IB Class Notes

Exercises: Linear Function

If we have two points \( A(x_1,y_1) and B(x_2,y_2) \) in a straight line then

  • \( d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \) ; where \( d \) = AB = length of line segment AB.
  • The mid-point is \( \left( \large \frac{x_2-x_1}{2}, \frac{y_2-y_1}{2} \right) \).
  • The slope/gradient: \( \begin{align} m &= \large \frac{y_2-y_1}{x_2-x_1} \\ m & = \frac{rise}{run} \end{align} \)
  • Linear equation: \( y - y_1 = m (x - x_1) \)
    Observe that this (point-gradient form) is directly derived from the gradient formula above.
  • If L1 and L2 are two parallel lines then \( m_1 = m_2 \).
  • If L1 and L2 are two lines that are perpendicular to each other then \( m_1 \times m_2 = -1 \)
  1. Find the distances between these two pairs of points. Report exact answers or accurate to 3 significant figures.
    1. (0,0) & (8,-6)
    2. (-1,-1) & (2,1)
    3. (-2,6) & (3,3)
    4. (2,5) & (-3,-2)
    5. (3,10) & (5,13)
  2. Find the mid-points of the line segments joining these pairs of points:
    1. (0,0) & (8,10)
    2. (-3,5) & (-4,-10)
    3. (9,12) & (-9,-12)
    4. (-4,-4) & (2,2)
    5. (3,10) & (5,13)
  3. Find the gradient of the line joining these pairs of points:
    1. (0,0) & (4, 6)
    2. (-1,0) & (2, 2)
    3. (3,3) & (-3,-3)
    4. (4,5) & (9,5)
    5. (0,0) and (0,9)
  4. A straight line L2 has a gradient of -2. Find the equation of line L1 that is parallel to L2 and passes through point (2,3).

  5. A straight line L2 has a gradient of -2. Find the equation of line L1 that is perpendicular to L2 and passes through the origin.

  6. L1 is a straight line that passes through (2,3) and (-3,-2).
    1. Find the equation of L1.
    2. L1 and L2 are perpendicular. L2 is a straight line that passes through (4,4). Find the linear equation of L2.

  7. L1 is a straight line passes through the origin. L1 has a gradient of 3. L2 is a straight line that is perpendicular to L1. L2 also passes through the origin.
    1. Find the gradient of L2.
    2. Find the linear equation of L2.

  8. A plane is monitored through a radar screen. The screen has Cartesian axes. The plane moves in a straight line through points (8,2) and (2,6). The plane also passes through the point (0,b).
    1. Find the gradient of the straight line joining (8,2) and (2,6).
    2. Find the equation of the plane's path.
    3. Hence or otherwise, find the value of b on the radar screen.

  9. The line L1 has equation \(9x + 12 y - 108 =0\). L1 passes through the x-axis at point A and passes through the y-axis at point B.
    1. Find the coordinates of point A and point B.
    2. M is the mid-point of the line segment [AB]. Find the coordinates of point M.
    3. Find the lengths of line segment [AB].
    4. L2 is a straight line that passes through the point M above and (0,-6). Find the equation of L2.

  10. A straight line L1 passes through P(0,2) and Q(2,0). Point M is the midpoint of line segment [PQ].
    1. Write down the equation of line L1 in the form of \(ax + by + c =0 \).
    2. A straight line L2 is perpendicular to L1 and passes through point M. Write down the equation of line L2 in the form of \(ax + by + c =0 \).

  11. For each of the line shown below, find its equation
    1. in the gradient/intercept form of \(y = mx + c\)
      and
    2. rewrite your answers above in the general form of
      \(ax + by + c =0 \).

Answers: (1i)10, (1ii) 3.61, (1iii) 5.83, (1iv) 8.60, (1v) 3.61
(2i) (4,5), (2ii) (-3.5,-2.5), (2iii) (0,0), (2iv) (-1,-1), (2v) (4,11.5)
(3i) 3/2, (3ii) 2/3, (3iii) 1, (3iv) 0 , (3v) undefined. This is a vertical line.
(4) y = -2x + 7
(5) y = (1/2)x
(6)(i) y = x + 1 ; (ii) y = - x + 8
(7i) -1/3, (7ii) y =(-1/3)x
(8i) slope=-2/3,(8ii) y = (-2/3)x + (22/3) or 3y+2x-22=0
(8iii) b = 22/3 this is the y-intercept
(9i) A(12,0) B(0,9), (9ii) M(6, 4.5), (9iii)15, (9iv) y = (7/4)x - 6
(10i) x+y-2 = 0 , (10ii) -x + y = 0 or y = x.
(11) L1: y = (2/3)x and -(2/3)x + y = 0
L2: y = -3x and 3x + y = 0
L3: y =x-2 and -x + y + 2 =0
L4: y = -x+2 and x + y - 2 = 0
L5: y = (3/2)x + 3 and (-3/2)x + y -3 = 0
L6: y = -x + 3 and x + y -3 = 0

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