Midpoints n The midpoint of the line AB is basically the averages of the x-coords & the y-coords of the points A and B, so n If A is (x 1,y 1 ) and B is.

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Presentation transcript:

Midpoints n The midpoint of the line AB is basically the averages of the x-coords & the y-coords of the points A and B, so n If A is (x 1,y 1 ) and B is (x 2,y 2 ) n Then M = ( (x 1 + x 2 ), (y 1 + y 2 ) ) n 2 2 ie A(x 1,y 1 ) B(x 2,y 2 ). M \\

Example 23 n P is (9,-107) & Q is (-3, 13) n Find the midpoint of PQ n midpoint = ( (9-3) / 2, ( ) / 2 ) n = ( 6 / 2, -94 / 2 ) n = (3, -47)

Special Lines & Triangles n There are three special types of lines which are often associated with triangles. These are; n Perpendicular Bisectors n Medians n Altitudes

Perpendicular Bisectors n This line does as it says - n It cuts through the middle of another line….. n at right angles to it. A B Perpendicular bisector of AB. \\

Example 24 n If E is (5, -4) and F is (-11,8) then find the equation of the perpendicular bisector of EF. n m EF = = 12 = -3 / 4 n n so gradient of bisector = 4 / 3 (m 1 m 2 = -1) n and midpoint = ((5-11)  2,(-4+8)  2) = (-3, 2) n now using y - b = m(x - a) n we get y - 2 = 4 / 3 (x + 3) n or y - 2 = 4 / 3 x + 4 n or y = 4 / 3 x + 6

Medians of Triangles n The median in a triangle is a straight line from any vertex to the middle of the opposite side. ie. \\ median. _ _. / /

Example 25 n In triangleJKL n J is (5, 8), K is (11, 1) and L is (-1, -2). n Find the equation of the median through K. n NB: diagram is roughly like J K L. \\ M

Ex25ctd n M = ((-1+5)  2, (-2+8)  2) = (2, 3) n m KM = (1-3) / (11 - 2) = -2 / 9 n now using y - b = m(x - a) n we get y - 3 = -2 / 9 (x - 2) (X9) n or 9y - 27 = -2(x - 2) n or 9y - 27 = -2x + 4 n or 2x + 9y = 31

Altitudes of Triangles n Altitude is simply height which is taken at right-angles to the base n There are three possible altitudes. A C B

Example 26 n In triangleEFG n E is (-1, 7), F is (5, 2) and G is (0, -3). n Find the equation of the altitude through E. n NB: diagram is roughly like E F G

Ex26ctd n m FG = (2 + 3) / (5 - 0) = 5 / 5 = 1 n required gradient = -1 (since m 1 m 2 = -1) n now using y - b = m(x - a) n we get y - 7 = -1( x + 1) n or y - 7 = -x - 1 n or y = -x + 6