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Introduction to Cartesian System

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Title: Introduction to Cartesian System


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Introduction to Cartesian System Session 3
3
Session Opener
4
Session Opener
  1. 7 (b) 6 ( c) 9 (d) 8

5
Session Objectives
  1. Transformation of one form of line to other Form
  1. Intersection of lines
  1. Concurrency of three lines
  1. Shortest distance of a point from a given line
  1. Distance between two parallel lines
  1. Position of a point w.r.t a given line

6
General Form - Equation of Line
This form can be transformed to other standard
forms to simplify problems
7
General to Slope- Intercept Form
General form of equation of a line axbyc 0
On rearranging,
8
Illustrative Problem
Find the slope and y-intercept for line given by
2x 4y 10 0.
Solution
On rearranging,
Comparing with y mx c
9
General to Intercepts Form
General form of equation of a line axbyc 0
On rearranging,
Put y 0 in general equation , Solve for x
10
Illustrative Problem
Find the x-intercept and y-intercept for line
given by 2x 4y 10 0. Find the area formed
by this line and the coordinate axis.
Solution
11
General to Normal Form
General form axbyc 0
Normal form xcos? ysin? - p 0
OP p
12
General to Normal Form
General form axbyc 0
Normal form xcos? ysin? - p 0
c is ve
13
Illustrative Problem
Show that the origin is equidistant from the
lines 4x3y10 0, 5x-12y26 0and 7x24y
50
14
Illustrative Problem
Show that the origin is equidistant from the
lines 4x3y10 0, 5x-12y26 0 and 7x24y 50
Solution
Transforming the equations to normal form, we
have,
Length from origin 2 units
15
Class Exercise - I (8 Min.)
  • Find the slope of line L 5x12y26 0. Convert
    the given equation in to normal form.
  • Find length of perpendicular and the slope of
    the perpendicular.
  • The area of triangles made by the lines L1 2x
    3y k 0, L2 x 0, L3 y 0 is 48
    sq. units. The values of k could be
  • (a) 48 (b) 24 (c) ?24 (d) None of these
  • The straight line passing through the point of
    intersection of the lines L1 x 3y 1 0 and
    2x 5y 9 0 and having infinite slope is
  • (a) y 1 (b) 3x y 1 0 (c) x 4 (d)
    None of these

16
Solution 1-Class Exercise - I
Find the slope of line L 5x12y26 0. Convert
the given equation in to normal form.Find length
of perpendicular from origin and the slope of the
perpendicular.
Solution
Rearranging the given equation,
5x 12y 26
Length of perpendicular from origin 2 units
17
Solution 2 - Class Exercise - I
The area of triangles made by the lines L1
2x 3y k 0, L2 x 0, L3 y 0 is 48
sq. units. The values of k could be
(a) 48 (b) 24 (c) ?24 (d) None of these
Solution
L2 and L3 are Y-axis and X-axis, triangle is
as
18
Solution 3 - Class Exercise - I
The straight line passing through the point of
intersection of the lines L1 x 3y 1 0 and
2x 5y 9 0 and having infinite slope is (a)
y 1 (b) 3x y 1 0 (c) x 4
(d) None of these
Solution
Slope is infinite, equation of line will be x
k
Point of intersection of L1 and L2 (2,1)
Equation of required line is x 2
19
Intersection of Two Lines
Consider two lines a1xb1yc1 0 and a2xb2yc2
0
Point of intersection will lie on both lines.
Let the point of intersection be P (x1, y1) ?
a1x1b1y1c1 0 a2x1b2y1c2 0 Solving by
cross-multiplication,
No need to memorise
Solve two simultaneous equations
20
Illustrative Problem
Find the point of intersection of lines x 2y
9 and 2x y 3 0.
Solve two simultaneous equations
Solution
x 3 , y 3
21
Concurrency of Three Lines
Three concurrent lines pass through a common
point.
Let the three lines be L1 ? a1xb1yc1 0, L2 ?
a2xb2yc2 0, L3 ? a3xb3yc3 0
P.O.I will satisfy L3
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Concurrency of Three Lines
L1 ? a1xb1yc1 0, L2 ? a2xb2yc2 0, L3 ?
a3xb3yc3 0
23
Illustrative Problem
If the three lines x y 1 0, y 5 and 2x
3y k are concurrent, the value of k is
(a) 1 (b) 1 (c) 0 (d) None of these
Solution
If the lines are concurrent
24
Family of Lines Passing through Intersection of
two lines
Let the to intersecting lines be L1 ? a1xb1yc1
0, L2 ? a2xb2yc2 0,
Family of lines passing through point of
intersection of L1 and L2 is
Why ?
L1k L2 0
Problem from booklet
25
Concurrency of Three Lines
Alternate condition for concurrency of three
lines
Let the three lines be L1 ? a1xb1yc1 0, L2 ?
a2xb2yc2 0, L3 ? a3xb3yc3 0
Iff there exist three non-zero constants p, q, r
such that pL1qL2rL3 0.
26
Illustrative Problem
If the three lines x y 1 0, y 5 and 2x
3y k are concurrent, the value of k is
(a) 1 (b) 1 (c) 7 (d) None of these
Solution
p(xy-1) q(y-5) r(2x3y-k) 0
x(p2r) y(pq3r) - (p5qrk) 0
p -2r , pq -3r , p5q - r k
p -2r , q -r
? -7r -rk
? 7 k
27
Class Exercise - II ( 5 Min.)
  • Prove that all lines represented by the equation
    (2cosA3sinA)x (3cosA 5sinA )y ( 5 cosA
    2 sinA) pass through a fixed point for all
    values of A .
  • Find the coordinates of fixed point.

28
Solution 1 - Class Exercise - II
The point through which the family of lines (a
2b)x (a 3b)y (a b) 0 passes for all
values of a and b is
(a) (2,1) (b) (1,2) (c) (2,1) (d)
(2,1)
Solution
Rearranging the given equation
a(x y 1) b(2x 3y 1) 0
The above lines passe through the p.o.i. of the
lines
x y 1 0 ...(i) and 2x 3y 1
0 ...(ii)
Solving (i) and (ii), we get x 2 and y 1
29
Solution 2 - Class Exercise - II
Prove that all lines represented by the equation
(2cosA3sinA)x (3cosA 5sinA )y ( 5 cosA
2 sinA) pass through a fixed point for all
values of A . Find the coordinates of fixed
point.
Solution
Rearranging the given equation
cosA(2x 3y 5) sinA(3x 5y 2) 0
The above lines passe through the p.o.i. of the
lines
2x 3y 5 0...(i) and 3x 5y 2 0
...(ii)
Solving (i) and (ii), we get x 1, y 1
30
Distance of a point from a line
Consider line L, xcos?ysin? p
and point P (x1, y1).
Draw a Line L1 passing through P but parallel to
L
L1 xcos?ysin? p1
? p1 x1cos?y1sin?
PQ CD OD OC p1 - p
PQ x1cos?y1sin? - p
31
Distance of a point from a line
Consider line L, xcos?ysin? p
and point P (x1, y1).
PQ x1cos?y1sin? - p
32
Distance of a point from a line
Algorithm
Step I Convert given equation to axbyc
0
Step II Put cordinates of the given point
(x1,y1) in place of x and y. ax1by1c
Note Absolute value
Step III
Problem from booklet
33
Distance between Two Parallel Lines
L1 a1 x b1 y c1 0 L2 a2 x b2 y c2
0
Distance between L1 and L 2 OM ON
Make coefficient of x as ve.
34
Illustrative Problem
Find the distance between the parallel lines
3x-4y9 0 and 6x-8y-15 0
Solution
35
Position of points relative to a line
Let the given line be L ? axbyc 0 Let the
given points be P1 ? (x1, y1) and P2 ? (x2,
y2) Let a point Q (x3, y3) divide line segment
P1P2 in ratio mn
If Q lies on L,
36
Positions of Points Relative to a Line
Ratio positive ? ax1by1c and ax2by2c have
opposite signs ? points are on opposite sides of
the line
37
Positions of Points Relative to a Line
Ratio negative ? ax1by1c and ax2by2c have
same sign ? points are on same side of the line
38
Positions of Points Relative to a Line
Algorithm
Step I Write the line in general
form axbyc 0 Step II Determine sign of
E1 ax1by1c and E2 ax2by2c
If E1 and E2 are of same sign, the point lie on
the same side of the line If E1 and E2 are of
opposite sign, the point lie on the opposite
sides of the line
39
Illustrative Problem
Are the points (3, -4) and (2, 6) on the same or
opposite sides of the line 3x-4y8?
Solution
Given equation in general form, 3x4y8 0
E1 3(3) 4(4)8 14
E2 3(2) 4(6)8 -26
Opposite side
40
Class Exercise - III ( 8 Min.)
  1. Find the distance between lines y mx c and 2y
    2mx d
  1. Check whether the origin and (2, 3) lies on the
    same side of 2x 5y 10 or not.
  1. If p and p1 are the two perpendiculars from
    origin on the line x sec? y cosec? a and x
    cos? - y sin? a cos 2? respectively. Prove that
    4p2 p12 a2.

41
Solution 1 -Class Exercise - III
Find the distance between lines y mx c and 2y
2mx d
Solution
The equation of lines can be written as
mx y c 0 ...(i) and mx y d/2
0 ...(ii)
Distance between parallel lines
42
Solution 2 -Class Exercise - III
Check whether the origin and (2, 3) lies on the
same side of 2x 5y 10 or not.
Solution
Equation of line is 2x 5y 10 0
E(0, 0) 2(0) 5(0) 10 10
E(2, 3) 2(2) 5(3) - 10 9
Opposite Side
43
Solution 3 -Class Exercise - III
If p and p1 are the two perpendiculars from
origin on the line x sec? y cosec? a and x
cos? - y sin? a cos 2? respectively. Prove that
4p2 p12 a2.
Solution
4 a2
44
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