Equations of Straight Lines Questions and Answers - Form 2 Topical Mathematics

Questions

1. A solid right pyramid has a rectangular base 10cm by 8cm and slanting edge 16cm.
   calculate:
   1. The vertical height
   2. The total surface area
   3. The volume of the pyramid
2. The line passing through the points A (-1, 3K) and B (K, 3) is parallel to the line whose
   equation is 2y + 3x = 9. Write down the co-ordinates of A and B
3. Find the value of a if the gradient of the graphs of the function y = x² - x³ and y = x - ax are equal at x = ¹/₃
4. Two perpendicular lines meet at the point (4,5). If one of the lines passes through the point (-2,1), determine the equation of the second line in the form ax + by + c =0.
5. Find the equation of the line passing through (-5, 2) and with X-intercept as 3. Leave your answer in the form of Y = mX + C.
6.  
   1. copy and complete the table below:
      

      x	0	1	2	3	4	5	6
      y=2x-4
      y = 12 – 2x

   2.  
      1. On the grid provided and using the same axes, draw the lines y = 2x + 4 and y = 12 – 2x
      2. Hence use your graphs to solve the simultaneous equations
         ½ x – ¼ y = 1
         x + ½ y = 6
   3. By use of substitution method, solve the simultaneous equations;
      6x + 4y = 36
      x + 3y = 13
7. Find the equation of a line through point -2, 4 which is parallel to 3y =- 2x + 8. Express your answer in the form y = mx + c.
8. Determine the equation of a line passing through (-1, 3) and parallel to the line whose equation is 3x -5y = 10
9. A straight line passing through point (-3,4) is perpendicular to the line whose equation is 2y-5x=11 and intersects the x-axis and y-axis at the points P and Q respectively. Find the co-ordinates of P and Q.
10. A triangle ABC is formed by the points A(3, 4), B(-7, 2) and C(1, -2)
    1. Find the co-ordinates of the mid-points K of AB and P of AC
    2. Find the equation of the perpendicular bisector of the KP
11. The equation of line L₁ is ^-3/₅x + 3y = 6. Find the equation of a line L₂ passing through point T (1, 2) and perpendicular to line L₁.
12. Determine the equation of a line passing through (-1, 3) and parallel to the line whose equation is 3x -5y = 10.
13. A straight line through the points A (2, 1) and B (4, m) is perpendicular to the line, whose equation is 3y = 5-2x. Determine the value of m.
14. Determine the equation of a line which is perpendicular to the line 2x + 3y + 4 = 0 and passes through P(1,1)
15. Koech bought 144 pineapples at shs.100 for every six pineapples. She sold some of them at shs.72 for every three and the rest at shs.60 for every two. If she made a profit of 40%; Calculate the number of pineapples sold at 72 for every three.
16. Solve the equation
    x + 2  - x – 1 = 5
       3         2

Answers

1.  
   1. Length of diagonal = √(10² + 8²)
      = √164
      Vertical height = √16² – (√¹⁶⁴/₂)²
      = 14.66cm
   2. Height of the slant surfaces
      √16² – 4² = √240
      √16² – 5² = √231
      Area of slant surfaces
      ( ½ x 8 x √240 x 2) = 124.0 cm2
      (½ x 10x √231 x 2) = 152.0cm2
      Area of the rectangular base= 8 x 10 = 80cm²
      Total surface area = 356cm²
   3. Volume
      = ( ¹/₃ x 80 x 14.66) = 391.0cm³
2. Gradient of line AB =
   3 - 3k
   K +1
   Equation of other line can be written as
   Y = ^-3x/₂ + ⁹/₂
    its gradient = ^-3/₂
   Hence
   3 - 3k = ^-3/₂
   K +1 
   6-6K = -3k- 3
   -3K = -9
   K= 3
   A(-1, 9), B (3,3)
3. M₁ = 2x – 3x²
   M₂ = 1 – 2ax
   M₁ = M₂ at x = ¹/₃
   2x – 3x² = 11 – 2ax
   ²/₃ – 3 (¹/₃)² = 1 – 2ax¹/₃
   ²/₃ – ¹/₃ = 1 – ²/₃a
   -³/₂ = -²/₃a
   ⁹/₄ = a
4. M₁ =
   5 – 1 = ⁴/₆ = ²/₃
   4 --2 
   M₂ = -³/₂
   i.e. -³/₂ = ^{(y – 5)}/_(x-4)
   2 (y – 5) = -3 (x – 4)
   2y – 10 - -3x + 12
   3x + 2y – 22
5. Points (3, 0) and (-5, 2)
   M = - ¼
   y – 0 = - ¼
   x – 3
   y = - ¼ x + ¾
6.  
7. Grad = ²/₃
   y – 4 = ²/₃
   x + 2 
   y = ²/₃x + ¹⁶/₃
8. 3y – 5x = 4 or equivalence
   5y = 3x – 10
   y = ³/₅x – 2
   Gradient = ^-5/₃
   5 = y – 3
         x + 1
   3y – 9 = 5x – 5
9. 2y – 5x = 11
   Y= ⁵/₂ x + ¹¹/₂
   g = ⁵/₂
   ⁵/₂m = -1
   M= -²/₅
   Y – 4 = -²/₅
   X + 4
   5y + 2x = 14
   P(x,o)
   5 X o + 2x = 14
   X = 7
   Q(o, y)
   5y + 2 X o = 14
   Y = 2.8
   P (7,0)
   Q (0, 2.8)
10.   
    1. K ( ^(3-7)/₂, ⁽⁴⁺²⁾/₂) (-2, 3)
       P (⁽³⁺¹⁾/₂, ^(4–2)/₂) = (2,1)
    2. K₁ =
       3 – 1 = - ½
       -2 – 2
       = 2
11. Gradient of L₁ = ¹/₅
    Gradient of L₂ = -5
    Y= mx + c
    2 = -5 (1) t c
    2 = -5tc
    C = 7
    Epuding L2
    Y= -5x + 7
12. 3y – 5x = 4 or equivalence
    5y = 3x – 10
    y = ³/₅ x – 2
    Gradient = -⁵/₃
    5 = y – 3
           x + 1
    3y – 9 = 5x – 5
13. Gradient = g = ^(m–1)/_(4-2)= ^{(m – 1)}/₂
    3y = 5 – 2x
    y = ⁵/₃ – ^2x/₃ g₁ = ^-2/₃
    g x g1= m – 1 – 2 = -1
    2 3
    -2(m – 1) = -6
    -2m + 2 = -6
    -2m = -8
    M = 4
14. L₁y = -²/₃x – ⁴/₃
    M₁ = -²/₃
    M₂ = ³/₂
    L₂y = ³/₂x + c  x = 1, y = 1
    1 = ³/₂ + c
    c = - ½
    L₂y = ³/₂x - ¹/₂
15.  BP = shs. ¹⁴⁴/₆ x 100
    SP = shs. ¹⁴⁰/₁₀₀ x ¹⁴⁴/₆ x 100
    Let pineapples sold at shs. 72 for every shs. 3 be x
    ∴ At shs. 60 for every 2 will be 144 – x
    ^x/₃ x 72 + ^{(144 – x)}/₃ = 3360
    24x + 30 (144 – x) = 3360
    -6x = -960
    x = 60
16. ^{(x + 2)}/₃ – ^{(x – 1)}/₂ = ⁵/₁
    2(x + 2) – 3(x – 1) = 30
    22x + 4 – 3x + 3 = 30
    -x + 7 = 30
    -x = 23
    x = -23