Download - Geometry 2 Tutorial Solutions Q1
Internal
Geometry 2 Tutorial β Solutions
Q1.
a) Find the slope using slope formula
π =(π¦2βπ¦1)
(π₯2βπ₯1) =
(β6β2)
(6β2) =
β8
4 = β2
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 β 6 = 0
π«πππππππ =|πππ+πππ+π|
β(ππ+ππ) =
|π(βπ)+π(βπ)βπ|
β(ππ+ππ)=
|βππ|
βπ =
ππ
βπ
d) Area of Triangle is Β½ x base x perpendicular height
Perpendicular height is 13
β5 from last question
Distance between A(2,2) and D(0,6) is the base
β(0 β 2)2 + (6 β 2)2
β4 + 16 = β20
Area = Β½ * β20* 13
β5 = 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
β(π β π)π + (π β π)π
β(βπ)π + (βπ)π = βππ + ππ=βππ
c) Equation of circle with centre (h,k) and radius r is: (π₯ β β)2 + (π¦ β π)2 = π2
(π₯ β 2)2 + (π¦ β 3)2 = β722
=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
β(β5 β 0)2 + (3 β 0)2
β52 + 32
β34
Internal
d) π₯Β² + π¦Β² = πΒ²
π₯Β² + π¦Β² = (β34)Β²
π₯Β² + π¦Β² = 34
e) Area = ΟπΒ²
Οβ342
34 Ο
106.81 units squared
f)
Side of square is equal to the diameter, area of square is diameter Β²
Diameter = 2r = 2β34
Area of square = (2β34 ) Β²
136 units squared
Internal
Q4.
a) Solution circle c1
π₯2 + π¦2 + 2ππ₯ + 2ππ¦ + π = 0
π₯2 + π¦2 β 6π₯ β 10π¦ + 29 = 0 (Formulae p. 19)
2g = -6
-g = 3
2f = -10
-f = 5
centre (3, 5)
Radius formula (p. 19)
β(g2 + f2 β c), c = 29
β32 + 52 β 29
β9 + 25 β 29
β5
Solution circle c2
π₯2 + π¦2 β 2π₯ β 2π¦ β 43 = 0
2g = -2
-g = 1
2f = -2
-f = 1
centre (1,1)
Radius formula
β12 + 12 + 43
β45
β9 β 5
3β5
Internal
b) Distance between centres:
β(3 β 1)2 + (5 β 1)2
β20
2β5
The distance between the centres is the difference of the radii => circles touch (internally).
c) Substitute (4,7) into cβ and cβ
42+ 72 β 6(4) β 10(7) + 29 = 0
(4, 7) is point on cβ
42+ 72 β 2(4) β 2(7) β 43 = 0
(4, 7) is point on cβ
Circles touch internally at (4,7)
d)
Internal
Slope from (3, 5) to (4, 7) is: 7β5
4β3 = 2
Slope of tangent = β1
2 (as perpendicular)
Equation of tangent with slope β1
2 and point(4,7)
π¦ β 7 = β1
2 (π₯ β 4)
2π¦ β 14 = βπ₯ + 4
π₯ + 2π¦ β 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 β 2 =1
2(π₯ β 3.5)
2y β 4 = π₯ β 3.5
x β 2y = β0.5
x + 3y = 12 point of intersection is centre of circle
Internal
β5y = β12.5 subtracting
y = 2.5; π₯ = 4.5
Centre is (4.5,2.5)
Now, need to find the radius
Distance between (4.5,2.5) and (3,3)
r = β(3 β 4.5)2 + (3 β 2.5)2
r = β2.5
Equation of the circle:
(π₯ β 4.5)2 + (π¦ β 2.5)2 = 2.5
π₯2 + π¦2 β 9π₯ β 5π¦ + 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 = ΒΎ
slope of perpendicular = -(4/3)
Equation of perpendicular: slope -4/3, point (-2,2)
y β 2 = β(4
3)(x β (β2))
3y β 6 = β4x β 8
4x + 3y = β2
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 β 1.5 = 7(x β 1.5)
7x β y = 9
Simultaneous equations
7x β y = 9
4x + 3y = β2
21x β 3y = 27
Adding: 25x = 25;
x = 1 and y = β2
centre (1, β 2)
So, centre (1,-2)
Internal
Radius is distance between (1,-2) and (5,1)
r = β(5 β 1)2 + (1 β (β2))2
r = β16 + 9
r = 5
Equation of circle: (π₯ β 1)2 + (π¦ β (β2))2 = 52
π₯2 + π¦2 β 2π₯ + 4π¦ β 20 = 0
Q7.
a)