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Intersecting Lines

Intersecting Lines. To find where two lines meet we can use SIMULTANEOUS EQUATIONS. For this reason it is useful to write the equations in the form Ax + By = C. Consider the following examples…. Example 27. Triangle PQR has vertices P(-3,6) , Q(5,12) & R(5,2).

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Intersecting Lines

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  1. Intersecting Lines To find where two lines meet we can use SIMULTANEOUS EQUATIONS. For this reason it is useful to write the equations in the form Ax + By = C. Consider the following examples…...

  2. Example 27 Triangle PQR has vertices P(-3,6) , Q(5,12) & R(5,2). The median from Q meets the altitude from P at point K. Find the equations of the median and altitude. Hence find the co-ordinates of K. Diagram roughly like median Q P K altitude R

  3. Midpoint of PR is (1,4). Gradient of median = (12 - 4) / (5-1) = 8/4 = 2 Equation is y - b = m(x - a) ie y - 4 = 2(x - 1) or y - 4 = 2x - 2 or 2x - y = -2 Gradient of side QR = (12 - 2) / (5 - 5) = 10 / 0 which is undefined Side QR is vertical so the required altitude is horizontal ie its gradient is zero Using y - b = m( x - a) We get y - 6 = 0(x + 3) or y - 6 = 0 or y = 6

  4. To find the co-ordinates of K we solve 2x - y = -2 y = 6 Simply put y = 6 in the top equation this gives us 2x - 6 = -2 or 2x = 4 or x = 2 Hence K is the point (2,6)

  5. Example 28 The points E(-1,1), F(3,3) and G(6,2) all lie on the circumference of a circle. Find the equations of the perpendicular bisectors of EF and FG. Hence find the co-ordinates of the centre of the circle, C. F Diagram something like G E C

  6. Middle of EF is (1,2). Gradient of EF = (3-1)/(3+1) = 1/2 Gradient of bisector = -2 (m1m2 = -1) Equation is y - b = m(x - a) ie y - 2 = -2( x - 1) or y - 2 = -2x + 2 or 2x + y = 4 Middle of FG is (4.5,2.5). Gradient of FG = (2-3)/(6-3) = -1/3 Gradient of bisector = 3 (m1m2 = -1) Equation is y - b = m(x - a) ie y - 2.5 = 3( x - 4.5) or y - 2.5 = 3x - 13.5 or 3x - y = 11

  7. Finding where the bisectors meet gives us the centre of the circle Solving 2x + y = 4 3x - y = 11 Adding 5x = 15 x = 3 Put x = 3 into 2x + y = 4 or 6 + y = 4 y = -2 Hence centre of circle at (3,-2)

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