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Modelling with Straight Lines Simplified Revision Notes

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3.1.4 Modelling with Straight Lines

Straight Lines

Gradient of a Line Segment

infoNote
  • The gradient mm of a line segment can be calculated using the formula:
m=ΔyΔx=y2y1x2x1m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}

where (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are two points on the line.

image
infoNote

Example: Given points (1,2)(1, 2) and (6,10)(6, 10):

Midpoint of a Line Segment

infoNote
  • The midpoint of a line segment is given by the mean of the xx and yy coordinates of its endpoints:
(12(x1+x2),12(y1+y2))\left( \frac{1}{2} (x_1 + x_2), \frac{1}{2} (y_1 + y_2) \right)
  • where (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the endpoints.
image

Length of a Line Segment

infoNote
  • The length of a line segment can be found using the distance formula:
  • Length=(x2x1)2+(y2y1)2\text{Length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
  • where (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the endpoints.
image

Example Problem: Finding Coordinates of Points on a Line

infoNote

Problem: The line ll has gradient 2-2 and passes through the point A(3,5).  BA(3, 5). \ \ B is a point on the line ll such that the distance ABAB is 656\sqrt{5}. Find the coordinates of each of the possible points BB.

Solution:

  1. Equation of Line ll:
  • The line has gradient 2-2 and passes through (3,5)(3, 5). y5=2(x3)y - 5 = -2(x - 3) y5=2x+6y - 5 = -2x + 6 y=2x+11y = -2x + 11
  1. Using Pythagoras' Theorem:
  • Let BB be (x,2x+11)(x, -2x + 11).
  • Distance AB=65AB = 6\sqrt{5}: (62x)2+(x3)2=65\sqrt{(6 - 2x)^2 + (x - 3)^2} = 6\sqrt{5}
image
  1. Solve the Distance Equation: (62x)2+(x3)2=180(6 - 2x)^2 + (x - 3)^2 = 180 3624x+4x2+x26x+9=18036 - 24x + 4x^2 + x^2 - 6x + 9 = 180 5x230x+45=1805x^2 - 30x + 45 = 180 5x230x135=05x^2 - 30x - 135 = 0 x26x27=0x^2 - 6x - 27 = 0 (x9)(x+3)=0(x - 9)(x + 3) = 0 x=9orx=3x = 9 \quad \text{or} \quad x = -3
  2. Find Corresponding yy -Coordinates:
  • For x=9x = 9: y=2(9)+11=18+11=7y = -2(9) + 11 = -18 + 11 = -7 B(9,7)\therefore B(9, -7)
  • For x=3x = -3: y=2(3)+11=6+11=17y = -2(-3) + 11 = 6 + 11 = 17 B(3,17)\therefore B(-3, 17)

Final Answers:

  • The coordinates of the possible points BB are (9,7)(9, -7) and (3,17)(-3, 17).
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