Algebra and Co-ordinate Geometry

**Category : **9th Class

**Algebra and Co-ordinate Geometry**

In this chapter, we will learn about polynomials, linear equations in two variables and co-ordinate geometry.

**Polynomials**

Polynomials are those algebraic expressions in which the variables involved have only non-negative integral powers. In other words, a polynomial p(x) in one variable x is an algebraic expression in x of the form,

\[P(x)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+......+{{a}_{3}}{{x}^{3}}+{{a}_{2}}{{x}^{2}}+{{a}_{1}}x+{{a}_{0}}.\]

Where \[{{a}_{n}}\],\[{{a}_{n-1}}\],…..\[{{a}_{3}},{{a}_{2}},{{a}_{1}},{{a}_{0}}\]are \[{{a}_{n}}\]\[\ne \]0.

Here, \[{{a}_{n}}\],\[{{a}_{n-1}}\],….,\[{{a}_{3}},{{a}_{2}},{{a}_{1}},{{a}_{0}}\] are respectively the coefficients of \[{{x}^{n}},{{x}^{n-1}}\],….,\[{{x}^{3}},{{x}^{2}},x,{{x}^{0}}\] and n is called the degree of the polynomial.

Each of \[{{a}_{n}}{{x}^{n}},{{a}_{n-1}},{{x}^{n-1}}\],…..,\[{{a}_{3}}{{x}^{3}},{{a}_{2}}{{x}^{2}},ax,{{a}_{0}}\],is called a term of the polynomial p(x). The degree of the polynomial in one variable is the highest index of the variable in that polynomial.

** **

**Note:**

(i) A non zero constant polynomial is a polynomial of degree 0. For example \[-3,\frac{2}{3,}\sqrt{5}\] etc are constant polynomials.

(ii) Constant polynomial 0 is called the zero polynomial. In such a polynomial all the constants are zero so degree of a zero polynomial is not defined.

(iii) For a polynomial p(x), a real number k is called a root (or zero) of the equation p(x) = 0 if p(k) =0.

**Types of Polynomials**

The following are the types of polynomials:

- Linear Polynomials (Polynomials of degree 1)
- Quadratic Polynomials (Polynomials of degree 2)
- Cubic Polynomials (Polynomials of degree 3)
- Biquadratic Polynomials (Polynomials of degree 4)

** **

**Remainder Theorem:**

If p (x) is a polynomial of degree greater than or equal to 1 and p(x) is divided by the linear polynomial\[x-r\], then the remainder so obtained is p(x).

** **

**Factor Theorem:**

For a polynomial p(x),

(i) If p(r) = 0 \[\Rightarrow \]( \[x-r\]) is a factor of p(x)

(ii) If (\[x-r\]) is a factor of p(x) \[\Rightarrow \] p(r) = 0

** **

**Algebraic Identities:**

**\[{{(x+y)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy\]****\[{{(x-y)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy\]****\[{{x}^{2}}-{{y}^{2}}=(x+y)(x-y)\]****\[(x+a)(x+b)={{x}^{2}}+(a+b)x=ab\]****\[{{(x+y+z)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx\]****\[{{(x+y)}^{3}}={{x}^{3}}+{{y}^{3}}+3xy(x+y)\]****\[{{(x-y)}^{3}}={{x}^{3}}+{{y}^{3}}+3xy(x-y)\]****\[{{x}^{3}}-{{y}^{3}}=(x-y)({{x}^{2}}+{{y}^{2}}+xy)\]****\[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=(x+y+z)\]****\[({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)\]**

** **

**Linear Equation in two variables**

An equation is a statement in which one expression equals to another expression. An equation of the form \[ax+by+c=0\] where a, b and c are real numbers, such that a and b are both non-zero, is called a linear equation in two variables.

** **

**Solution of a Linear Equation in two Variables**

A linear equation in two variables has infinitely many solutions.

The solution of a linear equation is not affected on

(i) Adding (or subtracting) the same number in both sides of the equation.

(ii) Multiplying (or dividing) the same non-zero number in both sides of the equation.

**Graph of a Linear Equation in two Variables**

General form of linear equation in two variables is \[ax+by+c=0\]

** \[\Rightarrow by=-ax-c\] \[\Rightarrow y=\left( \frac{-a}{b} \right)x-\frac{c}{b}\]**

The following steps are followed to draw a graph:

**Step 1: **Express x in terms of y or y in terms of x.

**Step 2:** Select at least three values of y or x and find the corresponding values of x or y respectively, which satisfies the given equation, write these values of x and y in the form of a table.

**Step 3:** Plot ordered pair (x, y) from the table on a graph paper.

**Step 4:** Join these points by a straight line.

Thus we see that the graph formed by every linear equation in two variables is a straight line and every point on the graph (straight line) shows a solution of the linear equation. Also every solution of the linear equation can be represented by a unique point on the graph of the equation.

** **

**Note:**

(i) The equation of \[x-a\] is \[y=0\] and the equation of \[y-a\] is \[x=0\]

(ii) The graph of \[(x,y)\ne (y,x)\] is a straight line parallel to the \[y-a\] is.

(iii) The graph of \[y=k\] is a straight line parallel to the \[x-a\] is

(iv) An equation of the type \[y=kx\] represents a line passing through the origin.

** **

**Conditions of Consistency**

For a pair of linear equation in two variables:

**\[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,\,\,{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\]**

If\[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\], then lines are intersecting.

If\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\], then lines are coincident.

If\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\], then lines are parallels.

**Co-ordinate Geometry**

It is the branch of Mathematics in which we solve geometrical problems with the help of algebra. To solve such types of problems we need a cartesian co-ordinate system.

** **

**Cartesian coordinate system**

Two perpendicular lines are required for locating a point in a plane, one of them is horizontal and other is vertical. The horizontal and vertical lines are called x-axis and y-axis respectively. Suppose X’OX and Y’OY be two mutually perpendicular lines through any point ‘O’. The point ‘O’ is known as origin. Taking a convenient unit of length and starting from the origin mark off a number scale on horizontal as well as on vertical line. The line X'OX is known as X-axis and the line Y'OY is known as Y-axis. Two axis taken together are known as coordinate axis. The coordinates of a point are always written in ordered pair.

**Note:**

(i) Co-ordinates of a point on x-axis are of the form \[(x,\,\,0)\] and that of the point on the y-axis is of the form\[(x,\,\,0)\].

(ii) If \[x\ne y\], then the position of (x, y) in the cartesian plane is different from the position of (y, x).

(iii) If \[x\ne y\], then\[(x,y)\ne (y,x)\]

** **

**Quadrants**

Let X’OX and Y’OY are the coordinate axes and it divides the given plane into four regions. Every region is known as quadrants.

** **

**Sign Convention of a Point in Quadrants**

Sign |
Quadrants |

\[(+,\,\,+)\] |
I |

\[(-,\,\,+)\] |
II |

\[(-,\,\,-)\] |
III |

\[(+,\,\,-)\] |
IV |

Here, \['+'\] refers to positive real number and \['-'\] refers to negative real number for tie coordinates of a point (x, y).

** **

**Distance Formula**

If A \[({{x}_{1}},{{y}_{1}})\] and B\[({{x}_{2}},{{y}_{2}})\] are the two points in the coordinate system, then the distance AB is given by

\[AB=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]

In various geometrical problems we need to use distance formula. Before using this formula the following points must we kept in your mind:

- The distance of point (x, y) from origin is \[\sqrt{{{x}^{2}}+{{y}^{2}}}\].
- To show three points P, Q and R are collinear, find PQ, QR and PR and then show that the greatest of these three is equal to the sum of other two.

** **

**Section Formula**

Let the coordinates of end points of a line segment be \[({{x}_{1}},{{y}_{1}})\]and \[({{x}_{2}},{{y}_{2}})\], then the coordinates of point C (x, y) which divides the line segment in the ratio m : n internally are:

** \[x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\] \[y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\]**

Note:

(i) If point C is the mid-point of AB, then the coordinates of C are:

\[\left( \frac{{{x}_{1}}+{{x}_{2}}}{2}\,\,,\,\,\frac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]

(ii) lf A \[({{x}_{1}},{{y}_{1}})\], B \[({{x}_{2}},{{y}_{2}})\] and C \[({{x}_{3}},{{y}_{3}})\] are the three vertices of a triangle then its centroid G is given by:

** \[\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3}\,\,,\,\,\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)\]**

**Example:**

Find the remainder when \[2{{y}^{3}}-3{{y}^{2}}-y+7\] is divided by \[y\text{ }-\text{ }4\].

(a) 83 (b) 135

(c) 36 (d) \[-83\]

(e) None of these

Answer (a)

**Explanation:** Let p(y) = \[2{{y}^{3}}-3{{y}^{2}}-y+7\], then according to the factor theorem the remainder will be p (4).

Now, \[p(4)=2{{(4)}^{3}}-3{{(4)}^{2}}-4+7=128-48+3=83\]

**Example:**

Write the quadrants of the following points in which they lie.

(a) \[\left( 1,2 \right)\] (b) \[\left( 1,2 \right)\]

(c) \[\left( -5,-2 \right)\] (d) \[\left( 5,-2 \right)\](5,-2)

**Solution:**

(a)\[\left( 1,2 \right)\], Here, x > 0 and y > 0, therefore, it lies in I quadrant.

(b)\[\left( 0,\text{ }2 \right)\], Here, x = 0 and y > 0, therefore it lies on \[y-axis\]

(c)\[\left( -5,-2 \right)\], Here x < 0 and y < 0, therefore it lies in III quadrant.

(d)\[\left( -5,-2 \right)\], Here x > 0 and y < 0, therefore it lies in IV quadrant.

** **

**Example:**

Draw the graph of \[4x-y+3=0.\]

**Solution:**

**Step 1:** Here it is easy to write y in terms of \[-\text{ }12y\text{ }=\text{ }-\text{ }192\text{ }=>\text{ }y\text{ }=\text{ }16\]

**Step 2: \[y\text{ }=\text{ }4x\text{ }+\text{ }3,\text{ }if\text{ }x\text{ }=\text{ }0\text{ }\Rightarrow \text{ }y\text{ }=\text{ }3\]**,\[If,\text{ }x\text{ }=\text{ }1\text{ }\Rightarrow \text{ }y\text{ }=\text{ }7\text{ }and\text{ }if\text{ }x\text{ }=\text{ }-\text{ }1,\text{ }y\text{ }=\text{ }-1\]

X |
0 |
1 |
-1 |

Y |
3 |
7 |
\[-1\] |

Step3:

**Example:**

‘In a fraction, if unity is added to the numerator and 2 is subtracted from the denominator then the number becomes .Find the fraction if its denominator is 2 more than the numerator.

** **

**Solution:** Let the numerator is x and the denominator is y.

\[\therefore \] Fraction \[=\frac{x}{y}\]

Now as per the given condition, \[\frac{x+1}{y-2}=\frac{4}{3}\]

**\[\Rightarrow \text{ }3x\text{ }+\text{ }3\text{ }=\text{ }4y\text{ }-\text{ }8\]**

**\[\Rightarrow \,\,3x-4y=-11\]**

**\[\Rightarrow \,\,4y-3x=11\] **… (i)

Since \[y\text{ }=\text{ }x\text{ }+\text{ }2\]

\[\therefore \] putting this value in equation (i), we get

**\[4\left( x+2 \right)-3x=11\]**

\[4x\text{ }+\text{ }8\text{ }-\text{ }3x\text{ }=\text{ }11\Rightarrow x=3\], so,

\[y\text{ }=\text{ }3\text{ }+\text{ }2\text{ }=5\]

\[\therefore \] Required fraction is \[\frac{3}{5}\]

**Example:** Before 14 years, Rahul’s age was 14 times that of Shubham. After 10 years, Rahul’s age will be twice that of Shubham. Find the present ages of Rahul and Shubham.

**Solution:** Let Rahul's present age is x years and Shubham’s present age is y years.

Now from first condition,

\[x-14=14\left( y-14 \right)\]

\[\Rightarrow x-14=14y-196\]

\[\Rightarrow x\text{ }-\text{ }14y\text{ }=\text{ }-\text{ }182\] ... (i)

and from second condition,

\[x\text{ }+\text{ }10\text{ }=\text{ }2\text{ }\left( y\text{ }+\text{ }10 \right)\]

\[\Rightarrow \text{ }x\text{ }+\text{ }10\text{ }=\text{ }2y\text{ }+\text{ }20\]

\[\Rightarrow x\text{ }-\text{ }2y\text{ }=\text{ }10\] ... (ii)

Subtracting equation (ii) from (i), we get

\[-12y=-192\Rightarrow y=16\]

\[~\therefore \,\,X=-182+14\times 16\text{ }=-182+224\text{ }=42\]

\[\therefore \] Rahul’s present age is 42 years

and Shubham’s present age is 16 years.

**Example:** Find the area bounded by the line \[3x\text{ }+\text{ }2y\text{ }=6\] with both of the axis.

**Solution:** The given linear equation is \[3x\text{ }+\text{ }2y\text{ }=6\]

The graph of this line will cuts the x-axis at \[\left( 2,\text{ }0 \right)\]\[\left\{ \because \text{ }3x+2y=6\,\,\,~\,\,=>\text{ }3x\text{ }+2\times 0=6~\,\,~=>\text{ }x\text{ }=\text{ }2 \right\}\]

Also, the graph of this equation will cut the y-axis at \[\left( 0,\text{ }3 \right)\]

\[\left\{ \because \,\,3\times 0+2y=6 \right\}\]

\[\therefore \] The graph of the equation \[3x\text{ }+\text{ }2y\text{ }=\text{ }6\] is shown in the figure.

The figure so obtained is of right A with both the axes.

\[\therefore \] Area of the triangle OPQ is

\[A=\frac{1}{2}\,\,\times base\times height\]

\[=\frac{1}{2}\times 2\times =3\,\,uni{{t}^{2}}\]

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