Transcript SLine8c.AT

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).
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
Midpoint of PR is (1,4).
Gradient of median = (12 - 4) / (5-1) = 8/4 = 2
Equation is
ie
y - b = m(x - a)
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
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)
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.
Diagram something like
F
G
E
C
Gradient of EF = (3-1)/(3+1) = 1/2
Middle of EF is (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 bisector = 3
Gradient of FG = (2-3)/(6-3) = -1/3
(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
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)