Coordinate Geometry Please choose a question to attempt from the following: 12345

Coordinate Geometry

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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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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 .


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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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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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.


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.


We get y – (-3) = -2(x – 5)

y + 3 = -2x + 10

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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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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


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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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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X

Y

P(4,-5)

Q(2,3)

R(10,-1)

Find


(b) the equation of the line f, the perpendicular bisector of QR.


y = -1(a)

y = 2x – 11(b)

(5,-1)(c)

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Question 4 (a)

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X

Y

P(4,-5)

Q(2,3)

R(10,-1)

Find


(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


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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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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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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Find




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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Find


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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Find

XY

G(2,-5)

E(6,-3)

F(12,-5)


(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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Find

XY

G(2,-5)

E(6,-3)

F(12,-5)


(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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• 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:

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Coordinate Geometry Please choose a question to attempt from the following: 12345