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 = tan θ :- tan α = ½ & tanβ = 3 α = tan -1 (½) β = tan -1 (3) α = 71.6 o β = 26.6 o Thus angle between 2 lines is α – β =θ = 71.6 – 26.6 = 45 o
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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
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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
Q10(a) Find the coordinates of the Centroid P, where medians meet. If D(1 , 2) ; E(-1 , 1) & F(3 , -1).
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
Q10(a) Find the coordinates of the Centroid P, where medians meet. If D(1 , 2) ; E(-1 , 1) & F(3 , -1).
Median from F
Midpoint of DE, = (1 + (-1) , 2 + 1) = (0 , 1½) 2 2