Coordinate geometry
The Basics:
Find the distance between two points using Pythagoras' theorem. The midpoint is the average (mean) of the coordinates.
The gradient =
Parallel lines have the same gradient. The gradients of perpendicular lines have a product of -1. Straight Lines:
Equation of a straight line is y = mx + c, where m = gradient, c = y-intercept. The equation of a line, if we know one point and the gradient is found using: (y - y1) = m(x - x1)
(If given two points, find the gradient first, and then use the formula.) Two lines meet at the solution to their simultaneous equations. Note: When a line meets a curve there will be 0, 1, or two solutions. 1. Use substitution to solve the simultaneous equations

2. Rearrange them to form a quadratic equation
3. Solve the quadratic by factorising, or by using the quadratic formula. 4. Find the y-coordinates by substituting these values into the original equations. Other Graphs (also in Functions):

Sketch the curve by finding:
1. Where the graph crosses the y-axis.
2. Where the graph crosses the x-axis.
3. Where the stationary points are.
4. Whether there are any discontinuities.
5. What happens as
Circles:
Cartesian equation for a circle is (x - a)2 + (y - b)2 = r2 , where (a, b) is the centre of the circle and r is the radius. Parametric Equations:
Sketch the graph by substituting in values and plotting points. Find the cartesian form by either using substitution (use t = ...), or by using the identity, . Find the gradient using the chain rule:

...
Sixth unit: operations with fractions and radicals. In this unit factor theorems and residue are addressed, and are operated with synthetic division by simplifying fractions to their lowest terms. Operations with radicals are discussed. At the end of this unit the student will be in a position to apply the acquired knowledge in algebraic problems that model various situations.
Seventh unit: equations and inequalities. In this unit, methods for solvingequations and inequalities are studied. Problems as an equation or an inequality of first or second degree with one variable are resolved, pretending that the students infer that there are situations in your environment that are expressed in terms of one variable with one or more possible solutions, but also that there are events that need to be represented, with more than one variable as it is in the next unit.
Eighth unit: systems of equations and inequalities. In this unit algebraically systems two and three linear equations in three variables are resolved, also expressed as such problems. Systems two and three linear equations with three variables resolve themselves, as well as such problems expressed. Resolve two first-degree inequalities with two variables and problems expressed as a system of inequalities.
Structuring the program listed.
First Unit: relations and functions. In...

...Analyticgeometry, or analytical geometry, has two different meanings in mathematics. The modern and advanced meaning refers to the geometry of analytic varieties. This article focuses on the classical and elementary meaning.
In classical mathematics, analyticgeometry, also known as coordinate geometry, or Cartesian geometry, is the study of geometry using a coordinate system and the principles of algebra and analysis. This contrasts with the synthetic approach of Euclidean geometry, which treats certain geometric notions as primitive, and uses deductive reasoning based on axioms and theorems to derive truth. Analyticgeometry is widely used in physics and engineering, and is the foundation of most modern fields of geometry, including algebraic, differential, discrete, and computational geometry.
Usually the Cartesian coordinate system is applied to manipulate equations for planes, straight lines, and squares, often in two and sometimes in three dimensions. Geometrically, one studies the Euclidean plane (2 dimensions) and Euclidean space (3 dimensions). As taught in school books, analyticgeometry can be explained more simply: it is concerned with defining and representing geometrical shapes in a numerical way and...

...the axes is bisected at
A . It’s equation is
1.
(a) 4 x 3 y 24 Ans: a
(b) 3x 4 y 25
(c) x y 7
(d) 3x 4 y 7 0
Sol: By formula required equation is given by
x y 2 4 x 3 y 24 3 4
2. The equation of the line which is the perpendicular bisector of the line joining the points 3, 5 and 9,3 is (a) 4 x 3 y 14 0 Ans: d Sol: A 3, 5 B 9,3 Midpoint of AB 6, 1 . Slope of AB Slope of perpendicular line (b) 3x 4 y 14 0 (c) 3x 4 y 14 0 (d) 3x 4 y 14 0
8 4 6 3
3 4 3 x 6 4
Equation of required line y 1 i.e., 3.
3x 4 y 14 0
The radius of the circle 16x2 16y 2 8x 32y 239 0 is (a) 32 Ans: c (b) 16 (c) 4 (d) none of these
4.
The equation of the circle with centres at (2, -3) and touching y axis is (a) x2 y 2 4 x 6 y 4 0 (c) x2 y 2 4 x 6 y 9 0 Ans: c (b) x2 y 2 4 x 6 y 4 0 (d) none of these
5.
The equation of the line whose intercepts on x -axis and y -axis are 3 and 4 is (a) 4 x 3 y 12 0 Ans: b Sol: Points on required line 3,0 Equation of line (b) 4 x 3 y 12 0 (c) 4 x 3 y 12 0 (d) x y 1 0
0, 4
x y 1 4 x 3 y 12 0 3 4
3 is given by, 2
6.
The equation of the line whose y intercept is 2 and slope is
1
CET11 Mathematics Question Bank –...

...13-14:
y = x – 3
13. Test the equation for symmetry with respect to the x-axis, the y-axis, and the origin.
A) Symmetric with respect to the x-axis
B) Symmetric with respect to the y-axis
C) Symmetric with respect to the origin
D) No symmetry with respect to x-axis, y-axis, or origin
Ans: D Section: 2.1
14. Sketch the graph of the equation.
A) C)
B) D)
Ans: A Section: 2.1
15. Test the equation for symmetry with respect to the x-axis, the y-axis, and the origin. State your results and sketch the graph of the equation.
y = –3x
Ans: Symmetric with respect to the origin.
Section: 2.1
Use the following to answer questions 16-21:
Use the graph to estimate to the nearest integer the missing coordinate of the point. (Be sure you find all possible answers.)
16. (3, ?)
Ans: 2
Section: 2.1
17. (–3, ?)
A) –2 B) 2 C) –4 D) –1, –4
Ans: C Section: 2.1
18. (0, ?)
Ans: –1
Section: 2.1
19. (?, –2)
A) –1 B) –4 C) –4, –1 D) –4, –1, 6
Ans: D Section: 2.1
20. (?, 4)
Ans: –6
Section: 2.1
21. (?, 0)
Ans: –5, 1, 5
Section: 2.1
22. A portion of a graph is shown. Extend the graph to one that exhibits y-axis symmetry.
Ans:
Section: 2.1
23. A portion of a graph is shown. Extend the graph to one that exhibits origin symmetry.
Ans:
Section: 2.1...

...C OORDINATE GEOMETRY
COORDINATE GEOMETRY
155
7
7.1 Introduction
In Class IX, you have studied that to locate the position of a point on a plane, we
require a pair of coordinate axes. The distance of a point from the y-axis is called its
x-coordinate, or abscissa. The distance of a point from the x-axis is called its
y-coordinate, or ordinate. The coordinates of a point on the x-axis are of the form
(x, 0), and of a point on the y-axis are of the form (0, y).
Here is a play for you. Draw a set of a pair of perpendicular axes on a graph
paper. Now plot the following points and join them as directed: Join the point A(4, 8) to
B(3, 9) to C(3, 8) to D(1, 6) to E(1, 5) to F(3, 3) to G(6, 3) to H(8, 5) to I(8, 6) to
J(6, 8) to K(6, 9) to L(5, 8) to A. Then join the points P(3.5, 7), Q (3, 6) and R(4, 6) to
form a triangle. Also join the points X(5.5, 7), Y(5, 6) and Z(6, 6) to form a triangle.
Now join S(4, 5), T(4.5, 4) and U(5, 5) to form a triangle. Lastly join S to the points
(0, 5) and (0, 6) and join U to the points (9, 5) and (9, 6). What picture have you got?
Also, you have seen that a linear equation in two variables of the form
ax + by + c = 0, (a, b are not simultaneously zero), when represented graphically,
gives a straight line. Further, in Chapter 2, you have seen the graph of
y = ax2 + bx + c (a ≠ 0), is a parabola. In fact, coordinate geometry has been developed
as an algebraic tool for studying...

...to gravity.
Equation (1) is and equation of straight line slope of this line is given by-
Slope =
k
g
=> k =
slope
g
(2)
We can plot l vs. m0 graph and determine its slope to determine k.
If the load is slightly pulled down and released, the spring will oscillate simple harmonically. Suppose, at time
t the velocity of the load is v and the spring is compressed by a distance y above the point C.
As you know from your earlier schools if the mass of the spring were negligible then the period of oscillation
would be given by
k
m
T
0
= 2π
Due to the mass, m of the spring an extra term m′ will be added with the mass of the load m0 in the above
mentioned equation. So, the period of oscillation is,
k
mm
T
+ ′
=
0
2π (3)
m′ is called to be the effective mass of the spring. It can be showed that m′ is related with the mass of the
spring by following equation:
3
m
m′ = (4)
Please see appendix A (provided in the soft copy of this script in the server) to learn...

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