Co-ordinate geometry Flashcards
Find the midpoint of (4,10) (-2,6)
= (4-2/2, 10 +6/2) = (1, 8)
Midpoint
= (x1 + x2/2, y1+y2/2)
(average of the two co-ordinates)
Find the midpoint of (-3,7/2) (-5/2, - 9)
= (-3 - 5/2/2, 7/2 - 9/2) = (-11/4, - 11/4
Finding the distance between two points.
= √(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)
Find the distance between (-3,7) (9,12)
d = √(-3 - 9)^2 + (7 - 12) ^6 = √169
= 13
Find the distance between (3,6) (4,7)
d = √(3-4)^2 + (6-7)^2 = √2
Find the distance between (3/7,4) (2,- 5/7)
d = √(3/7 - 2)^2 + (4+5/7)^2
= √1210/4 = 11√10/7 = 11/7 √10
Find the gradient of a line
m = y2 - y1/x2 - x1
or y1 - y2/x1 - x2(unsure though ask miss if you can do it this way)
Find the gradient of the points (2, - 3) (5,12)
12 - -3/5 -2 = -15/-3 = 5
find the gradient of the points (3,8) (1,1)
m = 1 -8/1-3 = -7/-2 = 7/2
The equations of a line
y = mx + c and y - y1 = m(x - x1)
Method for y - y1 = m(x -x1)
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.
Find the equation of the line passing through (8,2) and (5, -7)
m = 2 - -7/8 -5 = 9/3 = 3
y -2 = 3(x-8) => y = 3x -22
Find the equation of the line passing through (-4,7) (5,10)
m = 7 -10/-4 - 5 = -3/-9 = 1/3
y -7 +1/3(x+4) => y =1/3x + 25/3
Finding the equation of a line in the form ax+by+c = 0
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.
Find the equation of the line passing through (3, - 5) (-7,2) in the form ax + by + c =0
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
Are these two lines Parallel, perpendicular or Neither?
y = 3x - 7
6x - 2y =5
(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,
Parallel, Perpendicular or Neither?
y = 8 - 5x
10 + 5y = x
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.
Negative reciprocal
m2 = - 1/m1
Find the negative reciprocal of 3
= - 1/3
Find the negative reciprocal of -5
= 1/5
Find the negative reciprocal of 1/7
= - 7/1 = -7
Find the negative reciprocal of
- 1/11
= 11/1 = 11
Find the negative reciprocal of -0.6
= 5/3
Find the negative reciprocal of 3/8
= - 8/3
Find the equation of the line that is
parallel to the line 3x + 2y + 7 = 0, passing through ( -2,4)
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
Find the equation of a line that is perpendicular to the line 3x + 4y = 5, passing through (3, - 2)
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
sketch y = 3x - 7
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.
Sketch 2x + 7y = 14
When x = 0 = > 7y = 14, y =2
When y = 0, 2x = 14, x = 7
Then sketch the graph
Perpendicular bisector
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.
Perpendicular bisector of A(3,8) and B(-9,4)
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
Find the equation of the perpendicular bisector of (2,7) (-6,9)
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
Find the equation of the perpendicular bisector of
(5, -3) (8, - 18)
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
Find the equation of perpendicular bisector of (-2, - 9) (1/2, 1/3)
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
Finding where two lines intersect
Solve the two equations of the lines simultaneously
Find where the line 2x + 7y = 28 and y =3x -1 intersect.
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
Types of quadrilaterals and there properties
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.
The equation of a circle
(x - h)^2 + (y - k)^2 = r^2
Circle centre (h,k)
with radius r
Identify the centre and radius of x^2+y^2 = 9
Centre (0,0) radius √9 = 3
Identify the centre and radius of
x^2 + (y-3)^2 = 16
centre(0,3)
radius √16 = 4
Identify the centre and radius of
(x-2)^2 + y^2 = 81
Centre (2,0) radius √81 = 9
Identify the centre and radius of
(x-3)^2 + (y+5)^2 = 144
Centre(3,-5) radius 12
Sketch the circle x^2 + (y-2)^2 =4
Centre (0,2)
Radius =2
the radius is also the centre so the circle must be touching the origin.
Sketch the circle (x-5)^2 + (y-5)^2 = 25
Centre(5,5)
Radius = 5
The Centre is both Radius so the circle must be touching the x and y axis at 5.
Sketch (x+6)^2 + (y -2)^2 = 5
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
solving equations to find the centre and radius of the circle
We can use completing the square to do this
Find the centre and radius of the circle with equation x^2 + y ^2 - 6x +2y + 9 =0.
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.
Find the centre and radius of
x^2 + 6x + y^2 - 4y -12 = 0
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
Find the centre and radius of
x^2 - 12x + y^2 - 8y + 49 = 0
(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
Find the centre and radius of
2x^2 + 36x + 2y^2 - 4y + 163 = 0
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)
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
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)
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.
(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)
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.
(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.
Determine the coordinates of intersection of the circles
(x-3)^2 + (y+8)^2 = 100 and
(x+2)^2 + (y-3)^2 = 121
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)
A(-4, 5), B(2,-1), C(8,5) all lie on a circle . Prove that AC is a diameter
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.
Find the centre and radius of the circle with points A(-2,1), B(2,5), C(-10,17)
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
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
(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
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.
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
Sketch y = x^3
Cant draw the answer
Sketch y = x^3 + 1
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.
Sketch y = x
straight line through origin
Sketch y = x^2
The wide bucket shape
Sketch y = x^4
bucket shape
Sketch y = 1/x
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.
Sketch y = 1/x^2
This is the volcano shape
(remember that the + or - number if there is one is the translation otherwise known as asymptotes)
