Download 08 Intersecting Lines

Applied Mathematic
(Preliminary Extension 1)
Intersecting
Lines
Stage 6 - Year 11
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1
Concurrent Lines
Intersecting
Lines
meet
point
meetat
ataP(x,y)
P(x,y)
y
To find the point P(x,y)
we solve the equations
simultaneously.
P(x,y)
x
Press
End of
Slide
2
Simultaneous Equation - Substitution
y = 3x + 2
y - 2x - 4 = 0
y
(1)
(2)
Substitute (1) into (2).
P(x,y)
P(2,8)
(3x +2) - 2x - 4 = 0
x-2=0
x=2
Substitute x = 2 into (1).
x
Press
Solution (2,8)
y = 3(2) + 2
y=8
End of
Slide
3
Simultaneous Equation – Elimination (1/2)
x – 2y – 5 = 0
x – 3y - 2 = 0
y
(1)
(2)
(1) - (2)
P(x,y)
P(11,3)
y -3=0
y= 3
Substitute y = 3 into (1).
x - 2(3) – 5 = 0
x - 11 = 0
x
x= 11
Press
Solution (3,11)
End of
Slide
4
Simultaneous Equation – Elimination (2/2)
3x – 2y – 4 = 0
-2x + 3y + 1 = 0
y
P(x,y)
P(2,1)
(1) x 2
(2) x 3
(1)
(2)
6x – 4y – 8 = 0
-6x + 9y + 3 = 0
(1) + (2)
Press
5y - 5 = 0
5y = 5
Sub x=1 in (1)
y=1
3x – 2(1) – 4 = 0
x
3x – 6 = 0
3x = 6
Solution (1,2)
x= 2
End of
Slide
5
Equation of a line through the intersection
of two other lines.
(a1x + b1y + c1) + k(a2x + b2y + c2) = 0
y
Proof
P(x1,y1) satisfies both equations.
P(x1,y1)
Press
 (a1x + b1y + c1) = 0 & (a2x + b2y + c2) = 0
Sub in (a1x + b1y + c1) + k(a2x + b2y + c2) = 0
x
0 + k(0) = 0
P(x1,y1) satisfies
(a1x + b1y + c1) + k(a2x + b2y + c2) = 0
so passes through P(x1,y1).
End of
Slide
6
Example: Find the equation of the line through
3x – 2y - 4 = 0, -2x + 3y + 1 = 0
and the point (1,2).
(a1x + b1y + c1) + k(a2x + b2y + c2) = 0
y
(3x – 2y – 4) + k(-2x + 3y + 1) = 0
P(x1,y1)
P(1,2)
Press
Find k by substituting x = 1 and y = 2
(3(1) – 2(2) – 4) + k(-2(1) + 3(2) + 1) = 0
-5 + 5k= 0 (3x – 2y – 4) + k(-2x + 3y + 1) = 0
5k= 5 3x – 2y – 4 + 1(-2x + 3y + 1) = 0
3x – 2y – 4 -2x + 3y + 1 = 0
k= 1
x+y– 3 =0
End of
Slide
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