Q3 Calculate the size of the angle between y = ½x + 2 & y = 3x - 1 Answer: (Let angles be α and β for each gradient) If m = tanθ:- tan α = ½&tanβ = 3 α

Q3 Calculate the size of the angle between y = ½x + 2 & y = 3x - 1

Answer: (Let angles be α and β for each gradient)

If m = tanθ:- tan α = ½ & tanβ = 3

α = tan-1(½) β = tan-1(3)

α = 71.6o β = 26.6o

Thus angle between 2 lines is

α – β =θ = 71.6 – 26.6 = 45o

Q4 Find the equation of the line passing thro’ (1, 2) perpendicular to y – 2x = -5

y – 2x = -5 must change to y = mx + c

y = 2x – 5 m = 2

As m1 x m2 = -1 perp gradient is m = -½

Thus if passes thro (1, 2) and m = -½

(y – 2) = -½(x – 1)

2y – 4 = -x + 1

x + 2y – 5 = 0 [or an alternative equation]

Q5. Where do y = 2x + 7 & y = -3x - 3 intersect?

If y = …. & y = … y = y

2x + 7 = -3x – 3

5x = - 10

x = -2

If x = -2 subst to find y:(either equation is fine)

Eq1 y = 2x + 7 or Eq2 y = -3x – 3

= -4 + 7 = 6 – 3

= 3 = 3

Thus point of intersection is at (-2, 3)

Q6 Where does 3y – 2x – 12 = 0 cut the x & y-axis?

Cuts x-axis when y = 0: 3y – 2x – 12 = 0

0 – 2x – 12 = 0 Thus cuts

– 2x = 12 x-axis at

x = -6 (-6 , 0)

Cuts y-axis when x = 03y – 2x – 12 = 0

3y – 0 – 12 = 0 Thus cuts

3y = 12 y-axis at

y = 4 (0, 4)

Q7 (a) Find altitude from C to ABGiven A(3 , 1); B(11 , 5) & C(2 , 8)?

mAB = 5 – 1 = 4 = 1 11 – 3 8 2

If mAB = ½ mc = -2

Altitude from C(2, 8) with mc = -2:- y – 8 = -2(x – 2)

y – 8 = -2x + 42x + y – 12 = 0

C

A B

Q7 (b) Find altitude from A to BCGiven A(3 , 1); B(11 , 5) & C(2 , 8)?

mBC = 8 – 5 = 3 = -1 2 – 11 -9 3

If mBC = -1 mA = 3 3

Altitude from A(3, 1) with mc = 3:- y – 1 = 3(x – 3)

y – 1 = 3x - 9 y = 3x - 8

A

C B

Q7 (c) Find the coordinates of T, the point of intersection of the 2 altitudes.

Altitude from A y = 3x – 8

Altitude from C 2x + y – 12 = 0

Substituting Equation 1 into 2 gives:

2x + y – 12 = 0

2x + (3x – 8) – 12 = 0

5x – 20 = 0

5x = 20

x = 4

Using y = 3x – 8 :

y = 12 – 8 = 4 T is ( 4 , 4 )

A

C B

T( 4 , 4 )

Q8 (a) Find line perpendicular to y = ⅓x + 1 which passes thro P(4 , 10) ?

From the above equation the gradient is

m = ⅓ mperp = -3

Perp line thro P(4, 10) with mp = -3:-

y – 10 = -3(x – 4)

y – 10 = -3x + 12

3x + y – 22 = 0 (3x + y – 21 = 0 if used (4 , 9))

Q8 (b) Find the coordinates where both lines meet?

y = ⅓x + 1 & 3x + y – 22 = 0

3x + y – 22 = 0 y = ⅓x + 1 3x + (⅓x + 1) - 22 = 0 y = ⅓(6.3) + 1 3x + ⅓x + 1 - 22 = 0 y = 2.1 + 1 3⅓x = 21 y = 3.1

10x = 21 3 Thus both lines meet at

10x = 63 (6.3 , 3.1) x = 6.3 (If used 3x + y – 21 = 0 meet at (6 ,

3))

Q9 (a) Find altitude from C to ABGiven A(2 , -4); B(14 , 2) & C(10 , 10)?

B(14 , 2)

A(2, -4)

C(10,10)

Q9 (b) Find median thro A given A(2 , -4); B(14 , 2) & C(10 , 10)?

Midpoint of BC = (10+14 , 2+10) = (12, 6)

2 2

If mAD = 6 –(-4) = 10 = 1

12 – 2 10

Median from A with mAD = 1:-

y – ( -4)= 1(x – 2)

y + 4 = x - 2

y = x - 6

C

A B

Q9 (c) Find the perpendicular bisector of AB?

Midpoint of AB, say E = (2+14 , -4+2) = (8 , -1) 2 2

If mAB = 2–(-4) = 6 = 1 mperp = -2 14 – 2 12 2

Perpendicular Bisector from AB with mAD = -2:-

y – ( -1)= -2(x – 8)y + 1 = -2x + 16

2x + y = 15

C

A B

Q9 (d) Find the coordinates of W, the point of intersection of the 2 lines?

2x + y = 15 -----1 y = x - 6 -----2

2x + (y) = 152x + (x – 6) = 15

3x = 21 x = 7

If y = x – 6 = 7 – 6

y = 1 W( 7 , 1)

C

A B

In Q10(a) depending on which 2 of the 3 equations you find will result in your working being laid out differently.

So to make it easier I have called each possibility “Options 1, 2 and 3” for you to view your own particular choice.

Q10(a) Find the coordinates of the Centroid P, where medians meet. If D(1 , 2) ; E(-1 , 1) & F(3 , -1).

Median from D

Midpoint of EF = (-1 + 3 , 1+(-1)) =(1 , 0 ) 2 2

If mD = 0 – 2 = -2 = ∞ vertical line 1 - 1 0

Median from D ( 1 , 2 ) with undefined gradient is therefore x =1

D

E

F


Median from E

Midpoint of DF, = (1 + 3 , 2+(-1)) = (2 , ½) 2 2

If mE = ½ – 1 = - ½ = -1 2 –(-1) 3 6

Median from E ( -1 , 1 ) with mE = -1/6:-

y – 1 = -1/6(x – (-1))6y - 6 = -x - 1

x + 6y – 5 = 0

D

E

F


Median from F

Midpoint of DE, = (1 + (-1) , 2 + 1) = (0 , 1½) 2 2

If mF = -1 – 1½= -2½ = -5 3 - 0 3 6

Median from F ( 3 , -1 ) with mF = -5/6:-

y – (- 1) = -5/6(x - 3)6(y + 1) = -5x + 15 6y + 6 = -5x + 15

5x + 6y – 9 = 0

D

E

F

Q10(a) Find the coordinates of the Centroid P (where medians meet) of triangle DEF?

x = 1 -----1 x + 6y – 5 = 0 -----2

5x + 6y – 9 = 0 -----3

Choose any 2 of 3 possible medianequations to solve for the point P

D

E

F

Q10(a) OPTION 1 Find the coordinates of the Centroid P (where medians meet) of triangle DEF?

x = 1 -----1 Choose any x + 6y – 5 = 0 -----2 of 2

median5x + 6y – 9 = 0 -----3 eqns to

solve

x = 1 -----1 x + 6y – 5 = 0 -----2

1 + 6y – 5 = 0 6y = 4 y = ⅔

If x = 1 and y = ⅔ Centroid P must be ( 1 , ⅔)

D

E

F

Q10(a) OPTION 2Find the coordinates of the Centroid P (where medians meet) of triangle DEF?

x = 1 -----1 Choose any

x + 6y – 5 = 0 -----2 of 2 median

5x + 6y – 9 = 0 -----3 eqns to solve

x = 1 -----1

5x + 6y – 9 = 0 -----3

5 + 6y – 9 = 0

6y = 4

y = ⅔

If x = 1 and y = ⅔ Centroid P must be ( 1 , ⅔)

D

E

F

Q10(a) OPTION 3 Find the coordinates of the Centroid P (where medians meet) of triangle DEF?

x = 1 -----1 Choose any x + 6y – 5 = 0 -----2 of 2 median5x + 6y – 9 = 0 -----3 eqns to

solve

x + 6y – 5 = 0-----2 5x + 6y – 9 = 0 -----3

4x – 4 = 0 4x = 4 x = 1

If x = 1 then 5x + 6y – 9 = 0 5 + 6y – 9 = 0

6y = 4 y = ⅔

Centroid P must be ( 1 , ⅔)

D

E

F

Q10(b) Find the coordinates of the Orthocentre Q of triangle DEF (where altitudes meet)? Given D(1 , 2) ; E(-1 , 1) & F(3 , -1).

Altitude from D

If mEF = 1 – (-1)= 2 = -½ mPerpD = 2 -1 – 3 -4

Altitude from D( 1 , 2 ) with mPerpD = 2

(y – 2) = 2(x – 1) y – 2 = 2x – 2

y = 2x

D

E

F


Altitude from E

If mDF = 2 – (-1)= 3 mPerpE = ⅔ 1 – 3 -2

Altitude from E( -1 , 1 ) with mPerpE = ⅔

(y – 1) = ⅔(x – (-1)) 3y – 3 = 2x + 2

2x – 3y + 5 = 0

D

E

F


Altitude from F

If mDE= 2 – 1 = 1 mPerpF = -2 1 – (-1) 2

Altitude from F( 3 , -1 ) with mPerpF = -2

(y – (-1)) = -2(x –3) y + 1 = -2x + 6

2x + y – 5 = 0

D

E

F

Again in Q10(b) depending on which 2 of the 3 equations you find will result in your working being laid out differently.


Q10(b) Find the coordinates of the Orthocentre Q (where altitudes meet) of triangle DEF?

y = 2x -----1

2x - 3y + 5 = 0 -----2

2x + y – 5 = 0 -----3

Choose any 2 of 3 possible altitude

equations to solve for the Orthocentre Q

Q10(b) OPTION 1 Find the coordinates of the Orthocentre Q (where altitudes meet) of triangle DEF?

y = 2x -----1 Choose any2x - 3y + 5 = 0 -----2 of 2 altitude 2x + y – 5 = 0 -----3 eqns to

solve

Substituting 1 into 2 gives: 2x - 3y + 5 = 02x – 3(2x) + 5 = 0 2x – 6x = -5

-4x = -5 x = 5/4 (or 1.25)

If x = 5/4 and y = 2x = 2(5/4) y = 5/2 Orthocentre Q must be ( 5/4 , 5/2)

D

E

F

Q10(b) OPTION 2Find the coordinates of the Orthocentre Q (where altitudes meet) of triangle DEF?

y = 2x -----12x + y – 5 = 0 -----3

Substituting y = 2x into eqn 3 gives: 2x + y - 5 = 02x + (2x) - 5 = 0 4x = 5

x = 5/4 (or 1.25)

If x = 5/4 and y = 2x = 2(5/4) = 5/2Orthocentre Q must be ( 5/4 ,

5/2)

D

E

F

Q10(b) OPTION 3 Find the coordinates of the Orthocentre Q (where altitudes meet) of triangle DEF?

2x - 3y + 5 = 0 -----2 2x + y – 5 = 0 -----3

Equation 3 - 2 gives: 4y - 10 = 0 4x = 10

x = 10/4 x = 5/4 (or 1.25)

If x = 5/4 and 2x + y - 5 = 0 5/2 + y = 5

y = 5/2 Orthocentre Q must be ( 5/4 , 5/2)

D

E

F

Q10(c) Find the coordinates of the Circumcentre R (where perpendicular bisectors meet) of triangle DEF?

D

E

F

Perp Bisector of DE

Midpoint of DE = (1 + (-1) , 2 + 1) = (0 , 3/2)or (0,1.5) 2 2

If mDE = 2 - 1 = 1 mperp = -2 1 – (-1) 2

Perpendicular Bisector from DE & mperp = -2 y – 1.5 = -2(x – 0)

y – 1.5 = -2x 4x + 2y – 3 = 0


D

E

F

Perp Bisector of DF

Midpoint of DF = (1 + 3 , 2 + (-1)) = (2 , ½) or (2,0.5) 2 2

If mDF = 2 – (-1) = 3 mperp = ⅔ 1 – 3 -2

Perpendicular Bisector from DF & mperp = ⅔ y – 0.5 = ⅔(x – 2)

3y – 1.5 = 2x - 4 2x - 3y – 2.5 = 0 4x – 6y – 5 = 0


D

E

F

Perp Bisector of EF

Midpoint of EF = (-1 + 3 , 1+ (-1)) = (1 , 0) 2 2

If mEF = 1 – (-1) = 2 = - ½ mperp = 2 -1 – 3 -4

Perpendicular Bisector from EF & mperp = 2 y – 0 = 2(x – 1)

y = 2x - 2

Again in Q10(c) depending on which 2 of the 3 equations you find will result in your working being laid out differently.


Q10(c) Find the coordinates of the Circumcentre R where perpendicular bisectors meet.

4x + 2y – 3 = 0 -----1

4x – 6y – 5 = 0 -----2

y = 2x – 2 -----3

Choose any 2 of 3 possible altitude

equations to solve for the Circumcentre R

Q10(c) OPTION 1 Find the coordinates of the Circumcentre R, where perpendicular bisectors meet.

4x + 2y – 3 = 0 -----1 4x – 6y – 5 = 0 -----2 y = 2x – 2 -----3Subtracting Equation 1 - 2 gives:

8y + 2 = 08y = -2 y = -¼

If y = -¼ and 4x + 2y – 3 = 0 4x + 2(-¼ )- 3 = 0 4x - ½ - 3 = 0

4x = 3 ½ 8x = 7 x = 7/8

Circumcentre R must be ( 7/8 , -¼)

D

E

F


4x + 2y – 3 = 0 -----1 4x – 6y – 5 = 0 -----2 y = 2x – 2 -----3Substituting Equation 3 into Equation 1 gives:

4x + 2y – 3 = 04x + 2(2x – 2) – 3 = 0 4x + 4x – 4 – 3 = 0

8x = 7 x = 7/8

If x = 7/8 and y = 2x – 2y = 2(7/8) – 2y = 7/4 - 2y = -¼


D

E

F


4x + 2y – 3 = 0 -----1 4x – 6y – 5 = 0 -----2 y = 2x – 2 -----3Substituting Equation 3 into Equation 2 gives:

4x - 6y – 5 = 04x - 6(2x – 2) – 5 = 0 4x - 12x + 12 – 5 = 0

-8x = -7 x = 7/8

If x = 7/8 and y = 2x – 2y = 2(7/8) – 2y = 7/4 - 2y = -¼


D

E

F

Q10(d) Show that P, Q, R are Collinear.

Centroid P ( 1 , ⅔) Orthocentre Q ( 5/4 , 5/2)Circumcentre R ( 7/8 , -¼)

mpq = 5/2 - ⅔ = 15/6 – 4/6 = 11/6 = 44 = 22 5/4 – 1 ¼ ¼ 6 3

mqr = 5/2-( - ¼) = 10/4 + ¼ = 11/4 = 88 = 22 5/4 – 7/8 10/8 – 7/8 3/8 12 3

As mpq & mqr have equal gradients and a common point exists at Q => points P, Q & R are collinear

Documents

Q3 Calculate the size of the angle between y = ½x + 2 & y = 3x - 1 Answer: (Let angles be α and β for each gradient) If m = tanθ:- tan α = ½&tanβ = 3 α