Co-ordinate Geometry , Linear Equation in Two Variables

Co-ordinate Geometry , Linear Equation in Two

Variables 4
Question
Suppose you want to correspond a straight line (infinite in both directions) with some sort of a number system.
i.e. you have one-to-one correspondence for every point on a line to some set of numbers. How can you create
this sort of one-to-one correspondence?

Question
Next, how could we correspond each point on a plane with a number system?

Plotting a point on the Cartesian System

Suppose you were told to locate "(5, 2)" on the plane. Where would you look? To understand the meaning of
"(5, 2)", you have to know the following rule:
Finding the location of (5, 2) and then drawing its dot is called "plotting the point (5, 2)".

So, for the point (5, 2), you would start at ...then count over to "five" on the x-
the "origin", the spot where the axes cross: axis:

...then count up to "two", moving ...and then draw in the dot:

parallel to the y-axis:
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Co-ordinate Geometry , Linear Equation in Two Variables

Quadrants
The two axes divide the plane into four sections called "quadrants". The quadrants
are labeled with Roman numerals (not Arabic numerals), starting at the positive x-
axis and going around anti-clockwise:

Equation of a straight line:

Let u be any line in the XY-plane. We will find the equation of u.
One of our axioms is that there exist three non collinear points.
Hence we can choose A1(x1, y1) such that A1 ∈/ u.Let M ∈ u be the foot of the
perpendicular from A1 to u.Let A2(x2, y2) be the reflection of A1 in u.

Clearly, u is the perpendicular bisector of (1)

As A1 ∈ u we deduce that, A1 ≠ A2.
Thus, at least one of x2 − x1, y2 − y1 is non zero. (2)
We will now find the equation of u.

Part I: Let ( , ) ∈ .
Hence by (1) |PA1| = |PA2|
⇒ ( − ) +( − ) = ( − ) +( − )

⇒( − ) +( − ) = ( − ) +( − 2)

⇒ −2 + + − 2 + = −2 + + − 2 +

⇒ 2( − ) + 2( − ) +( + − )=0

Hence ( , ) satisfies the question.

2( − ) + 2( − ) +( + − ) = 0 … . (3)

Part II : Let ( , ) satisfy (3).

⇒ 2( − ) + 2( − ) +( + − − )=0

⇒ −2 + + − 2 + = −2 + + − 2 +

⇒( − ) +( − ) = ( − ) +( − 2)

⇒ =

⇒ ( , ) lies on the perpendicular bisector of

⇒ ( , ) ∈ by (1)

Hence we have proved that the equation of u is

2( − ) + 2( − ) +( + − + )=0

We note that by (2) at least one of x2 − x1, y2 − y1 is nonzero. Hence equation of u is a first degree equation in x
and y.We formulate above observation as:

Theorem 1: Equation of a straight line in the XY plane is a first degree equation in x and y.
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Co-ordinate Geometry , Linear Equation in Two Variables

A natural question arises:

Does every first degree equation in x and y represent a straight line in the XY-plane?Indeed, this is also true.
Consider a line L in the XY plane.
Let its equation be ax + by + c = 0.
Note that Lǁ Y-axis ⇔ b = 0.
Let L be not parallel to Y-axis. ∴ b ≠ 0.
Let A1(x1, y1) and A2(x2, y2) be two distinct points on L.
∴ ax1 + by1 + c = 0 and ax2 + by2 + c = 0.
This gives a(x2 − x1) + b(y2 − y1) = 0 · · · (∗)
If possible let x2 = x1. This gives b(y2 − y1) = 0. But b≠0
∴ y2 = y1. ∴ A1 ≡ A2. This is a contradiction.
∴ x2 ≠ x1.
Now (∗) gives a(x2 − x1) = −b(y2 − y1) y2 − y1
A ≠ ≠0 ℎ =− .
ℎ ℎ = .
This shows that choice of labeling of points does not matter.
Hence we note that if A1, A2 are two distinct points on any line L and L is not parallel to Y axis,
then as we move from A1 to A2, the ratio of change in y co-ordinate to change
in x co-ordinate is always constant = − .
Can we deduce the same thing geometrically?
Let A1(x1, y1), A2(x2, y2), A′ (x′ , y′ ) and A′ (x′ , y′ ) be points on line L which is not parallel to Y axis.

Note that, as shown in the figure △A1BA2 ∼ △A1′ B′A2′ .

Giving us = . This leads to = .

Hence we obtain = .

Hence we see that, as one moves from one point on a line to other
point on it the ratio of is constant.
In other words it gives the change in coordinate per unit changein coordinate. This value is called the
slope of line .
Remark:
If L ∥ Y axis, then − = 0. In this case we do not define slopeof .

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Co-ordinate Geometry , Linear Equation in Two Variables

Theorem1:
Let ∈ . Consider the point ( 0, 0).
Then there exists exactly one line passing through ( , ) and having slope .

Proof:
First will show that, there exists a line through ( , ) havingslope .
Consider point ( + 1, + )
( )
Slope of = ( )
= =
Hence is such a line.
Let be any line through whose slope is .
Hence it is not parallel to Y-axis.
Hence must intersect line given by = + 1, say at ( + 1, ).
Now slope of = slope of = = = .
∴ = + ∴ ≡ ∴ ≡

This completes the proof of our theorem. Hence we have deduced that, there exists unique line through
( , ) having slope
Having established the existence and uniqueness let us find the equation of the line passing through A( , )
having slope m.
Let u denote this line.
Let u∗ = u − {A( , )}.
We say that u∗ denotes the line u punctured at A( , ).
We will first find the equation of u∗.

Part I:
Let P ( , ) ∈ u∗.
−
⇒ =
−

⇒ − = ( − ).

Hence P ( , ) satisfies the equation y − = m(x − ) · · · (1).

We note that A( , ) also satisfies (1).
Hence each point of u satisfies (1).

Part II:
Let ( , ) be any point satisfying (1) such that ≠ .
⇒ − = ( − )
⇒ = as − ≠0
⇒ slope of = = slope of .
∴ by uniqueness part of Theorem 1, ( , ) ∈ .
As ≠ , ( , ) ∈ ∗. ∴ ( , ) ∈ …... (2).
Further, ( , ) satisfies (1) and ( , ) ∈ . ⋯ (3).
By (2) and (3) every point satisfying (1) is on . ⋯ (4).
Hence by (1) and (4) the equation of is − = ( − ).

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Co-ordinate Geometry , Linear Equation in Two Variables
Homework:
≡ (10, −1), ≡ (4, 1) and ≡ (−2, 5)
1) Find slope of .
2) Find slope of .
3) Deduce that , , are collinear.
4) Find equation of .
5) Check if satisfies the equation of .
6) Which amongst , , is between other two points? Justify.
7) Let ∩ − axis = { }. Find .
8) Let ∩ − axis = { }. Find .
9) Find slope of .
10) Find the equation of using slope point form.
Theorem 1:
If two distinct lines in the XY-plane have equal slopes then they are parallel.

Proof:
Let and be two distinct lines in the XY-plane ...(1)
Let slope of = slope of … (2)
If possible, let ∩ ≠∅
Therefore there exists A( , ) ∈ ∩
Hence two distinct lines and pass through A( , ) and have same slope. This contradicts our previously
proved theorem.
Hence ∩ ≠ ∅. Hence ∥ .
This completes the proof of our theorem.

Theorem 2:
If two lines in the XY-plane are parallel, then either
(1) they are both parallel to Y-axis or,
(2) their slopes are equal.

Proof:
Let : + = and : + = be the two lines such that ∥ … (1) .
Case (i) : ∥ Y=axis. It is given that ∥ .∴ ∥ − .
⇒ ≠ 0 and slope of = - .
As ∥ , is not parallel to Y-axis.
⇒ ≠ 0 and slope of = - .
If possible let their slopes not be equal.
∴ − ≠ − ⇒ − ≠ 0.

Consider P ( )= ,

Note that as − ≠ 0, ( ) satisfies equation of and equation of .

∴ ∩ ≠ ∅.
This contradicts (1).
∴ slope of = slope of .
This completes the proof of our theorem.

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Co-ordinate Geometry , Linear Equation in Two Variables
You must really be wondering how did we know that
P( )= , is going to satisfy equation of and .
You are given 10 minutes to discover this independently.

Theorem 3:
Let : + = …. (1)
and : + = … (2)
and − ≠ 0 then,
∩ = , … (*).

Remark:
In other words, if − ≠ 0 then, and have exactly one common point given by (*).

Proof :
If possible, let P ( ) satisfy (1) and (2).
+ = …. (3) and
+ = …. (4)
We will eliminate from (3) and (4).
Compute (3) − (4). We get,
( − ) = −

As − ≠ 0, we get = … (5).

We will now eliminate from (3) and (4).

Compute (4) − (3). We get,
( − ) = −

As − ≠ 0, we get = … (6).

Hence, if ( , ) is a common solution of (1) and (2),

then it must be , .

Of course, we need not be the common solution. Hence we must check

that , satisfies each and .

This is left to you.This completes the proof of our theorem.

Homework:
H1: Consider points A1(x1,y1) and A2 (x2, y2).
Let M = ,
(1) Find the slope of .
(2) Find the slope of .
(3) Find .
(4) Find .
(5) Hence deduce that A1–M–A2 and M is the mid point of .

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Co-ordinate Geometry , Linear Equation in Two Variables

H2: Consider points A1(x1,y1) , A2 (x2, y2) and A3 (x3, y3)

(1) Find co-ordinates of mid-point M of .
(2) Find co-ordinates of mid-point N of .
(3) Find A 2A3.
(4) Find MN.
(5) Find the slope of .
(6) Find the slope of .
(7) What can you deduce from (1) to (6)?

H3: Project: Circumcircle of a given triangle.

Consider points A(7, 1), B (5, 5) and C9-2, 4)
Let u be the perpendicular bisector of .
Let v be the perpendicular bisector of .
Let w be the perpendicular bisector of .
(1) Find the equation of u.
(2) Find the equation of v.
(3) Find the equation of w.
(4) Find u ∩ v .
(5) Find u ∩ w .
(6) Note that u ∩ v = u ∩ w {M} say.
(7) Find MA, MB and MC.

Two Points Form of the equation of a line:

Let ( , ) and ( , ) be two distinct points in the XY plane.
To find the equation of =

Case (i) =
In this case ∥ axis.
Hence equation of = is = .

Case (ii) ≠
Then slope of is .
Hence using the slope point form we deduce that the equation of is − = ( − ).
Remark:

In each case we note that final equation is first degree equation in and .
Two lines that are mutually perpendicular:

Let and be two mutually-perpendicular lines.

Case (i):
One of or is parallel to coordinate axes.
This means that one of them is parallel to the X-axis and other is parallel to the Y-axis.
Hence the slope of one of them is zero and the slope of the other line is not defined.
Case (ii):
None of , is parallel to any coordinate axes.
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Co-ordinate Geometry , Linear Equation in Two Variables

>0 <0 <0 >0

Let slope of be and slope of be . Let ⊥ at ( , ).

Let ( + 1, ). Let the line through and parallel to -axis meet in .
Hence ≡ ( + 1, + ).
Let the line through and parallel to -axis meet in .
Hence ≡ ( + 1, + ).
It is important to note that and will have opposite signs.
We note that | | = | | and | |=| |
Applying Complete Pythagoras Theorem (CPT) to the right angled △ we get,
| |⋅| |=| | ∴| |⋅| |=1

But as and have opposite signs, we deduce that ⋅ = −1.

We have proved the following theorem.

Theorem 1: Let and be two lines in the XY plane such that ⊥ .

None of , is parallel to any coordinate axes.
If is slope of and is slope of then = −1.
It is a natural question to ask, whether the converse of above result is true.
Converse: Let and be two lines in the XY-plane.
Let be slope of and be slope of .
If = −1 then ⊥ .

Proof:
We will first prove that the lines and are not parallel.
If possible let ∥ .⇒ =
∴ = −1 ⇒ = −1
But the square of a real number cannot be negative.
Hence ∩ ≠ .
Let ( , ) ∈ ∩
We know that there exists a unique line through ( , ) and perpendicular to .
Let us denote it by . We will show that ≡ .
Let be slope of . ⊥ by construction = −1
But it is given that = −1
∴ =
Hence, cancelling from both sides, we get = .
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Co-ordinate Geometry , Linear Equation in Two Variables
Note: =0⇒ = 0 ≠ −1 contradiction. ∴ ≠0
∴ and both have same slope and pass through ( , ).
But there exists only one such line.
∴ ≡

As ⊥ by construction ∴ ⊥ .
This completes the proof.

Aim: To find slope of the line : + + = 0. Case (i) = 0.

Then : + = 0.
∴ = − . ∴ is parallel to Y-axis.
∴ slope of is not defined.

Case (ii) ≠ 0.
Let ( , ), ( , ) be two distinct points on .
∴ + + = 0 and + + = 0.
∴ ( − )+ ( − )=0

If possible let − = 0 ∴ ( − ) = 0.
As ≠ 0, − = 0
∴ ≡ , hence contradiction.
∴ − ≠ 0. ∴ =− .
Hence slope of the line =− .
Problem:
Consider the line given by 3 + 4 − 10 = 0
(1) Find slope of .
(2) Find the equation of line passing through (2,3) and parallel to .
(3) Find the equation of line passing through (2,3) and perpendicular to .

Competition:
In the next class, you will be given a similar problem, of course with different numerical values. The winner
will be the student who will complete the problem earlier than anybody else. How will you plan for the
competition?
Homework:
1. To find the distance of a given point from a given line. Consider the line : 5 − = 14 and point (2,4).
To find the ( , ).
(1) First find slope of .
(2) Let ⊥ . And be the feet of perpendicular. Find the slope of .
(3) Find equation of .
(4) Find coordinates of .
(5) Find .

Simulation: Playing with slopes

Do activity 4 given in the workbook and play with the simulation to get familiar with the concept of slopes.
Note how the slope changes in different lines

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Co-ordinate Geometry , Linear Equation in Two Variables
Simulation: Parallel and Perpendicular lines
Play the ‘parallels’ and ‘perpendiculars’ games in the Pythagorea app and try to reach the maximum level!
1. What do you notice in parallel lines? What happens when two lines are parallel? State your observation
below.
2. State your observation in terms of what you have learnt in this chapter about slopes etc.
3. What do you notice in perpendicular lines? What happens when two lines are parallel? State your
observation below.
4. State your observation in terms of what you have learnt in this chapter about slopes etc.
equation would sink an enemy ship without graphing

Higher Level Reading:

Prove the below statement
Theorem 4
Every first-degree equation in x and y represents a straight line in the XY plane.

Homework
1. Find the equation of the line:
a) passing through (3, 2) and having slope – 1/3
b) making intercepts – 2/3 and – 4/3 on the axes.
c) passing through (– 1, 6) and making an angle of 150⁰ with the positive x – axis.

2. Find the value of p such that the line passing through (– 4, p) and (1, 3) is:
a) parallel
b) perpendicular to the line passing through the points (– 2, 5) and (8, 7).

3. Show that the points (– 1, 2), (5, 0) and (2, 1) are collinear by using slope formula.
4. Find the value of m and c so that the line with the equation y = mx + c may pass through the points
(– 2, 3) and (4, – 3).
5. Find the equation of the line passing through (– 4, – 5) and perpendicular to the line passing through the
points (– 2, 3) and (4, – 3).
6. Write the equation of a line which has the y – intercept 2 and slope 7.
7. Write the equation of a line which has the y – intercept – 1 and is parallel to y = 5x – 7.
8. Write the equation of a line which has the y – intercept – 5 and is equally inclined to the axis.
9. What will be the value of m and c, if the straight line y = mx + c passes through the points (3, -4)
and (-1, 2)?
10. Find the equation of a line whose y – intercept is – 4 and which is parallel to the line joining the points (-3,
4) and (2, -5).
11. Find the equation of a line whose y – intercept is 5 and which is perpendicular to the line joining the points
(1, -2) and (-2, -3).
12. Write an equation for a line passing through the points (c, 2b) and (c, 3b).
13. Are the lines with equations 2x - 3y = 8 and –x + 4y = 2 parallel, perpendicular or neither?
14. Are the lines with equations 2x = 8 and –3y = 15 parallel, perpendicular or neither?
15. Rewrite the equation |y| = |x| as two equations of two lines.

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Co-ordinate Geometry , Linear Equation in Two Variables
Linear Equations in Two Variables In
last chapter we plotted some points in xy-plane, then we
connected them to form lines, then we wrote equations
representing those line. If you look closely at the equations
most of them have 2 variables ‘x’ and ‘y’. Also neither of the
variables are raised to fractional or integer powers.
So does these equations represents linear equations
in two variables?
Picture shown below represent graph of linear equation
3x + 4y = 12.
How did we plot it?
We know that linear equation in 2 variables represents a straight line. So all we need is 2 points which satisfies
above equation and when we connect them we get a straight line.
So let’s put x = 0 in above equation we get
4y = 12 -> y = 3. So (0,3) is a point on the line.
Let’s put y=0 in above equation, then we get
3x = 12 -> x =4 .So (4,0) is a point on the line.
Plot these 2 points on the graph paper and then connect them.
We have a problem now. When we were solving linear equation in one variable we used get only one solution.
Now we have 2 solutions (x, y) = (0, 3) or (x, y) = (4,0). So which one is correct? Do we have any order pair of
co-ordinates which satisfies above equations? Do we have multiple solutions?
If so what is the final solution?
Answer is yes, we do have multiple solutions. All the points on the line are the solutions. For example in graph
you can see that (8,-3) is on the line. And if you substitute x=8 and y= -3 in the equation, you will see that (8,-3)
is a solution.
So when do we get single solution? Is it really possible to finalise upon single ordered pair?
In next example let’s plot 2 lines in a single graph paper.
Let’s plot 4x+3y= 6 and 2x-5y=16

As you can see points (0,2), (1.5,0) (3,-2) all satisfies 4x + 3y = 6 and points (8,0) , (3,-2) and (0, -3.2) satisfies 2x
-5y =16.
But (3, -2) satisfies both the equations. So (3,-2) is the common solution for both the equation. It is the point of
intersection of above lines when they are plotted in a graph sheet.

So to find single solution we need at least 2 lines. Or is it so?

What happens if the 2 lines we took are parallel? Will they ever meet? Will they have a point of intersection?

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Co-ordinate Geometry , Linear Equation in Two Variables
Ok. So we need 2 non-parallel lines to find single common solution.
Above method is called Graphical way of solving pair of linear equations.

Algebraic Method of solving pair of linear equations

What will you do if don’t have a graph sheet and someone asks you to solve pair of linear equations?
You solve it algebraically.
There are 2 methods. Both are one and the same. You can follow any method.

Type 1: Substitution Method

Find the common solution to following pair of linear equations.
y  2 x  7
x  4y  0
Original System y  2 x  7
x  4y  0

Step 2: Substitute the expression for y into the x  4(2 x  7)  0

second equation.

Step 3: Solve the equation. x  4(2 x  7)  0

x  8x  28  0
 7 x  28  0
 7 x  28  28  0  28
 7 x  28
 7 x  28

7 7
x4
Step 4: Substitute the value into one of the original y  2 x  7
equations and solve.
y  2(4)  7

y  8  7

y  1

Solution (4, -1)

Step 5: Check y  2 x  7 x  4y  0

 1   2( 4)  7 4  4(1)  0

 1  8  7 44  0

 1  1 00

Type 2: Elimination Method

Use elimination method to find all possible solutions of following pair of linear equations :
2 x  3 y  8        (1)
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Co-ordinate Geometry , Linear Equation in Two Variables
4 x  3 y  1        (2)
Step 1: Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficients of x equal. Then we
get the equations as:
4 x  6 y  16        (3)
4 x  3 y  1        (4)
Step 2: Subtracting Equation (4) from Equation (3),
(4 x  4 x)  (6 y  3 y )  16  1
3 y  15,
i.e., y5
Step 3: Putting the value of y in equation (1) we get
2 x  3(5)  8  2 x  8  15  7  7/2

Inconsistent, Dependent & Independent Systems

We have discussed that 2 linear equation might not have common solution always. Let’s see different system
of equations bases on number of solutions they give.

We can mainly categories them into 2 categories. Consistent system and Inconsistent system

Consistent system of equation will have at least one common solution. Inconsistent system of equation will
never have common solution.

Inconsistent System
Two linear equations are inconsistent if lines representing them never meet. i.e lines are parallel to each to
other. And 2 lines are parallel and non-overlapping if their slopes are equal and y-intercepts are different.
For example a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are two lines

We can rewrite them in slope intercept form (y = mx +c , where m is slope of the line and c is y-intercept)

y =- - and y =- -

OR

y = m1x + c1 and y = m2x + c2

We want m1 = m2

- =- → = → =

also c1 ≠ c2 → ≠ → ≠

Consistent System
System of linear equations is called consistent if they have at least one common solution. In consistent system
we have 2 types, namely dependent and independent system.

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Co-ordinate Geometry , Linear Equation in Two Variables
1. Dependent equations
In this type all the equations will have infinite common solution. And this is possible only when all the lines
overlap. For 2 lines to overlap they should have same slope as well as same y-intercept.
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 will overlap if (m1 = m2 and c1 = c2)
= =

2. Independent Equations
In this type all the equations will have single common solution. And it is possible only when all the lines
have different slope. If all the lines have same y-intercept , then that y-intercept will be common solution.
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are independent if (m1≠m2)
i.e - ≠- → ≠ → ≠

one solution Infinite number of solutions no solutions

Consistent system Consistent system inconsistent system

independent equations dependent equations

Exercise:
Homework
1. Solve the following pair of linear equations by the substitution method:
a. 7x – 15y = 2 & x + 2y = 3
b. 2x + 3y = 9 & 4x + 6y = 18
c. x + 2y = 5 & 2x + 3y = 8
d. 0.2x + 0.3y = 1.3 & 0.4x + 0.5y = 2.3

2. Solve the following equations by the method of elimination by equating the coefficients.
a. 17x + 12y = – 2 & 15x + 8y = 6
b. 23x + 17y = 6 & 39x – 19y = 58
c. 43x – 37y = 31 & 13x + 23y = – 59
d. a(x + y) + b(x – y) = a2– ab + b2 &
a(x + y) – b(x – y) = a2 ab + b2

3. For what value of k will the following system of equations have a unique solution?
a. (2x + 5y = 7 and 3x – ky = 5
4. For what value of k will the following system of equations have infinitely many solutions?
b. kx + 4y = k – 4 and 16x + ky = k
5. Find the conditions so that the following systems of equations have infinitely many solutions.
a. 3x – (a + 1) y = 2b – 1 and 5x + (1 – 2a) y = 3b, find a and b
b. 2x + 3y = 7 and (p + q) x + (2p – q) y = 3(p + q + 1), find p and q
c. 2x – (2a + 5) y = 5 and (2b + 1) x – 9y = 15, find a and b
6. Show that the following systems of equation are inconsistent.
a. 2x – y = 9 & 4x – 2y = 15
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Co-ordinate Geometry , Linear Equation in Two Variables
7. For what value of k the following systems of equations have no solution.
a. 8x + 5y = 9 and kx + 10y = 8
b. x – 4y = 6 and 3x + ky = 5
c. kx – 5y = 2 and 6x + 2y = 7
d. 4x + 6y = 11 and 2x + ky = 7
e. 2x + ky = 11 and 5x – 7y = 5
8. Solve the following pair of linear equations.
a. 3(a + 3b) = 11ab & 3(2a + b) = 7ab
b. 5x + = 9 & 7x - = = 5; y ≠0
c. - = , x ≠ −1, ≠1& = +3
d. ax + by = c & bx + ay = 1 + c
( )
e. ax + by = 1 & bx + ay = –1
f. = = & + = ; x ≠ 0, ≠0
Solve the following problems
1. Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give Rs. 50 and Rs. 100 notes
only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received?
2. Yash scored 40 marks in a test, receiving 3 marks for each right answer and losing 1 mark for each
wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for
each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the
test?
3. The difference between two numbers is 26 and one number is three times the other. Find them.
4. Father’s age is three times the sum of ages of his two children. After 5 years his age will be twice the
sum of ages of two children. Find the age of father.
5. A Taxi charges consist of fixed charges and the remaining depending upon the distance travelled in
kilometers. If a person travels 10 km, he pays Rs. 68 and for travelling 15 km, he pays Rs.98. Express
the above statements with the help of simultaneous equations and hence, find the fixed charges and
the rate per km.
6. Places A and B are 100 km apart on the highway. One car stars from A and another from B at the
same time. If the cars travel in the same direction at a different speed, they meet in 5 hours. If they
travel towards each other, they meet in 1 hour. What is the speed of the two cars?
7. A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km
away in time, it has to increase the speed by 250 km/hr from the usual speed. Find its usual speed.
8. A person rows downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find man’s speed of
rowing in still water and the speed of the current.
9. The larger of two supplementary angles exceeds the smaller by 18o. Find them.
10. Two cars are 500 miles apart and moving directly towards each other. One car is moving at a speed of
100 mph and the other is moving at 70 mph. Assuming that the cars start moving at the same
time how long does it take for the two cars to meet?(Distance=Rate x time)
11. If p + q = 1 and the ordered pair (p, q) satisfies 3x + 2y = 1, then it also satisfies–
(A) 3x + 3y = 35 (B) x + 4y = 45 (C) x + 5y = 4 (D) None of thes

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Co-ordinate Geometry , Linear Equation in Two Variables
12. The cost of 9 chairs and 3 tables is Rs 306, while the cost of 8 chairs and 2 tables is Rs. 246. Then the
cost of 6 chairs and 1 table is–
(A) Rs. 164 (B) Rs. 165 (C) Rs. 166 (D) Rs. 186
13. If 2x– 2x–1 = 4, then xx is equal to –
(A) 7 (B) 3 (C) 27 (D) none of these
14. The equations 3x – 5y + 2 = 0 and 6x + 4 = 10y have–
(A) No solution (s) (B) A single solution (C) Two solutions (D) An infinite number of solutions
Higher Level Problems:
1. If the numbers 2a + 2 and 2b +2 add up to 2004, find the sum of the numbers − 2 and − 2.
2. Bill’s age is one third larger than Tracy’s age. In 30 years Bill’s age will be one eighth larger than
Tracy’s age. How many years old is bill?
3. The product of two positive numbers is equal to 50 times their sum and 75 times their difference. Find
their sum.
4. When Tim was Jim’s age, Kim was twice as old as Jim. When Kim was Tim’s age, Jim was 30. When
Jim becomes Kim’s age, Tim will be 88. When Jim becomes Tim’s age, what will be the sum of the ages
of Tim, Jim and Kim?
5. A baker uses 6 cups of flour when she prepares recipes of rolls. She will use 9 cups of flour when
she prepares recipes of rolls where and are relatively prime positive integers. Find + .
6. Let and be nonzero real numbers such that + = 2011 and + = 1. What is the quotient when
+ is divided by ?
7. The sum of the numbers 3 − 4, 3 − 4 and 3 − 4 is 2016. Find the sum of the numbers 4 − 3, 4 −
3, and 4 − 3.
8. For some > 0 the lines 50 + = 1240 and = 8 + 544 intersect at right angles at the point
( , ). Find + .
9. In eight years Henry will be three times the age that Sally was last year. Twenty five years ago their
ages added to 83. How old is Henry now?
10. A tailor met a tortoise sitting under a tree. When the tortoise was the tailor’s age, the tailor was only a
quarter of his current age. When the tree was the tortoise’s age, the tortoise was only a seventh of its
current age. If the sum of their ages is now 264, how old is the tortoise?
11. In five years, Tom, will be twice as old as Cindy. Thirteen years ago, Tom was three times as old as
Cindy. How many years ago was Tom four times as old as Cindy?
12. Boll bought 13 notebooks, 26 pens, and 19 markers for 25 dollars. Paula bought 27 notebooks, 18 pens
and 31 markers for 31 dollars. How many dollars would it cost Greg to buy 24 notebooks and 120pens
and 52 markers?In 10 years the product of Melanie’s Age and Phil’s age will be 400 more than it is
now. Find what the sum of Melanie’s age and Phil’s age will be 6 years from now.
13. In four years Kay will be twice as old as Gordon. Four years after that Shaun will be twice as old as
Kay. Four years after that Shaun will be three times as old as Gordon. How many years old is Shaun
now?
14. On the table was a pile of 135 chocolate chips. Phil ate of the chips, Eric ate of the chips, and
Beverly ate the rest of the chips. How many chips did Beverly eat?

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Co-ordinate Geometry , Linear Equation in Two Variables
15. A store had 376 chocolate bars. Min bought some of the bars and Max bought 41 more of the bars
than Min bought. After that, the store still had three times as many chocolate bars as Min bought.
Find the number of chocolate bars that Min bought
16. One afternoon at the park there were twice as many dogs as there were people, and there were twice
as many people as there were snakes. The sum of the number of eyes plus the number of legs on all of
these dogs, people and snakes was 510. Find the number of dogs that were at the park.
17. Last month a pet store sold three times as many cats as dogs. If the store had sold the same number of
cats but eight more dogs, it would have sold twice as many cats as dogs. How many cats did the pet
store sell last month?
18. Andrea is three times as old as Jim was when Jim was twice as old as he was when the sum of their
ages was 47. If Andrea is 29 years older than Jim, what is the sum of their ages now?

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