## What is a straight line?

**A straight line is a curve such that every point on the line segment joining any two points on it lies on it.**

Let ax + by + c = 0 be a first degree equation in x, y where a, b, c are constants. Let P (x_{1}, y_{1}) and Q(x_{2}, y_{2}) be any point on the curve represented by ax+ by + c = 0. Then,

ax_{1} + by_{1} + c = 0 and ax_{2} + by^{2} + c = 0

When we say that the first degree equation is x, i.e. ax + by + c = 0 represents a line, it means that all points (x, y) satisfying ax + by + c = 0 lie along a line. Thus, a line is also defined as the locus of a point satisfying the condition ax + by + c = 0, where a, b, c are constants.

It should be noted that there are only two unknowns in the equation of a straight line because the equation of every straight line can be put in the form ax + by + c = 0, where a and b are two unknowns. It is important to note here that x and y are not unknowns. In fact, these are the coordinates of any point on the line and are known as current coordinates. Thus, **to determine a line we will need two coordinates to determine the two unknowns**.

## Slope of a Line

**The trigonometrical tangent of the angle that a line makes with the positive direction of the x-axis in an anticlockwise sense is called the slope or the gradient of a line.**

The slope of a line is generally denoted by *m*. Thus m = $\tan \Theta$

**Since a line parallel to the x-axis makes an angle of 0 ^{o} with the x-axis, therefore, its slope is tan 0^{0} = 0**

A line parallel to the y-axis, i.e. a line that is perpendicular to the x-axis makes an angle of 90^{o} with the x-axis, so its slope is tan $\frac{\pi }{2}$ = ∞. Also, the slope of a line equally inclined with axes is 1 or -1 as it makes an angle of 45^{o} or 135^{o} with the x-axis.

The angle of inclination of a line with the positive direction of the x-axis in an anticlockwise sense always lies between 0^{0} and 180^{0}.

Let us now understand the slop using some examples.

**Example**

What can be said regarding a line is its slope is a) positive b) zero c) negative?

**Solution**

Let $\Theta$ be the angle of inclination of the given line with the positive direction of the x-axis in an anticlockwise sense. Then, its slope is given by m = tan $\Theta$.

a) If the slope of the line is positive, then,

m = tan $\Theta$ > 0 ⇒ $\Theta$ lies between 0^{0} and 90^{0}

This means that $\Theta$ is an acute angle.

**Thus the line of positive slope makes an acute angle with the parallel direction of the x-axis.**

b) If the slope of a line is zero, then,

m = tan $\Theta$ = 0 ⇒ $\Theta$ = 0^{0}

This means that either the line is x-axis or it is parallel to the x-axis.

**Thus the line of zero slope is parallel to the x-axis.**

c) If the slope of the line is negative, then,

m = tan $\Theta$ < 0 ⇒ $\Theta$ lies between 90^{0} and 180^{0}

This means that $\Theta$ is an obtuse angle.

**Thus the line of negative slope makes an obtuse angle with the parallel direction of the x-axis in an anticlockwise direction.**

### Slope of a line in Terms of Coordinates of any Two Points on it

What would be the slope of a line in terms of coordinates of any two points on it? Let us find out.

Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) be two points on a line making an angle q with the positive direction of the x-axis. Draw Pl, QM perpendicular on x-axis and PN_QN on QM. Then,

PN = LM = OM – OL = x_{2} – x_{1} and QN = QM – NM = QM – PL = y_{2} – y_{1}

In △ PQN, we have,

tan $\Theta =\frac{QN}{PN}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Thus, if (x_{1}, y_{1}) and (x_{2}, y_{2}) are coordinates of any two points on a line then its slope is given by

**m = ** **$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{Difference\, in\, ordinates}{Difference\, in \, adbscissae}$**

Let us understand it through an example.

**Example**

Find the slope of a line that passes through the points (3, 2) and (-1, 5)

**Solution**

We know that the slope of a line passing through two points (x_{1}, y_{1}) and (x_{2}, y_{2}) is given by

m = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Here, y_{1} = 2, y_{2} = 6, x_{1} = 3, x_{2} = -1

Substituting these values in the given equation, we have

m = $\frac{5-2}{-1-3}=\frac{-3}{4}$

**Hence, the slope of a line that passes through the points (3, 2) and (-1, 5) is** $\frac{-3}{4}$.

## Point-Slope Form

The equation of a straight line can be written in different forms depending on the data give. One of such forms is the point-slope form. The point-slope form includes the slope of the straight line and a point on the line as the name suggests. There can be infinite lines with a given slope, but when we specify that the line passes through a given point then we get a unique straight line. Thus, only a point on the line and its slope are required to represent a straight line in the point-slope form. Let us now understand what is a point-slope form?

The theorem of the point-slope form states that:

**The equation of a line that passes through the point (x _{1}, y_{1}) and has the slope “m” is given by:**

**y – y _{1} = m (x – x_{1})**

How did we reach upon this formula? Let us check out.

Let the straight line passing through the point Q(x_{1}, y_{1}) and let P (x, y) be any point on the line.

Then, we know that the Slope of a line = $\frac{y-y_{1}}{x-x_{1}}$

But, we also know that the slope of the line is “m”

Therefore,

m = $frac{y-y_{1}}{x-x_{1}}$

⇒ $y-y_{1}=m(x-x_{1})$

**Hence,** **$y-y_{1}=m(x-x_{1})$ is the required equation of the line.**

## Solving Point-Slope Form Equation

The following steps should be followed for finding the equation of a given line of for solving the point-slope form:

- Make a note of the coordinate values of (x
_{1}, y_{1}) and the value of m of the straight line. - In the formula, $y-y_{1}=m(x-x_{1})$, substitute the values given.
- Simplify the equation to obtain the standard form of the equation.

## Solved Examples

Find the equation of a line passing through (2, -3) and inclined at an angle of 135^{o} with the positive direction of the x-axis.

**Solution **

To find the equation first we need to find the slope of the line using the given inclination.

We have been given that

m = Slope of the line = tan 135^{o}

Now we can write tan 135^{o} as tan (90^{o} + 45^{o})

Further, tan (90^{o} + 45^{o}) = -cot 45^{o} = -1 ( because tan (90 + q) = -cot q and cot 45^{o} = 1)

Also, we have bene given that x_{1} = 2 and y_{1} = -3

Now, we know that the equation of the line in point-slope forms I

$y-y_{1}=m(x-x_{1})$

Substituting the given values in the above equation, we have

$y-(-3)=-1(x-2)$

y + 3 = -x + 2

⇒ x + y +1 = 0

**Hence, the equation of a line passing through (2, -3) and inclined at an angle of 135 ^{o} with the positive direction of the x-axis will be x + y +1 = 0**

Determine the equation of the line through the point (4, -3) and parallel axis.

**Solution**

We have been given that the straight line passes through the point ( 4, -3 ).

This means we have been given that x_{1} = 4 and y_{1} = -3

Also, the straight line is parallel to the x-axis.

Now, recall that we have learnt that the line of zero slope is parallel to the x-axis. This means that is a line is parallel to the axis if the slope of a line is zero, or,

m = tan $\Theta$ = 0 ⇒ $\Theta$ = 0^{0}

So, we have m = 0

We also know that,

Now, we know that the equation of the line in point-slope forms I

$y-y_{1}=m(x-x_{1})$

Substituting the given values in the above equation, we have

$y-(-3)=0(x-2)$

⇒ y + 3 = 0

**Hence, the equation of the line through the point (4, -3) and parallel axis will be given by y + 3 = 0**

Find the equation of a straight line passing through (-9, 5) and inclined at an angle of 120° with the positive direction of the x-axis.

**Solution**

To find the equation first we need to find the slope of the line using the given inclination.

We have been given that slope of the line (m) = tan 120°

Now we can write tan 120° as tan (90° + 30°) = cot 30° = -√3. ( because tan (90 + q) = -cot q and cot 30^{o} = √3)

Also, we have bene given that x_{1} = -9 and y_{1} = 5

Now, we know that the equation of the line in point-slope forms I

$y-y_{1}=m(x-x_{1})$

Substituting the given values in the above equation, we have

Therefore, the required equation of the straight line is

= $y-(-5)=-\sqrt{3}(x-(-9))$

⇒ y – 5 = -√3 ( x+ 9 )

⇒ y – 5 = -√3x – 9√3

⇒ y + √3x + 9√3 – 5 = 0

**Hence, the equation of a straight line passing through (-9, 5) and inclined at an angle of 120° with the positive direction of the x-axis will be y + √3****x + 9√3 – 5 = 0**

Find the equation of the straight line with slope 3 and passing through (-1, 5).

**Solution**

We have been given that the slope of the line is 3 or m = 3 and the line passes through (-1, 5),

We have also been given that (x_{1}, y_{1}) = (-1, 5). This means that x_{1} = -1 and y_{1} = 5

Now, we know that the equation of the line in point-slope forms I

$y-y_{1}=m(x-x_{1})$

Substituting the given values in the above equation, we have

Therefore, the required equation of the straight line is

$y-5=3(x-(-1))$

y – 5 = 3 ( x + 1 )

y – 5 = 3x + 3

⇒ 3x – y + 8 = 0

**Hence, the equation of the straight line with slope 3 and passing through (-1, 5) will be given by 3x – y + 8 = 0**

Find the equation of the straight line with slope -2 and passing through (7, -4).

**Solution**

We have been given that the slope of the line is 2 or m = -2 and the line passes through ( 7, -4 ),

We have also been given that (x_{1}, y_{1}) = ( 7, -4) . This means that x_{1} = 7 and y_{1} = -4

Now, we know that the equation of the line in point-slope form is

$y-y_{1}=m(x-x_{1})$

Substituting the given values in the above equation, we have

Therefore, the required equation of the straight line is

$y-(-4)=-2(x-7)$

y + 4 = -2x + 14

⇒ 2x – y + 10 = 0

**Hence, the equation of the straight line with slope -2 and passing through (7, -4) is ****2x – y + 10 = 0**

Find the equation of the perpendicular bisector of the line segment joining the points A (2, 3) and B (6, -5 ).

**Solution **

To find the equation first we need to find the slope of the line

Recall that if (x_{1}, y_{1}) and (x_{2}, y_{2}) are coordinates of any two points on a line then its slope is given by

m = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

We have been given that x_{1} = 2 and y_{1} = 3. Also x_{2} = 6 and y_{2} = – 5

Substituting these values in the above equation we will get

m = $\frac{-5-3}{6-2}=\frac{-8}{4}=-2$

Now, we know that the slope of a line perpendicular to the line AB will be given by $\frac{-1}{m}$.

Therefore, we have

Slope of the line = $\frac{1}{2}$

Now, let P be the mid-point of AB. Then the coordinates of P will be given by

$(\frac{2+6}{2},\frac{3-5}{2})=( 4, -1 )$

Since now we have one point as well as the slope of the line, we can use the point-slope form to find the equation of the line.

We know that the equation of the line in point-slope form is

$$

Substituting the given values in the above equation, we have

Therefore, the required equation of the straight line is

$\frac{y-y_{1}}{x-x_{1}}$

$y+1=\frac{1}{2}(x-4)$

⇒ 2 ( y + 1 ) = 1 ( x – 4 )

⇒ 2y + 2 = x – 4

⇒ x – 2y – 6 = 0

**Hence, the equation of the perpendicular bisector of the line segment joining the points A (2, 3) and B (6, -5 ) will be given by ****x – 2y – 6 = 0**

A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1:2. Find the equation of the line.

**Solution**

We have been given that the points are(1, 0) and (2, 3) and the line perpendicular to the line segment divides it in the ratio of 1 : 2.

Let M divide the join of A(1,0) and B(2,3) in the ratio of 1:2

Then, the point M is $(\frac{1\times 2+2\times 1}{1+2},\frac{1\times 3+2\times 0}{1+2})$.

Hence, M= $(\frac{4}{3},1)$

Slope of the line AB= $\frac{3-0}{2-1}$ = 3

Let L be the required line. Then, the slope of L= $\frac{-1}{3}$ and it passes through M.

Now, let L be the required line. we have been given that L is perpendicular to AB.

Therefore,

Slope of L= $\frac{-1}{3}$ and it passes through M $(\frac{4}{3},1)$

Since now we have one point as well as the slope of the line, we can use the point-slope form to find the equation of the line.

We know that the equation of the line in point-slope form is

$y-y_{1}=m(x-x_{1})$

Substituting the given values in the above equation, we have

Therefore, the required equation of the straight line is

$y-1=\frac{-1}{3}(x-\frac{4}{3})$

9 ( y – 1 ) = -1 ( 3x – 4 )

⇒ 9y – 9 = -3x + 4

⇒ 3x + 9y – 13 = 0

Hence, the equation of the line that is perpendicular to the line segment joining the points (1, 0) and ( 2, 3 ) and divides it in the ratio 1:2 will be given by 3x + 9y – 13 = 0

# Remember

- A straight line is a curve such that every point on the line segment joining any two points on it lies on it.
- To determine a line we will need two coordinates to determine the two unknowns.
- The trigonometrical tangent of the angle that a line makes with the positive direction of the x-axis in an anticlockwise sense is called the slope or the gradient of a line.
- The slope of a line is generally denoted by
*m*. Thus m = tan$\Theta$ - Thus the line of positive slope makes an acute angle with the parallel direction of the x-axis.
- Thus the line of negative slope makes an obtuse angle with the parallel direction of the x-axis in an anticlockwise direction.
- Thus the line of zero slope is parallel to the x-axis
- The point-slope form includes the slope of the straight line and a point on the line
- Thus, if (x
_{1}, y_{1}) and (x_{2}, y_{2}) are coordinates of any two points on a line then its slope is given by**m = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{Difference\, in\, ordinates}{Difference\, in \, adbscissae}$** - The equation of a line that passes through the point (x
_{1}, y_{1}) and has the slope “m” is given by: y – y_{1}= m (x – x_{1})

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