co-ordinate geometry -...
TRANSCRIPT
Co-ordinate Geometry
Types of sums
1. Lines
2. Equations
3. Slopes
4. Parallel & Perpendicular lines
5. Circles
6. Areas
Importance
1.CAT
2.CMAT
3.NMAT
4.SNAP
5.XAT
Basics
Scope of the topic
Astronomy: Computing paths of celestial bodies like planets, comets, binary star systems.
Graphs analysis
Company performance Data Interpretation
Geometry (Area, distance, slope, rate of change) Missile Trajectory
Gyaan: Descartes is regarded as the father of analytical geometry. He founded the concept of coordinate plane(Can be asked in GK :D)
What is a Coordinate plane?
2 Dimentsional
Horizontal line:
X-axis
Vertical line: Y-axis
Intersection is origin which is
the reference point.
It is used to locate an object on a plane.
Quadrants
Plotting a point
P(4,3
)Q(-
6,2)
Single variable
x =5
y = -3
x = -2
Two variables X and Y
x – y =0i.e. x = y
Two variables X and Y
2x – y = 4y = 2x - 4
Degree 2: Quadratic Function
Shape:
Parabola
Parabola (Quadratic function)
Degree 2: Two variables
Circle Ellipse
LinesDistance Formula: d (PQ) = √(x2-x1)2 + (y2-y1)2
Find the distance between points (-1,2) and (2,6)
Find the distance between points (-1,2) and (2,6)
Distance between two points:
Distance between (4,1) and (7,5)
d
P(7,5)
Q(4,1)
4
3
Distance between points
Find the distance between (7, 5) & (4, 1)
a) 5
b) 5 √2
c) 7 √3
d) 10
Distance between points
Find the distance between (4, 2) & (6, 6)
a) 5.45
b) 4.45
c) 7.14
d) 5.65
Section of a line segment
Internal Division External division
m n
n
m
Equations• Find the equation of the line
passing through (-1,-2) & (-5,2)
• Point of intersection of 2x + y = 4 and x - y + 1= 0
Equation of line joining two points
Find the equation of the line passing
through (-1,-2) & (-5,2)
(a) 2x + y = 3
(b) 3x - 2y + 7 = 0
(c) x + y + 3 = 0
(d) None of these
(y + 2)/(-2-2) = (x + 1)/(-1+5)4(y + 2) = -4 (x + 1)
y + 2 = -x -1 => x + y = -3
Point of intersection of two lines
Point of intersection of
2x + y = 4 and x - y + 1= 0
a) (3, 1)
b) (1, 2)
c) (3, 2)
d) (1, 4)
Slopesy = mx + c
Slope of a line
x = y y = mx + c
Slope of a line
2x – y = 4
Find slope of line y = 4
a) 0b)2c) 4d)Cannot be det
Find slope of line x = 4
a) 0b)2c) 4d)Cannot be det
Find slope of line y = –2x + 3
a) 0b) -2c) 4d)Cannot be det
Finding the slope of a line given two points
Finding the slope from the equation of a line
Finding the equation of a line given a point and a slope
Finding the equation of a line given two points
Finding the equation of the line that is parallel to a given line and passing through a given point
Finding the equation of the line that is perpendicular to a given line and passing through a given point
ParallelPerpendicular
Parallelism
x – y =0
x –y +3 = 0
m1=m2
Perpendicularity
2x – y = 4
x + 2y = 4
One line passes through the points (–1, -
2) and (1, 2); another line passes through
the points (–2, 0) and (0, 4).
Are these lines..
a) Parallel
b) Perpendicular
c) Neither
One line passes through the points (0, –4)
and (–1, –7); another line passes through
the points (3, 0) and (–3, 2).
Are these lines..
a) Parallel
b) Perpendicular
c) Neither
Find the equation of the line, with the slope
m= -1 and passing through the point (3,0)
a) y = -x + 3
b) 2y = x + 3
c) 4y = 4x + 3
d) y = -2x + 3
Use ( y - y1 ) = m ( x - x1)
Find the equation of the line parallel to the
line 6x + 9y = -5, and passing through
the point ( 7, 4 )
a) y = -x + 11
b) 9x = - 6y + 3
c) 2x + 3y = 26
d) 6x + 9y = 11
m = -6/9 = -2/3
Use ( y - y1 ) = m ( x - x1)
Equation using slope and point
Find the equation of line passing through
point (2,-3) having slope 5/4.
(a) 4x - 5y = 20
(b) 3x – 2y = 5
(c) 5x - 4y = 22
(d) None of these
(y - y1) = m (x-x1)(y + 3) = 5/4 (x - 2)4y + 12 = 5x -10 => 5x - 4y = 22
Note: Sometimes angle will be given. Slope can be found as tanѲ
Find the equation of the line perpendicular
to the line -5x - y = 3, and passing
through the point ( 4, 3 )
a) y = -x + 7
b) 2y = x + 3
c) 5y = x + 3
d) -x + 5y = 11
m1 = 5/-1 = -5
So, use m2 = 1/5
Equation of perpendicular
Find the equation of the line passing through
point (2,7) and perpendicular to 2x + 3y +
8= 0
a) 5x – 2y = 17
b) 3x + 2y = 20
c) 3x – 2y = -4
d) 6x – 4y = -16
Equation of parallel
Find the equation of the line passing through
point (2,7) and parallel to 2x + 3y + 8= 0
a) 5x - 2y = 17
b) 2x + 3y = 25
c) 2x + 3y = -4
d) 4x + 6y = -16
Area
Centroid Incentre
A(x1,y1)
B(x2,y2)C(x3,y3)
c b
a
Area of Triangle
Area of triangle formed by points P1(x1,y1), P2(x2,y2) & P3(x3,y3)
Area of triangle
Find the area of triangle formed by points
(4,2), (-5,7) and (5,-3)
a) 10
b) 14.5
c) 22.5
d) 20
Formula substitution: [4(3-(-7)) – 2(-5 -5)+ 1(15-35)]/2= 40/2 = =20 Option (d)
Circles
Find the centre and radius of the
following circles:
1.
2.
3.
Find the centre and radius of the
following circle
Find the equation of the circle:
1.with centre (0,5) and radius 5
2.with centre (2, 0) and radius 4
3.with centre (5, 7) and radius 18
Find the equation of the circle with
centre (2, 1) which passes through (4,1).
Find the equation of the circle with
centre (-3, -2) which passes through
(1,-4).
Find the equation of a circle (centre O) with
a diameter between two points, P at and
Q at .
a) x2 + y2 = 30
b) x2 + y2 = 50
c) x2 + y2 = 80
Applications
Manufacturing of product includes some
fixed cost and variable cost A firm produces
50 units of a good for Rs. 320 and 80 units for Rs. 380. Estimate the cost for producing
110 units. (Assume cost curve to be linear)
a) Rs. 330
b) Rs. 365
c) Rs. 440
d) Rs. 1665
Manufacturing of product includes some fixed cost and variable cost A firm produces 50 units of a good for Rs. 320 and 80 units for Rs. 380. Estimate the cost for producing 110 units. (Assume cost curve to be linear)
a) Rs. 330b) Rs. 365c) Rs. 440d) Rs. 1665
We have two points (50,320) and (80, 380) lying on the cost line. Obtain the equation of straight line and substitute x = 110
(y-380)/(380-320) = (x-80)/(80-50)6y - 3y= -660y= 2x + 220Ans: 440/-