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#### Linear Equation Ax + By + C = 0

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(x1, y1), Q(x2, y2) and R(x3, y3) 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,
Ax1 + By1 + C = 0 . . . . . . . . . . . . . . (i)
Ax2 + By2 + C = 0 . . . . . . . . . . . . . . (ii)
Ax3 + By3 + 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(y1 - y2), B = k(x2 - x1) and C = k(x1y2 - x2y1)
Substituting the values of A, B and C in the third equation, we get
k(y1 - y2) x3 +k(x2 - x1) y3 + k(x1y2 - x2y1) = 0
or, k(x3y1 - x3y2 + x2y3 - x1y3 + x1y2 - x2y1) = 0
or, x1y2 - x2y1 + x2y3 - x3y2+ x3y1 - x1y3 = 0
Multiplying both sides by $$\frac{1}{2}$$, we get
$$\frac{1}{2}$$(x1y2 - x2y1 + x2y3 - x3y2+ x3y1 - x1y3) = 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.

#### Reduction of general equation of first degree to the three standard forms

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 = cos2α + sin2α = 1
or, k2 (A2 + B2) = 1
or, k= $$\frac{1}{A^2 + B^2}$$
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.

#### Length of the perpendicular from a point on a straight line

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(x1, y1) 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 (x1, x1)
So, x1 cosα + y1 sinα = p'
Hence, PM = ± (x1 cosα + y1 sinα - p) which is the length of the perpendicular drawn from (x1, y1) 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 (x1, y1) to the line x cosα + y sinα - p is

L = ± (x1 cosα +y1 sinα - p)

= ± (x1$$\frac{A}{\sqrt{A^2 + B^2}}$$ + y1$$\frac{B}{\sqrt{A^2 + B^2}}$$ -$$\frac{C}{\sqrt{A^2 + B^2}}$$)

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

Hence, length of the perpendicular drawn from the point (x1, y1) to the line
Ax + By + C = 0 is ± ($$\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}$$)

1. 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.
2. Ax + By + C = 0 is± ($$\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}$$)
3. Ax + By + C = 0
4. $$\frac{1}{2}$$(x1y2 - x2y1 + x2y3 - x3y2+ x3y1 - x1y3) = 0
.

#### Click on the questions below to reveal the answers

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) = $$\frac{-5}{4}$$

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

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

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, (x1,y1)= (-3,0)

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

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 (x1 , y1) 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

0%
• ### Find the length of the perpendicular drawn from (0,0) to the line 2x +3y+5=0.

(frac {8}{sqrt{29}})
(frac {8}{sqrt{29}})
(frac {5}{sqrt{13}})
(frac {2}{sqrt{30}})
• ### Find the length of perpendicular drawn from(1,2) to the line 2x + 5y -20=0

(frac {6}{sqrt{29}})
(frac {8}{sqrt{29}})
(frac {2}{sqrt{29}})
(frac {9}{sqrt{29}})

(sqrt {5})
(sqrt {3})
(sqrt {4})
(sqrt {2})
• ### Find the perpendicular drawn from(3, -2) to the line x - y- 6 = 0

(frac {2}{sqrt{4}})
(frac {1}{sqrt{2}})
(frac {1}{sqrt{3}})
(frac {3}{sqrt{2}})
• ### Find the perpendicular drawn from(-7, -2) to the line 6x -8y- 20 = 0

(frac {23}{5})
(frac {25}{2})
(frac {21}{5})
(frac {24}{8})
• ### find the distance between the parallel lines 3x +  5y = 11 and 3x +5y =-23

(frac {13}{sqrt{35}})
(frac {12}{sqrt{34}})
(frac {11}{sqrt{34}})
(frac {10}{sqrt{34}})
• ### find the distance between the parallel line 2x - y + 5 = 0 and 6x  -3y +20 = 0

(frac {sqrt5}{3})
• ### find the distance between the parallel line x+2y - 1 = and 5x + 10y + 7 = 0

(frac {2}{5sqrt{7}})
(frac {2}{5sqrt{5}})
(frac {2}{5sqrt{6}})
(frac {2}{5sqrt{8}})
• ### The straight line 4x  + 5y  -20 = 0 cuts  X- axis at A and Y axis at B. find X-intercept, Y-intercept hence find the area of ( riangle)AOB.

12 sq. units
;i:3;s:18:
10 sq. units
• ### The  line 4x  - y  +16 = 0 cuts  X- axis at P and Y axis at Q. find the area of ( riangle) POQ

34sq. units
32 sq. units
31 sq. units
33 sq. units