 18 it is necessary to have one apex at the origin.
Q2.
a)
Internal
b) Distance between (8,9) and (2,3) is radius
โ(๐ โ ๐)๐ + (๐ โ ๐)๐
โ(โ๐)๐ + (โ๐)๐ = โ๐๐ + ๐๐=โ๐๐
c) Equation of circle with centre (h,k) and radius r is: (๐ฅ โ โ)2 + (๐ฆ โ ๐)2 = ๐2
(๐ฅ โ 2)2 + (๐ฆ โ 3)2 = โ722
=72
Q3.
a)
b) Midpoint of A(-5,3) and B(5,-3) is the centre of the circle
Midpoint is (0,0)
c) Distance between (-5,3) and (0,0) is radius
โ(โ5 โ 0)2 + (3 โ 0)2
โ52 + 32
โ34
Internal
d) ๐ฅยฒ + ๐ฆยฒ = ๐ยฒ
๐ฅยฒ + ๐ฆยฒ = (โ34)ยฒ
๐ฅยฒ + ๐ฆยฒ = 34
e) Area = ฯ๐ยฒ
ฯโ342
34 ฯ
106.81 units squared
f)
Side of square is equal to the diameter, area of square is diameter ยฒ
Diameter = 2r = 2โ34
Area of square = (2โ34 ) ยฒ
136 units squared
Internal
Q4.
a) Solution circle c1
๐ฅ2 + ๐ฆ2 + 2๐๐ฅ + 2๐๐ฆ + ๐ = 0
๐ฅ2 + ๐ฆ2 โ 6๐ฅ โ 10๐ฆ + 29 = 0 (Formulae p. 19)
2g = -6
-g = 3
2f = -10
-f = 5
centre (3, 5)
Radius formula (p. 19)
โ(g2 + f2 โ c), c = 29
โ32 + 52 โ 29
โ9 + 25 โ 29
โ5
Solution circle c2
๐ฅ2 + ๐ฆ2 โ 2๐ฅ โ 2๐ฆ โ 43 = 0
2g = -2
-g = 1
2f = -2
-f = 1
centre (1,1)
Radius formula
โ12 + 12 + 43
โ45
โ9 โ 5
3โ5
Internal
b) Distance between centres:
โ(3 โ 1)2 + (5 โ 1)2
โ20
2โ5
The distance between the centres is the difference of the radii => circles touch (internally).
c) Substitute (4,7) into cโ and cโ
42+ 72 โ 6(4) โ 10(7) + 29 = 0
(4, 7) is point on cโ
42+ 72 โ 2(4) โ 2(7) โ 43 = 0
(4, 7) is point on cโ
Circles touch internally at (4,7)
d)
Internal
Slope from (3, 5) to (4, 7) is: 7โ5
4โ3 = 2
Slope of tangent = โ1
2 (as perpendicular)
Equation of tangent with slope โ1
2 and point(4,7)
๐ฆ โ 7 = โ1
2 (๐ฅ โ 4)
2๐ฆ โ 14 = โ๐ฅ + 4
๐ฅ + 2๐ฆ โ 18 = 0
Q5.
The perpendicular bisector of any chord is a line containing the centre of the circle
End points of chord: (3,3) and (4,1)
Midpoint of Chord: (3.5, 2)
slope of chord: (1-3)/(4-3) = -2
slope of perpendicular of chord: 1
2
Eqn of perpendicular of chord: slope = 1
2, point (3.5,2)
y โ 2 =1
2(๐ฅ โ 3.5)
2y โ 4 = ๐ฅ โ 3.5
x โ 2y = โ0.5
x + 3y = 12 point of intersection is centre of circle
Internal
โ5y = โ12.5 subtracting
y = 2.5; ๐ฅ = 4.5
Centre is (4.5,2.5)
Now, need to find the radius
Distance between (4.5,2.5) and (3,3)
r = โ(3 โ 4.5)2 + (3 โ 2.5)2
r = โ2.5
Equation of the circle:
(๐ฅ โ 4.5)2 + (๐ฆ โ 2.5)2 = 2.5
๐ฅ2 + ๐ฆ2 โ 9๐ฅ โ 5๐ฆ + 24 = 0
Q6.
a)
b) Need to find the centre of the circle to find the equation of the circle
Know: The radius is perpendicular to the tangent at the point of intersection
Know: The perpendicular bisector of any chord is a line containing the centre of the circle
Internal
The point of intersection of these two lines is the centre of the circle
Tangent: 3x -4y+14=0
Slope of tangent = ยพ
slope of perpendicular = -(4/3)
Equation of perpendicular: slope -4/3, point (-2,2)
y โ 2 = โ(4
3)(x โ (โ2))
3y โ 6 = โ4x โ 8
4x + 3y = โ2
To find perpendicular bisector of chord:
Midpoint (-2,2) and (5,1): (1.5, 1.5)
Slope of (-2,2) and (5,1): (1-2)/(5-(-2))
Slope: -(1/7)
Slope of perpendicular = 7
Equation of perpendicular: slope 7, point (1.5,1.5)
y โ 1.5 = 7(x โ 1.5)
7x โ y = 9
Simultaneous equations
7x โ y = 9
4x + 3y = โ2
21x โ 3y = 27
Adding: 25x = 25;
x = 1 and y = โ2
centre (1, โ 2)
So, centre (1,-2)
Internal
Radius is distance between (1,-2) and (5,1)
r = โ(5 โ 1)2 + (1 โ (โ2))2
r = โ16 + 9
r = 5
Equation of circle: (๐ฅ โ 1)2 + (๐ฆ โ (โ2))2 = 52
๐ฅ2 + ๐ฆ2 โ 2๐ฅ + 4๐ฆ โ 20 = 0
Q7.
a)