coordinate geometry please choose a question to attempt from the following: 12345
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Coordinate Geometry
Please choose a question to attempt from the following:
1 2 3 4 5
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STRAIGHT LINE : Question 1
Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4).
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STRAIGHT LINE : Question 1
Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4).
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Reveal answer only y = -5/3x - 6
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3x – 5y = 4
3x - 4 = 5y
5y = 3x - 4 (5)
y = 3/5x - 4/5
Using y = mx + c , gradient of line is 3/5
So required gradient = -5/3 , ( m1m2 = -1)
We now have (a,b) = (-6,4) & m = -5/3 .
Using y – b = m(x – a)
We get y – 4 = -5/3 (x – (-6))
y – 4 = -5/3 (x + 6)
y – 4 = -5/3x - 10
y = -5/3x - 6
Question 1
Find the equation of the
straight line which is
perpendicular to the line with
equation 3x – 5y = 4 and
which passes through
the point (-6,4).
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3x – 5y = 4
3x - 4 = 5y
5y = 3x - 4 (5)
y = 3/5x - 4/5
Using y = mx + c , gradient of line is 3/5
So required gradient = -5/3 , ( m1m2 = -1)
We now have (a,b) = (-6,4) & m = -5/3 .
Using y – b = m(x – a)
We get y – 4 = -5/3 (x – (-6))
y – 4 = -5/3 (x + 6)
y – 4 = -5/3x - 10
y = -5/3x - 6
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•An attempt must be made to put the original equation into the form y = mx + c to read off the gradient.
•State the gradient clearly.
• State the condition for perpendicular lines m1 m2 = -1.
•When finding m2 simply invert and change the sign on m1
m1 = 35 m2 =
-5 3
• Use the y - b = m(x - a) form to obtain the equation of the line.
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STRAIGHT LINE : Question 2
Find the equation of the straight line which is parallel to the line with equation 8x + 4y – 7 = 0 and which passes through the point (5,-3).
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STRAIGHT LINE : Question 2
Find the equation of the straight line which is parallel to the line with equation 8x + 4y – 7 = 0 and which passes through the point (5,-3).
y = -2x + 7
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Question 2
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8x + 4y – 7 = 0
4y = -8x + 7 (4)
y = -2x + 7/4
y = -2x + 7
Using y = mx + c , gradient of line is -2
So required gradient = -2 as parallel lines have equal gradients.
We now have (a,b) = (5,-3) & m = -2.
Using y – b = m(x – a)
We get y – (-3) = -2(x – 5)
y + 3 = -2x + 10
Find the equation of the
straight line which is parallel
to the line with equation
8x + 4y – 7 = 0 and which
passes through the point
(5,-3).
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• An attempt must be made to put the original equation into the form y = mx + c to read off the gradient.
• State the gradient clearly.
• State the condition for parallel lines m1 = m2
• Use the y - b = m(x - a) form to obtain the equation of the line.
8x + 4y – 7 = 0
4y = -8x + 7 (4)
y = -2x + 7/4
y = -2x + 7
Using y = mx + c , gradient of line is -2
So required gradient = -2 as parallel lines have equal gradients.
We now have (a,b) = (5,-3) & m = -2.
Using y – b = m(x – a)
We get y – (-3) = -2(x – 5)
y + 3 = -2x + 10
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STRAIGHT LINE : Question 3
In triangle ABC, A is (2,0),
B is (8,0) and C is (7,3).
(a) Find the gradients of AC and BC.
(b) Hence find the size of ACB. X
Y
A B
C
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STRAIGHT LINE : Question 3
In triangle ABC, A is (2,0),
B is (8,0) and C is (7,3).
(a) Find the gradients of AC and BC.
(b) Hence find the size of ACB. X
Y
A B
C
= 77.4°(b)
mAC = 3/5
mBC = - 3
(a)
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Question 3
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(a) Using the gradient formula:
mAC = 3 – 0
7 - 2 = 3/5
mBC = 3 – 0 7 - 8
= - 3
2 1
2 1
y ym
x x
In triangle ABC, A is (2,0),
B is (8,0) and C is (7,3).
(a)Find the gradients of AC
and BC.
(b) Hence find the size of ACB. (b) Using tan = gradient
If tan = 3/5 then CAB = 31.0°
If tan = -3 then CBX = (180-71.6)°
= 108.4 o
Hence :
ACB = 180° – 31.0° – 71.6°
= 77.4°
so ABC = 71.6°
X
Y
A B
C
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(a) Using the gradient formula:
mAC = 3 – 0
7 - 2
mBC = 3 – 0 7 - 8
= - 3
2 1
2 1
y ym
x x
(b) Using tan = gradient
= 3/5
If tan = 3/5 then CAB = 31.0°
then CBX = (180-71.6)°
= 108.4 o
Hence :
ACB = 180° – 31.0° – 71.6°
= 77.4°
If tan = -3
• If no diagram is given draw a neat labelled diagram.
• In calculating gradients state the gradient formula.
• Must use the result that the gradient of the line is equal to the tangent of the angle the line makes with the positive direction of the x-axis. Not given on the formula sheet.
A
B
Ø °
mAB = tanØ °
Ø ° = tan-1 mABso ABC = 71.6°
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STRAIGHT LINE : Question 4
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In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1).
X
Y
P(4,-5)
Q(2,3)
R(10,-1)
Find
(a) the equation of the line e, the median from R of triangle PQR.
(b) the equation of the line f, the perpendicular bisector of QR.
(c) The coordinates of the point of intersection of lines e & f.
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STRAIGHT LINE : Question 4
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In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1).
X
Y
P(4,-5)
Q(2,3)
R(10,-1)
Find
(a) the equation of the line e, the median from R of triangle PQR.
(b) the equation of the line f, the perpendicular bisector of QR.
(c) The coordinates of the point of intersection of lines e & f.
y = -1(a)
y = 2x – 11(b)
(5,-1)(c)
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Question 4 (a)
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In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1).
X
Y
P(4,-5)
Q(2,3)
R(10,-1)
Find
(a) the equation of the line e, the median from R of triangle PQR.
(a) Midpoint of PQ is (3,-1): let’s call this S
Using the gradient formula m = y2 – y1
x2 – x1
mSR = -1 – (-1)
10 - 3
Since it passes through (3,-1)
equation of e is y = -1
= 0 (ie line is horizontal)
Solution to 4 (b)
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Question 4 (b)
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(b) the equation of the line f,
the perpendicular bisector of QR.
In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). Find
X
Y
P(4,-5)
Q(2,3)
R(10,-1)
(b) Midpoint of QR is (6,1)
mQR = 3 – (-1)
2 - 10 = 4/-8 = - 1/2
required gradient = 2 (m1m2 = -1)
Using y – b = m(x – a) with (a,b) = (6,1)
& m = 2
we get y – 1 = 2(x – 6)
so f is y = 2x – 11
Solution to 4 (c)
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Question 4 (c)
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In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). Find
(c) The coordinates of the point of intersection of lines e & f.
X
Y
P(4,-5)
Q(2,3)
R(10,-1)
(c) e & f meet when y = -1 & y = 2x -11
so 2x – 11 = -1
ie 2x = 10
ie x = 5
Point of intersection is (5,-1)
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• If no diagram is given draw a neat labelled diagram.
Q
P
R
y
x
median
Perpendicular bisector
(a) Midpoint of PQ is (3,-1): let’s call this S
Using the gradient formula m = y2 – y1
x2 – x1
mSR = -1 – (-1)
10 - 3
Since it passes through (3,-1)
equation of e is y = -1
(ie line is horizontal)
Comments for 4 (b)
•Sketch the median and the perpendicular bisector
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Q
P
R
y
x
(b) Midpoint of QR is (6,1)
mQR = 3 – (-1)
2 - 10 = 4/-8
required gradient = 2 (m1m2 = -1)
Using y – b = m(x – a) with (a,b) = (6,1)
& m = 2
we get y – 1 = 2(x – 6)
so f is y = 2x – 11
= - 1/2
• To find midpoint of QR
2 + 10 3 + (-1) 2 2
,
• Look for special cases:
Horizontal lines in the form y = kVertical lines in the form x = k
Comments for 4 (c)
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(c) e & f meet when y = -1 & y = 2x -11
so 2x – 11 = -1
ie 2x = 10
ie x = 5
Point of intersection is (5,-1)
y = -1y = 2x - 11
• To find the point of intersection of the two lines solve the two equations:
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STRAIGHT LINE : Question 5
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In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5).
Find
(a) the equation of the altitude from vertex E.
(b) the equation of the median from vertex F.
(c) The point of intersection of the altitude and median.
X
Y
G(2,-5)
E(6,-3)
F(12,-5)
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STRAIGHT LINE : Question 5
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In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5).
Find
(a) the equation of the altitude from vertex E.
(b) the equation of the median from vertex F.
(c) The point of intersection of the altitude and median.
X
Y
G(2,-5)
E(6,-3)
F(12,-5)
x = 6(a)
x + 8y + 28 = 0(b)
(6,-4.25)(c)
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Question 5(a)
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In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5).
Find
(a) the equation of the altitude from vertex E.
XY
G(2,-5)
E(6,-3)
F(12,-5)
(a) Using the gradient formula 2 1
2 1
y ym
x x
mFG = -5 – (-5)
12 - 2 = 0
(ie line is horizontal so altitude is vertical)
Altitude is vertical line through (6,-3)
ie x = 6
Solution to 5 (b)
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Question 5(b)
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In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5).
Find
XY
G(2,-5)
E(6,-3)
F(12,-5)
(b) the equation of the median from vertex F.
(b) Midpoint of EG is (4,-4)- let’s call this H
mFH = -5 – (-4)
12 - 4 = -1/8
Using y – b = m(x – a) with (a,b) = (4,-4)
& m = -1/8
we get y – (-4) = -1/8(x – 4) (X8)
or 8y + 32 = -x + 4
Median is x + 8y + 28 = 0
Solution to 5 (c)
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Question 5(c)
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In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5).
Find
XY
G(2,-5)
E(6,-3)
F(12,-5)
(c) The point of intersection of the altitude and median.
(c)
Lines meet when x = 6 & x + 8y + 28 = 0
put x =6 in 2nd equation 8y + 34 = 0
ie 8y = -34
ie y = -4.25
Point of intersection is (6,-4.25)
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• If no diagram is given draw a neat labelled diagram.
• Sketch the altitude and the median.
y
x
F
E
Gmedian
altitude
(a) Using the gradient formula 2 1
2 1
y ym
x x
mFG = -5 – (-5)
12 - 2 = 0
(ie line is horizontal so altitude is vertical)
Altitude is vertical line through (6,-3)
ie x = 6
Comments for 5 (b)
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y
x
F
E
G
Comments for 5 (c)
(b) Midpoint of EG is (4,-4)- call this H
mFH = -5 – (-4)
12 - 4 = -1/8
Using y – b = m(x – a) with (a,b) = (4,-4)
& m = -1/8
we get y – (-4) = -1/8(x – 4) (X8)
or 8y + 32 = -x + 4
Median is x + 8y + 28 = 0
• To find midpoint of EG
2 + 6 -3 + (-5) 2 2
,H
Horizontal lines in the form y = kVertical lines in the form x = k
• Look for special cases: