Coordinates and Graphs Questions and Answers

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Coordinates and Graphs Questions and Answers - Form 1 Topical Mathematics

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Questions

Find the angle θ in degrees from the figure below
In the diagram below, determine the equation of the line XY in the form y = mx + c
Find the equation of a line which passes through the point (2, 3) and is perpendicular to y – 3x+ 1 = 0, giving your answer in the form y = mx + c
T is the mid-point of line XY where X is point (1,4) and Y is the point (-5, 10). Find the equation of a line, L₂ which is perpendicular to line XY and goes through point T
1. On the grid provided below, plot points A(2,1) B(-4,3) and C(2,5)
2. Given that the gradient of CD = -1 and CD =AD locate D and complete the quadrilateral ABCD
3. What name is given to quadrilateral ABCD?
In the figure below (not drawn to scale), PQRS is a rectangle and P and Q are the points (3, 2) and (1,4) respectively.

Given that the equation of the line PQ is y =3x -7, find:
1. The equation of line QR
2. The coordinates of point R
3. The coordinates of point S
OABC is a trapezium such that the coordinates of O, A , B and C are (0, 0), (2, -1), (4, 3) and (0, y)
1. Find the value of y
2. M is the mid-point of AB and N is the mid-point of OM. Find in column form
  1. the vector AN
  2. the vector
  3. Vector AC NC
3. Hence show that A, N and C are collinear
Use ruler and a pair of compasses only in this question.
1. Construct triangle ABC in which AB = 7 cm, BC = 8 cm and ∠ABC = 600.
2. Measure
  1. side AC
  2. ∠ ACB
3. Construct a circle passing through the three points A, B and C. Measure the radius of the circle.
4. Construct â PBC such that P is on the same side of BC as point A and ∠ PCB = ½ ∠ ACB, ∠ BPC = ∠ BAC measure ∠ PBC.
ABCD is a parallelogram with vertices A (1,1) and C(8,10). AB has the equation 4x -5y = -1 and BC has the equation 5x – 2y = 20. Determine by calculation;
1. the co-ordinates of the point M where the diagonals meet
2. The co-ordinates of the vertices B and D
3. the length of AB correct to 4 significant figures
The table shows corresponding values of x and y for a certain curve;
x 1.0 1.2 1.4 1.6 1.8 2.0 2.3
y 6.5 6.2 5.2 4.3 4.0 2.6 2.4

Using 3 strips and mid-ordinate rule estimate the area between the curve, x-axis

Answers

2x - 3y + 6 =0
-3y = -2x – 6
y = ^2x/₃ + 2
When y = 0 x = -3
x = 0 y = 2

∴ Co-ordinate of y – intercept is (0,2)
Coordinate of x – intercept is (-3,0)
∴ ∠CAO = tan^-1²/₃= 33.69^o∴∠θ = 180 – 33.69^o= 146.31^o
Point y(^{4 + -2}/₂, ^{7 + -1}/₂) = (1, 3)
grad AB = ^{7 + 1}/_{4 + 2} =⁸/₆grad xy = - ¾
grad xy = - ¾
^{y – 3}/_{x – 2} = - ¾
y = - ¾ x + ¹⁵/₄
Y = 3x – 1
M =3
M₁M₂= -1
M₂= -¹/₃^{y – 3}/_{x -2}= -¹/₃3y – 9 =-x +2
^3y/₃ = -^x/₃ + ^11/₃Y = -^x/₃+ ¹¹/₃
Pt T is ^{1 + 5}/₂, ^{4 + 10}/₂ = (-2, 7)
grad. of grid xy =^{10 - 4}/_{-5 -1} = ¹⁴/_-6 = ^-7/_-3∴ grad of L₂= ³/₇Take a general pt P(x,y) on L2
→^{y - 7}/_{x - 2} = ³/₇→7y – 49 = 3x + 6
7y = 3x + 55
Or y = 3x + 55 (equation of L₂
a, b

(c) Name : a kite
1. Grad of line QP = ^4-2/_1-3= ²/_-2= -1
  Grad of line QR = 1
  Take a pt Q(1,4) and T(x,y) on line QR
  ^{y- 4}/_{x – 1}= 1
  y – 4 = x -1
  y = x + 3 .....equ. of QR
2. y= x+3 …(i) Equ of QR
  y = 3x -7 ...(ii) Equ. of Pr
  Solving simultaneously ;:
  x +3 = 3x -7
  2x = 10
  x = 5
  Substituting ; y = 8
  ∴R is the pt (5,8)
1. Gradient OA = Gradient of CB
  ^{-1 – 0}/_{2 – 0} = - ½
  Gradient of CB
  ^{y – 3}/_{0 – 4}= - ½
  2y – 6 = 4
  2y = 10
  y = 5
3. 4AN = AC And A is a common point hence A, N, C lie on a straight line.
1. â ABC line AB = 7 cm and BC = 8 cm â¡ ABC = 60^o
  Construction of â¡60^o
2. AC = 7.6 + 0.1 and
  â¡ ACB = 53 + 1^o
3. 2 sides bisector 1
  Circle drawn radius 4.4. ± 0.1
4. Bisect â¡ ACB
  Bisection line to cut the circle to identify P
  â¡BPC = â¡BAC = 67^o
  â¡ PBC = 88 ± 0.1^o
1. M (¹⁺⁸/₂, ¹⁺¹⁰/₂) = M (4.5, 5.5)
2. AB: 4x – 5y = -1 x 2
  BC: 5x – 2y = 20 x 5
  8x – 10y = -2
  25x – 10y = 100
  -17x = -102
  x=¹⁰²/₁₇= 6.0
  24 – 5y = -1
  5y = -25
  Y = 5
  ∴B(6,5)
  ^{x+ 6.0}/₂ = 4.5 x = 3
  ^{y + 5}/₂ = 5.5 y = 6
  ∴D (3,6)
3. AB = √(16 – 1)²+ (5-1)²√25 + 16
  √41 = 6.40 (units)
Mid ordinate
Area = 1.2 (6.2 + 4.3 + 2.6)
= 15.72