Subject: Optional Mathematics

An equation between any two variables which straight line on a graph is known as a linear equation. In this equation, Ax + By + C = 0, where A, B, and C are neutrals and also A and B will not be together zero. This note gives the information about the standard forms of reducing general equation and to find a straight line from the perpendicular length.

An equation between any two variables which straight line on a graph is known as a linear equation. In this equation, Ax + By + C = 0, where A, B, and C are neutrals and also A and B will not be together zero.

To prove this statement, let P(x_{1}, y_{1}), Q(x_{2}, y_{2}) and R(x_{3}, y_{3}) be nay three points on the locus represent by the equation Ax + By + C = 0. The coordinates of the points must satisfy the equation.

Hence,

Ax_{1} + By_{1} + C = 0 . . . . . . . . . . . . . . (i)

Ax_{2} + By_{2} + C = 0 . . . . . . . . . . . . . . (ii)

Ax_{3} + By_{3} + C = 0 . . . . . . . . . . . . . . (iii)

Solving first two equations by the rule of cross multiplication, we get

\(\frac{A}{y_1 - y_2}\) = \(\frac{B}{x_1 - x_2}\) = \(\frac{C}{x_1 y_2 - x_2 y_1}\) = k (say)

∴ A = k(y_{1} - y_{2}), B = k(x_{2} - x_{1}) and C = k(x_{1}y_{2} - x_{2}y_{1})

Substituting the values of A, B and C in the third equation, we get

k(y_{1} - y_{2}) x_{3} +k(x_{2} - x_{1}) y_{3} + k(x_{1}y_{2} - x_{2}y_{1}) = 0

or, k(x_{3}y_{1} - x_{3}y_{2} + x_{2}y_{3} - x_{1}y_{3} + x_{1}y_{2} - x_{2}y_{1)} = 0

or, x_{1}y_{2} - x_{2}y_{1} + x_{2}y_{3} - x_{3}y_{2}+ x_{3}y_{1} - x_{1}y_{3} = 0

Multiplying both sides by \(\frac{1}{2}\), we get

\(\frac{1}{2}\)(x_{1}y_{2} - x_{2}y_{1} + x_{2}y_{3} - x_{3}y_{2}+ x_{3}y_{1} - x_{1}y_{3}) = 0

i.e. Area of ΔPQR = 0

This result sgows us that the points P, Q and R are collinear.

Thus, the general equation of first degree in x and y always represents a straight line.

There is three standard form to reduce the linear equation. They are given below:

**Reduction to the slope intercept form**

The first-degree general equation in x and y is

Ax + By + C = 0

This equation can be writtenas

By = -Ax - C

or, y = (-\(\frac{A}{B}\))x + (-\(\frac{C}{B}\)) which is of the form y = mx + c where

slope (m) = -\(\frac{A}{B}\) = -\(\frac{coefficient \: of \: x}{coefficient \: of \: y}\) and

Y-intercept (c) = -\(\frac{C}{B}\) = -\(\frac{content \: term}{coefficient \: of \: B}\)

**Reduction of the double intercept form**The first-degree general equation in x and y is

Ax + By + C = 0

This equation can be writtenas

Ax + By = -C

Dividing both sides by -C, we get

\(\frac{Ax}{-C}\) + \(\frac{By }{-C}\) = 1

or, \(\frac{x}{\frac{-C}{A}}\) + \(\frac{y}{\frac{-C}{B}}\)= 1 which is of the form \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 where

X-interccept (a) = -\(\frac{C}{A}\) = -\(\frac{constant \: term}{coefficient \: of \: x}\)

Y-intercept (b) = -\(\frac{C}{B}\) = -\(\frac{constant \: term}{coefficient \: of \: y}\)

**Reduction to the normal form**

The equation Ax + By + C = 0 and x cosα = p will represent one and ssame straight line if their corresponding coefficients are proportional.

∴ \(\frac{cosα}{A}\) = \(\frac{sinα}{B}\) = \(\frac{-p}{C}\) = k (say)

Then, cosα = Ak, sinα = Bk and -p = Ck

Now,

(Ak)^{2} + (Bk)^{2} = cos^{2}α + sin^{2}α = 1

or, k^{2} (A^{2} + B^{2}) = 1

or, k^{2 }= \(\frac{1}{A^2 + B^2}\)

or, k = ± \(\frac{1}{\sqrt{A^2 + B^2}}\)

∴ cosα = \(\frac{A}{±\sqrt{A^2 + B^2}}\), sinα = \(\frac{B}{±\sqrt{A^2 + B^2}}\) and p = \(\frac{-C}{±\sqrt{A^2 + B^2}}\)

Hence, the normal form is \(\frac{A}{±\sqrt{A^2 + B^2}}\)x + \(\frac{B}{±\sqrt{A^2 + B^2}}\)y = \(\frac{-C}{±\sqrt{A^2 + B^2}}\)

The + or - sign in the RHS being so chosen as to make the RHS positive.

**To find the length of the perpendicular from a point on the line x cosα + y sinα = p**

Let the equation of a line AB be x cosα + y sinα = p.

Then the length of a perpendicular from the origin on the line is p.

i.e. ON = p and∠AON =α

Let P(x_{1}, y_{1}) be any point and draw perpendicular PM from P to the line AB.

Through the point P, draw a line CD parallel to the given line AB. Let ON' be the perpendicular drawn from the origin to the CD such that ON' = p'.

Then,

PM = ON' - ON' = p' - p when p' > p and PM = ON - ON' = p - p' when p> p'.

Thus, PM = ± (p' - p). Here, the proper sign is taken so as make PM positive.

Now equation of CD is x cosα + y sinα = p'

But this line passes through the point (x_{1}, x_{1})

So, x_{1} cosα + y_{1} sinα = p'

Hence, PM = ± (x_{1} cosα + y_{1} sinα - p) which is the length of the perpendicular drawn from (x_{1}, y_{1}) on the line x cosα + y sinα = p.

**To find the length of the perpendicular from a point on the line Ax + By + C = 0**

Here, the general equation of the first degree in x and y in Ax + By + C = 0.

Changing the equation into perpendicular form we get,

\(\frac{A}{\sqrt{A^2 + B^2}}\)x + \(\frac{B}{\sqrt{A^2 + B^2}}\)y + \(\frac{C}{\sqrt{A^2 + B^2}}\) = 0

Comparing this equation with x cosα + y sinα - p = 0 we get,

cosα = \(\frac{A}{\sqrt{A^2 + B^2}}\), sinα =\(\frac{B}{\sqrt{A^2 + B^2}}\) and p = -\(\frac{C}{\sqrt{A^2 + B^2}}\)

Now, length of the perpendicular drawn from the point (x_{1}, y_{1})_{ }to the line x cosα + y sinα - p is

L = ± (x_{1} cosα +y_{1} sinα - p)

= ± (x_{1}\(\frac{A}{\sqrt{A^2 + B^2}}\) + y_{1}\(\frac{B}{\sqrt{A^2 + B^2}}\) -\(\frac{C}{\sqrt{A^2 + B^2}}\))

= ± (\(\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}\))

Hence, length of the perpendicular drawn from the point (x_{1}, y_{1}) to the line

Ax + By + C = 0 is ± (\(\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}\))

- An equation between any two variables which straight line on a graph is known as a linear equation. In this equation, Ax + By + C = 0, where A, B, and C are neutrals and also A and B will not be together zero.
- Ax + By + C = 0 is± (\(\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}\))
- Ax + By + C = 0
- \(\frac{1}{2}\)(x
_{1}y_{2}- x_{2}y_{1}+ x_{2}y_{3}- x_{3}y_{2}+ x_{3}y_{1}- x_{1}y_{3}) = 0

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Reduce 3x + 4y + 5 = 0 in slope intercept form and hence find slope and y-intercept.

Solution:

Here,

3x + 4y + 5 = 0

Reduce it into slope intercept from, 4y = -3x - 5

or, y = \(\frac{-3}{4}\) x - \(\frac{5}{4}\)

which is in the form y = mx + c,

so, on comparing, we get,

Slope (m) = \(\frac{-3}{4}\) and

y-intercept (c) = \(\frac{-5}{4}\)

Reduce 4x + 5y + 20 = 0 in double intercept form and hence find x and y-intercept.

Solution:

Here,

4x + 5y + 20 = 0

or, \(\frac{4x}{-20}\) + \(\frac{5y}{-20}\) = 1

or, \(\frac{x}{-20/4}\) + \(\frac{y}{-20/5}\)

or, \(\frac{x}{-5}\) + \(\frac{y}{-4}\) = 1

which is in the form \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1,

∴ x-intercept (a) = -5 and y-intercept (b) = -4

Reduce x - \(\sqrt{3}\) y - 6 = 0 in perpendicular form and hence obtain 'p' and 'α'.

Solution:

x - \(\sqrt{3}\) y - 6 = 0

or, x - \(\sqrt{3}\) y = 6

Now,

= \(\sqrt{(coefficient \: of \: x)^2 + (coefficient \: of \: y)^2}\)

= \(\sqrt{(1)^2 + (-\sqrt{3})^2}\)

= \(\sqrt{1 + 3}\)

= \(\sqrt{4}\)

= 2.

Dividing both sides by 2, we get,

\(\frac{1}{2}\) x - \(\frac{\sqrt{3}}{2}\) y = \(\frac{6}{2}\)

or, \(\frac{1}{2}\) x + (\(\frac{\sqrt{3}}{2}\)) y = 3

which is in form x cosα + y sinα = p

∴p = 3 units, cosα = \(\frac{1}{2}\) and sinα = -\(\frac{\sqrt{3}}{2}\)

Since, cosα is positive and sinα = -\(\frac{\sqrt{3}}{2}\), α lies in the 4^{th} quardrant.

∴ cosα = \(\frac{1}{2}\)

cosα = cosα (360° - 60°)

α = 300°

∴ x cos 300° + y sin 300° = 3

Reduce 2x + 5y - 20 = 0 in double intercept form and hence find the area of ΔOAB is the line cuts x-axis at A and Y-axis at B.

Solution:

Here,

2x + 5y - 20 = 0

or, 2x + 5y = 20

or, \(\frac{2x}{20}\) + \(\frac{5y}{20}\) = 1

or, \(\frac{x}{10}\) + \(\frac{y}{4}\) = 1

Which is in the form \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1

∴ X-intercept (a) = 10

y-intercept (b) = 4

∴ Area of ΔAOB = \(\frac{1}{2}\) OA × OB

= \(\frac{1}{2}\) × 10 × 4

= 20 sq. units

Find the length of perpendicular drawn from (-3, 0) to the line 3x + 4y + 7 = 0.

Solution:

The equation of line is 3x + 4y + 7 = 0

Perpendicular length (p) = |\(\frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}}\)|

we have, (x_{1},y_{1})= (-3,0)

So, p = |\(\frac{3\times(-3)+ 4\times 0}{\sqrt{3^2 + 4^2}}\)|= \(\frac{9}{5}\)

Find the length of perpendicular drawn from (-3, 0) to the line 3x + 4y + 7 = 0.

\(\vert \frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\vert\)

Solution:

The equation of line is 3x + 4y + 7 = 0

Perpendicular length (p) =\(\vert \frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\vert\)

Where (x_{1} , y_{1}) is the point from whhich the perpendicular is drawn on the line

or, p = \(\vert\frac{3 . (-3) + 4 . 0 + 7}{\sqrt{(3)^2 + (4)^2}}\vert\)

or, p = \(\vert\frac{-9 + 7}{\sqrt{9 + 16}}\vert\)

or, p = \(\vert\frac{-2}{5}\vert\)

∴ p = \(\frac{2}{5}\) units

If p is the length of the perpendicular dropped from origin on the line \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1, Prove that \(\frac{1}{a^2}\) + \(\frac{1}{b^2}\) = \(\frac{1}{p^2}\)

Solution:

Here,

Let \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 be the equation of a line peprendicular length frawn from origin (0, 0) to \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 and its equation is given by

or, p = \(\vert\frac{0 . \frac{1}{a} + 0 . \frac{1}{b} - 1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}\vert\)

or, p = \(\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}\)

Squaring on both sides we get,

or, p^{2} = \(\frac{1}{\frac{1}{a^2} + \frac{1}{b^2}}\)

or, \(\frac{1}{a^2}\) + \(\frac{1}{b^2}\) = \(\frac{1}{p^2}\)

∴ \(\frac{1}{p^2}\) = \(\frac{1}{a^2}\) + \(\frac{1}{b^2}\) proved.

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