 # General Equation of a Line

In simple terms, we can say the line is a figure which has no curvature. In two variables, the general equation of a line of the first degree is represented as

Ax + By +C = 0,

A, B ≠ 0 where A, B, and C are constants that belong to real numbers.

Coordinate of a point is denoted by writing the x-coordinate first and then y-coordinate, which is separated by a comma.

How to Find the Equation of a Straight Line that is Parallel to One of the Coordinate Axes.

To find the normal form of a straight line, consider we have two axes in a two-dimensional form that is x and y.

A straight line AB cuts the x-axis at point A. This line AB is parallel to axis y.

We can consider the distance OA = a. Let P be any point on line AB such that the coordinates of P are (x, y).

The abscissa of the point is always c.

x = c

This is true for every point on the line AB.

It should be noted that the equation does not contain the coordinate y.

Similarly, we can find the equation of a straight line parallel to the x-axis.

y = d

The equation of x-axis is y = 0

The equation of y-axis is x = 0.

How to Find the Equation of a Straight that Cuts Off a Given Intercept on the Axis of Y and is Inclined at an Angle to the X-axis.

Let the intercept be c and the angle be a.

Let c be a point on y-axis such that the distance  OC is c.

Draw a perpendicular PM to OX and a line parallel to the x-axis that is CN. See figure below

Let the coordinates of P be (x, y) such that OM = x and PM = y.

MP = NP + MN

By trigonometry, we get NP = CN. tan(a)

Therefore,

We know that tan a = m ( also called the slope of the line ).

So, y = CN.tan(a) + OC

= mx + c

Therefore,

We get the general equation of a line which is

y = MX +c

Example 1: Find the equation of the straight line cutting of an intercept 3 in the negative direction of the y-axis and inclined at 120° to the axis of x.

y = x.tan(120°) +(-3)

y +x√3 + 3 = 0.

Example 2: How to find the equation of a line that cuts off given intercepts a and b on x-axis and y-axis respectively.

Given is that OA = a and OB = b

Now join AB and extend it in both ways making it a line. Let P be any point which has coordinates (x, y) present on this straight line.

Draw PM perpendicular to OX and touching line AB at P as shown in fig below.

By geometry we get

OM/OA = PB/AB and MP/OB =AP/AB

Adding LHS for both of these we get

OM/OA + MP/OB =( PB + AP)/AB

= 1

Therefore we get

x/a + y/b =1

How to Find the Equation of a Straight Line in Terms of Perpendicular which Falls Upon it from the Origin and the Angle Which this Perpendicular Makes with the X-axis.

Step1: Let the given line be AB (for which we need to find it's equation).

Step 2: Let OR be perpendicular to AB. Its length is let's say p.

Step 3: Let a be the angle that OR makes with OX. Let P be any point whose coordinates are x and y lying on AB.

Step 4: Draw the ordinate PM and ML perpendicular to OR. All of these is given in the image below

OL = OM cos(a)...................(1)

LR = NP = MP sin (NMP)

Ang NMP =90° - ang NMO = ang MOL= a

LR = MP sin(a).......................(2)

By adding 1 and 2 we get

OM cos a + MP sin a = OL + LR = OR = p

Therefore, xcosa + ysina = p

In all the above forms of the line, we see that they are of first degree in x and y.

We now know that any equation of first degree in x and y always represents a straight line.

A general form of this equation is :

Ax + By + C = 0

Here A, B, C are constants ie these are quantities that don't contain x and y.

Ax + By + C = 0 it may also be written as

y = {-A/B}(x) - C/B

Comparing this above equation with y= mx + c we get

m =-A/B and c = -C/B

Therefore the equation Ax + By + C = 0 represents a straight line cutting off an intercept -C/B to the axis of y and inclined at an angle Tan^-1(-A/B).

How to Find the Equation of a Straight Line Which Passes Through Two Given Points (-1, 3) and (4,-2)

We know that equation to any straight line is y = mx + c ;..................(1)

for first point we get

3 = -m + c, such that c = m + 3

Hence (1) becomes

y = mx + m + 3……………(2)

For the second point, we have

-2 = 4m + m +3

So solving this we get m = -1,

Hence  (2) becomes

y = -x + 2 => x + y = 2

If two lines are parallel to each other we get their slopes to be the same.

That is m1 = m2.

If two lines are perpendicular we get

m1.m2 = -1

If two lines subtend an angle let's say "a" between them then

Tan(a) =( m1 - m2)/1 + m1m2

To fix definitely the position of a straight line we must always have two quantities given. Such that one is the point and the other is a quantity that gives the direction of a line.