Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

Maths > Co-ordinate geometry > Flashcards

Co-ordinate geometry Flashcards

1
Q

Find the midpoint of (4,10) (-2,6)

A

= (4-2/2, 10 +6/2) = (1, 8)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
1
Q

Midpoint

A

= (x1 + x2/2, y1+y2/2)
(average of the two co-ordinates)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Find the midpoint of (-3,7/2) (-5/2, - 9)

A

= (-3 - 5/2/2, 7/2 - 9/2) = (-11/4, - 11/4

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

Finding the distance between two points.

A

= √(x2 - x1)^2 + (y2 - y1)^2
(we want to find the hypotenuse D = distance, try drawing a triangle to help you, where we want to find bae and height)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Find the distance between (-3,7) (9,12)

A

d = √(-3 - 9)^2 + (7 - 12) ^6 = √169
= 13

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Find the distance between (3,6) (4,7)

A

d = √(3-4)^2 + (6-7)^2 = √2

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Find the distance between (3/7,4) (2,- 5/7)

A

d = √(3/7 - 2)^2 + (4+5/7)^2
= √1210/4 = 11√10/7 = 11/7 √10

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Find the gradient of a line

A

m = y2 - y1/x2 - x1
or y1 - y2/x1 - x2(unsure though ask miss if you can do it this way)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Find the gradient of the points (2, - 3) (5,12)

A

12 - -3/5 -2 = -15/-3 = 5

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

find the gradient of the points (3,8) (1,1)

A

m = 1 -8/1-3 = -7/-2 = 7/2

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

The equations of a line

A

y = mx + c and y - y1 = m(x - x1)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

Method for y - y1 = m(x -x1)

A

Find the gradient of the two points.
Place in the x1 y1 points.
Multiply out the bracket and solve to get it in the form y = mx + c.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Find the equation of the line passing through (8,2) and (5, -7)

A

m = 2 - -7/8 -5 = 9/3 = 3
y -2 = 3(x-8) => y = 3x -22

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Find the equation of the line passing through (-4,7) (5,10)

A

m = 7 -10/-4 - 5 = -3/-9 = 1/3
y -7 +1/3(x+4) => y =1/3x + 25/3

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Finding the equation of a line in the form ax+by+c = 0

A

Find the gradient of the points.
Use the equation y - y1 = m(x-x1).
Don’t put it in the form y = mx + c
Multiply out only the brackets.
Then place everything on the left side and boom done.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Find the equation of the line passing through (3, - 5) (-7,2) in the form ax + by + c =0

A

m = -5 - 2/3 - -7 = -7/10
y + 5 = -7/10(x-3)
(get rid of brackets x by 10)
10y + 50 = -7(x - 3)
10y + 50 = -7x + 21
=> 7x + 10y + 29 = 0

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Are these two lines Parallel, perpendicular or Neither?
y = 3x - 7
6x - 2y =5

A

(If the two lines are parallel they have the same gradient)
(If the two lines are perpendicular then there gradients multiplied together make -1)
y = 3x -7, m1 = 3
6x - 2y = 5 = -2y = 5 -6x
= y = -5/2 + 3x, m2 = 3
These lines are parallel, there gradients are the same,

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

Parallel, Perpendicular or Neither?
y = 8 - 5x
10 + 5y = x

A

y = 8 - 5x, m1 = -5
10 + 5y = x = 5y = x -10
=>y = 1/5x - 2, m2 = 1/5
m1 x m2 = -5 x 1/5 = -1
These lines are perpendicular, there gradients multiplied together make -1.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

Negative reciprocal

A

m2 = - 1/m1

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

Find the negative reciprocal of 3

A

= - 1/3

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

Find the negative reciprocal of -5

A

= 1/5

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

Find the negative reciprocal of 1/7

A

= - 7/1 = -7

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

Find the negative reciprocal of
- 1/11

A

= 11/1 = 11

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

Find the negative reciprocal of -0.6

A

= 5/3

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

Find the negative reciprocal of 3/8

A

= - 8/3

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

Find the equation of the line that is
parallel to the line 3x + 2y + 7 = 0, passing through ( -2,4)

A

Find the gradient of the line.
3x + 2y + 7 = 0
= 2y = - 3x - 7
= y = -3/2x - 7/2
The line we need is parallel to that line so it has the same gradient -3/2x then we place into the form
y - y1 = m(x-x1).
= y - 4 = -3/2(x + 2)
=y - 4 = -3/2x - 3
= y = -3/2x + 1

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

Find the equation of a line that is perpendicular to the line 3x + 4y = 5, passing through (3, - 2)

A

Rearrange 3x + 4y = 5 to get it in the from y =.
3x + 4y = 5
=> 4y = 5 - 3x
=>y = -5/4 - 3/4x
Since it is perpendicular and m1 x m2 = -1 then we the gradient of the line is the negative reciprocal of the gradient of the other line.
-3/4x = 4/3
Then place it in y - y1 = m(x-x1).
y + 2 = 4/3(x-3)
y+2 = 4/3x - 4
y = 4/3x - 6

26
Q

sketch y = 3x - 7

A

When sketching we need to find the x and y interecepts of the line then sketch the graph.
When x = 0(for y intercept), y = 3x0 - 7 = -7
When y = 0, 0 = 3x - 7
=> 7 = 3x => 7/3 = x
Then plot those points.

27
Q

Sketch 2x + 7y = 14

A

When x = 0 = > 7y = 14, y =2
When y = 0, 2x = 14, x = 7
Then sketch the graph

28
Q

Perpendicular bisector

A

Line that is perpendicular to the line joining AB that will cut the line in half cutting through the midpoint of AB, Its best to draw a diagram to help you.

29
Q

Perpendicular bisector of A(3,8) and B(-9,4)

A

First we find the midpoint
(3-9/2, 8+4/2) = (-3,6)
We need to find the equation of the perpendicular bisector, cause its perpendicular its gradient will be the negative reciprocal of the other line, as m1 x m2 = -1, m2 = - 1/m1
m = 8 - 4/ 3- -9 = 4/12 = 1/3 so the gradient of the perpendicular bisector is the negative reciprocal of 1/3 = - 3/1 = -3.
So place it in the equation of a line.
y- 6 = -3(x+3)
y - 6 = -3x - 9
y = -3x -3

30
Q

Find the equation of the perpendicular bisector of (2,7) (-6,9)

A

midpoint (2-6/2, 7+9/2) = (-2, 8)
gradient m = 7-9/2- -6 = -2/8 = -1/4
-1/4 = Negative reciprocal = 4
y - 8 = 4(x+2)
y = 4x + 16

31
Q

Find the equation of the perpendicular bisector of
(5, -3) (8, - 18)

A

midpoint(5+8/2, -3-18/2) =
(13/2, -21/2)
gradient m = -3 - -18/5-8 = 15/-3 = -5
Negative reciprocal of -5 = 1/5
y + 21/2 = 1/5(x - 13/2)
5y + 105/2 = x -13/2
5y = x - 59
y = 1/5x - 59/5

32
Q

Find the equation of perpendicular bisector of (-2, - 9) (1/2, 1/3)

A

midpoint (-2 + 1/2/2, -9 + 1/3)
= (-3/4, -13/3)
gradient m = -9 - 1/3/-2 -1/2 = 56/15
negative reciprocal of 56/15 = -15/56
y + 13/3 = -15/56(x + 3/4)
y + 13/3 = -15/56x - 45/224
y= - 15/56x - 3047/672

33
Q

Finding where two lines intersect

A

Solve the two equations of the lines simultaneously

34
Q

Find where the line 2x + 7y = 28 and y =3x -1 intersect.

A

2x + 7(3x - 1) = 28
2x + 21x - 7 =28
23x = 35
x= 35/23
sub into 2
y = 3(35/23) -1
y= 105/23 - 23/23 = 82/23

35
Q

Types of quadrilaterals and there properties

A

Trapezium properties -one pair of parallel sides.
Parallelogram properties-2 pairs of parallel sides.
Rhombus properties-all sides the same length.
Square-properties-all angles the same = 90 degrees.
A trapezium is a special case quadrilateral.
A parallelogram is a special case Trapezium.
A rhombus is a special case Parallelogram.
A square is a special case Rhombus.
or a rectangle all angles the same = 90 degrees is a special case parallelogram.
Square can be a special case Rectangle.
The kite is a special case Rhombus as they have the similar property of there diagonals meeting at right angels.

36
Q

The equation of a circle

A

(x - h)^2 + (y - k)^2 = r^2
Circle centre (h,k)
with radius r

37
Q

Identify the centre and radius of x^2+y^2 = 9

A

Centre (0,0) radius √9 = 3

38
Q

Identify the centre and radius of
x^2 + (y-3)^2 = 16

A

centre(0,3)
radius √16 = 4

39
Q

Identify the centre and radius of
(x-2)^2 + y^2 = 81

A

Centre (2,0) radius √81 = 9

40
Q

Identify the centre and radius of
(x-3)^2 + (y+5)^2 = 144

A

Centre(3,-5) radius 12

41
Q

Sketch the circle x^2 + (y-2)^2 =4

A

Centre (0,2)
Radius =2
the radius is also the centre so the circle must be touching the origin.

42
Q

Sketch the circle (x-5)^2 + (y-5)^2 = 25

A

Centre(5,5)
Radius = 5
The Centre is both Radius so the circle must be touching the x and y axis at 5.

43
Q

Sketch (x+6)^2 + (y -2)^2 = 5

A

Centre (-6,2)
Radius = √5
Since it crosses the x axis we need to find where it crosses so we substitute in y = 0 in circle equation to find x.
(x+6)^2 + (0-2)^2 = 5
(x+6)^2 + 4 =5
(x+6)^2 = 1
x+6 = + or - 1
x= + or -1 - 6
so it crosses the x axis at -7 and -5

44
Q

solving equations to find the centre and radius of the circle

A

We can use completing the square to do this

45
Q

Find the centre and radius of the circle with equation x^2 + y ^2 - 6x +2y + 9 =0.

A

To do this we can do completing the square.
First group the x together and the y together.
x^2 - 6x + y^2 +2y + 9 =0
For x^2 - 6x we complete the square so half the 6x to get 3.
(x-3)^2, then we - the - 3^2 to get -9
= (x-3)^2 - 9.
Then we do the same method for the y to get:
(y+1)^2 - 1 +9, then place the whole thing together to get.
(x-3)^2 -9 + (y+1)^2 - 1 +9 = 0.
The 9s cancel out and the one taken to the right side to become +1.
= (x-3)^2 + (y+1)^2 = 1.

46
Q

Find the centre and radius of
x^2 + 6x + y^2 - 4y -12 = 0

A

Complete the square.
(x+3)^2-9 + (y-2)^2-4 - 12 = 0
(x+3)^2 + (y-2)^2 = 25
Centre(-3,2)
Radius 5

47
Q

Find the centre and radius of
x^2 - 12x + y^2 - 8y + 49 = 0

A

(x-6)^2 - 36 + (y-4)^2 -16 +49 =0
(x-6)^2 + (y-4)^2 = 3 (remember radius is a length so it cannot be negative)
Centre(6,4)
Radius√3

48
Q

Find the centre and radius of
2x^2 + 36x + 2y^2 - 4y + 163 = 0

A

Divide through by 2 to get x.
x^2 + 18x + y^2 - 2y + 81.5 = 0
Complete the square.
(x+9)^2 - 81 + (y-1)^2 -1 + 81.5 = 0
(x+9)^2+ (y-1)^2 = 1/2
Radius = √1/2 = 1/√2 = √2/2
Centre(-9,1)

49
Q

Determine whether y = -6.1 intersects the circle (x -1)^2 + (y+3)^2 = 10
For any that do, find the coordinates of the points of intersection

A

Sub into equation of circle.
(x-1)^2 + (-61 +3)^2 = 10
(x-1)^2 + 961/100 = 10
(x-1)^2 = 39/100
x-1 = + or - √39/100 = + or - √39/10
x = 1 = + or - √39/10
Yes it intersects the circle at ( 1 - √39/10, -6.1) and(1+√39/10, -6.1)

50
Q

Determine whether y = 1/2x - 1 intersects the circle (x-1)^2 + (y+3)^2 = 10
For any that do, find the points coordinates of the points of intersection.

A

(x-1)^2 + (1/2x-1 + 3)^2 = 10
(x-1)^2 + (1/2x + 2)^2 = 10
x^2-2x+1 + 1/4x^2 + 2x + 4 = 10
5/4x^2 = 5
5x^2 = 20
x^2 =4
x = + or - 2
Intersects the circle at (2,0) and (-2, -2)

51
Q

Determine whether x = 4.4 intersects the circle (x-1)^2 + (y+3)^2 = 10.
For any that do, find the coordinates of the points of intersection.

A

(4.4 - 1)^2 + (y +3)^2 = 10
289/25 + (y+3)^2 = 10
(y+3)^2 = -39/25
no real solutions as we can not square root a negative number and end up with real values.

52
Q

Determine the coordinates of intersection of the circles
(x-3)^2 + (y+8)^2 = 100 and
(x+2)^2 + (y-3)^2 = 121

A

It might help with any question to first sketch it out.
Expand both circle equations.
x^2-6x+9 + y^2 + 16y + 64 = 100
= x^2 - 6x +y^2 + 16y - 27 = 0(1)

X^2 + 4x + 4 + y^2 - 6y + 9 = 121
= x^2 + 4x + y^2 - 6y - 108 = 0(2)
Solve simultaneously.
(1) - (2)
=> - 10x + 22y + 81 = 0
Rearrange to get y =
22y = 10x - 81
=> y = 5/11x - 81/22
Sub into either circle equations.
(x +2)^2 + (5/11x - 81/22 - 3)^2 = 121
(x+2)^2 + (5/11x - 147/22)^2 = 121
x^2 + 4x + 4 + 25/121 x^2 - 735/121x +21609/484 = 121
=> 146/121x^2 - 251/121x - 35019/484 = 0
x = 8.65, - 6.93
(8.65, 0.250) (3.sf) and (-6.93, -6.83) (3.sf)

53
Q

A(-4, 5), B(2,-1), C(8,5) all lie on a circle . Prove that AC is a diameter

A

If three points are on a circle then Points AB an BC will be at right angles which means they are perpendicular which means that AC has to be a diameter.
Mab = -1 - 5/2- -4 = - 6/6 = -1
Mbc = 5 - -1/ 8-2 = 6/6 = 1
Mab x Mbc = -1 x -1 = -1
So AB and BC are perpendicular , therefore AC must be a diameter with ABC on the circumference of a circle.

54
Q

Find the centre and radius of the circle with points A(-2,1), B(2,5), C(-10,17)

A

Find the two perpendicular bisectors and then find where they intersect.

Find the midpoint of Ab
AB midpoint = (-2 +2/2, 1+5/2) = (0,3)
Gradient = 1-5/-2 - 2 = -4/-4 = 1
Then find the perpendicular bisector (just place the negative reciprocal into equation of a line formula to find equation for perpindicular bisector line)
y - 3 = -1(x-0)
=>y = -x +3

BC midpoint (2-10/2, 5+17/2) = (-4,11)
Gradient = 5 - 17/2 - -10 = -12/12 = -1
Perpendicular bisector =
y - 11 = 1(x+4)
=>y = x+15

Find where they intersect.
-x + 3 = x +15
-12 = 2x
x = -6 => y = -6 + 15 = 9
centre(-6, 9)
Radius = √8^2 + 4^2 = √80
(x+6)^2 + (y-9)^2 = 80

55
Q

Show that y - 4-3x is a Tangent to the circle(x+5)^2 + (y-9)^2 = 10
The tangent meets the circle at A
Hence find the equation of the Normal to the circle at A

A

(x +5)^2 + (4-3x - 9)^2 = 10
(x +5)^2 + (-3x - 5)^2 = 10
x^2 + 10x + 25 + 9x^2 + 30x + 25 = 10
10x^2 + 40x + 40 = 0
x^2 + 4x + 4 = 0
(x+2)^2 = 0
x = -2 -> y = 4 -3(-2) = 10
The line intersects the circle once at (-2,10) and so it must be a tangent line.
(the normal is perpindicular to the tangent so its gradient is the negative reciprocal of the tangents.
y - 10 = 1/3(x+2)
3y - 30 = x + 2
3y = x +32
y = 1/3x + 32/3

56
Q

A circle has centre (3, - 7) The point P (2,8) lies on the circle, Find the equation of the Tangent and Normal to the circle at P.

A

Gradient of the normal.
m = -7 - 8/3 -2 = -15/1 = -15
Normal: y -8 = -15(x-2)
y = -15x + 38
Tangent: y - 8 = 1/15(x-2)
15y - 120 = x-2
15y = x + 118
y = 1/15x + 118/15

57
Q

Sketch y = x^3

A

Cant draw the answer

58
Q

Sketch y = x^3 + 1

A

The + 1 is the translation so we make that a dotted line then we use that as our base for our x axis and do it from there, also remember to not where it intercepts the x and y axis.

59
Q

Sketch y = x

A

straight line through origin

60
Q

Sketch y = x^2

A

The wide bucket shape

61
Q

Sketch y = x^4

A

bucket shape

62
Q

Sketch y = 1/x

A

it has curve in the positive x and y axis and a curve in the negative x and y axis, hard to explain look online.

63
Q

Sketch y = 1/x^2

A

This is the volcano shape
(remember that the + or - number if there is one is the translation otherwise known as asymptotes)

Maths (4 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions