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 ± (\(\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}\))
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, (x_{1},y_{1})= (-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 (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
Find the length of the perpendicular drawn from (0,0) to the line 2x +3y+5=0.
Find the length of perpendicular drawn from(1,2) to the line 2x + 5y -20=0
Find the perpendicular drawn from (-2, 5) to the line x +y -5=0
Find the perpendicular drawn from(3, -2) to the line x - y- 6 = 0
Find the perpendicular drawn from(-7, -2) to the line 6x -8y- 20 = 0
find the distance between the parallel lines 3x + 5y = 11 and 3x +5y =-23
find the distance between the parallel line 2x - y + 5 = 0 and 6x -3y +20 = 0
find the distance between the parallel line x+2y - 1 = and 5x + 10y + 7 = 0
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.
The line 4x - y +16 = 0 cuts X- axis at P and Y axis at Q. find the area of ( riangle) POQ
You must login to reply
shankar
how to find two lines are parallel
Mar 12, 2017
1 Replies
Successfully Posted ...
Please Wait...