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# Straight Lines Last updated date: 09th Dec 2023
Total views: 23.4k
Views today: 1.23k     ## Concept of Straight Lines

We come across many things daily which are completely straight and aligned along the similar lines or in parallel setting with the other line of the same length or different measurements but straight line. Now several examples of straight lines would be railway tracks, ruler (to measure), even parts of a mirror can be called a straight line. Now if we add those sides then we would be looking at full geometric relations where adding up lines will make up different shapes, that is a scientific notion but Mathematically any fixed line or curve on the line segment can be defined as a straight line.

### What is meant by a Straight Line?

A line with zero or no curves or a structure with infinite length devoid of any curves is defined as a straight line. The straight line extends to both sides.

A straight line can be defined as an endless one-dimensional figure with no width. It can also be said to be a combination of endless points joined on both sides of a point. A straight line is devoid of curves and does not have any curve on it. A straight line can be further differentiated into horizontal, vertical, or slanted.

three different types of straight lines:

• Horizontal lines: The lines drawn horizontally

• Vertical lines: The lines drawn vertically

• Oblique or slanting lines: The lines drawn in a slanting position are termed as oblique or slanting lines.

We define a straight line as a curve where every point on the line segment joining any two points lies on it.

Let’s throw light on the coordinate geometry to prove that every first-degree equation in x, y represents a straight line.

Let ax + by + c = 0 be a first-degree equation in x,y where a, b, and c are constant.

Let points P (x1, y1) and Q (x2 , y2 ) be any two points on the curve represented by ax + by + c = 0.

Then, ax1+ by1 + c = 0, and ax2  + by2  + c = 0 ….(1)

Let M be the point on the line segment joining P and Q. Let’s say this point M divides PQ in the ratio of n:1. Then the coordinates of M are:

$\frac{(nx2+x1)}{n+1},\frac{(ny2+y1)}{n+1}$

Now putting in the equation ax + by + c = 0, we have,

=$a\frac{(nx2+x1)}{n+1}+b\frac{(ny2+y1)}{n+1}+c$

=$\frac{n[(ax2+by2+c)+(ax1+by1+c)]}{n+1}$

From equation (1), we get,

= n x 0 + 0

= 0

This means point M $\frac{(nx2+x1)}{n+1},\frac{(ny2+y1)}{n+1}$ lies on the curve represented by ax + by + c = 0.

Thus, every point on the line segment joins P and Q lines on ax + by + c = 0.

Hence, ax + by + c = 0 corresponds to a straight line.

### General Equation of a Straight Line

The general equation of a straight line y = m * x + c.

m is the gradient (slope) of the line

We define the slope of a line as the trigonometrical tangent of the angle that a line makes with the positive direction of the x-axis in an anticlockwise sense.

c = intercept on the y-axis, and

∴ y  = c is the value where the line cuts the y-axis.

Now, let’s look at an example to understand this equation of a straight line.

∠MPQ = Ө

∴ The slope of a line l =  m = tan Ө = QM/PM =(y2−y1)/(x2−x1)

For example,

m  =$\frac{(y2-y1)}{(x2-x1)}$

= $\frac{(25-5)}{(30-15)}$

= $\frac{(4)}{(3)}$

Hence,  y = m * x + c is an equation of straight, which is also known as a slope-intercept form.

y = mx + b is the slope intercept form that can be used to derive the  equation of a straight line. So the place where the line intercepts the “y” axis and the “m” denotes the slope of the line.

### What is the formula y=mx+b ?

The equation that is derived for any straight line is called a linear equation and it can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept.

### What is mx+b in a slope?

In the above mentioned equation, the slope is the number “m” is multiplied on the x and similarly the “b” is the intercept for “y”( the actual point where the line crosses the vertical y-axis).

### Equation of Pair of Straight Lines

Let’s take two different straight lines intersecting at the origin, and the equation of each is:

L1 = ax1+ by1 +c, and L2 =  ax2  + by2  + c = 0

Now, let’s take any point P on the line L1.

The path it travels on these two lines is its locus. Now, we need to find the focus of this point that can travel on both these lines. This means coordinates of P (h,k) should satisfy an equation that shows it can be on L1 and L2.

Therefore, to find the equation of a pair of straight lines, we multiply the equations of L1 and L2, i.e.,

(ax1+ by1 +c) * (ax2  + by2  + c) = 0.

Since these two lines intersect at the origin, this equation is homogeneous.

If point P lies on line L1, then the product is zero, and if it is on L2, then also it is zero.

So, the point that satisfies the equation or not can be determined by the overall product, which should always be zero.

We have two equations y = m1x, and  y = m2x

Equation of a pair of straight lines = (m1x - y) (m2x - y) = m1m2x2 - (m1 + m2)xy + y2 , which is a homogeneous 2nd degree equation.

### Sum and Product of Slopes

If the equations y = m1x, and  y = m2x are represented by ,ax2+2hxy+by2=0

then , m1 + m2 = $-\frac{2h}{b}$, and m1m2 = $\frac{a}{b}$

From the equation , (a)x2+(2hy)x+by2=0 we can consider an equation to be quadratic to x, then we have:

$x=\frac{-2hy\pm \sqrt{4h^2y^2-4aby^2}}{2a}$

Here, we will get two lines on because we have a +, and a - sign.

### Pair of Straight Lines Formulas

1. We know that angle between two lines is:

$\mid tan\theta =\pm \frac{m2-m1}{1+m1m2}\mid$,

and on substituting the value of  m1 + m2 = - 2h/b, and m1m2 = a/b, we get, 2

$\Theta =tan^{-1}\mid {\frac{2\sqrt{2h^2-ab}}{(a+b)}}\mid$, which is the angle between the two lines expressed as ax2+2hxy+by2=0

2. If these two lines are parallel, then the angle between them, i.e., Ө = 0°, then

$\mid tan\Theta =\pm \frac{m2-m1}{1+m1m2}\mid =0\Rightarrow m_1=m_2$

This means when two lines are parallel; their slopes are equal.

3. If two lines are perpendicular to each other, the angle between them, i.e., Ө = 90°, then

$\frac{1}{tan90^{\circ}}=\frac{1}{\infty } = \mid \frac{1+m1m2}{m2-m1}\mid =0\Rightarrow m_1=m_2=-1$

This means when two lines are perpendicular to each other, the product of their slopes is -1, i.e., if m is the slope of L1, then the slope L2 ⏊ to it is (-1/m).

## FAQs on Straight Lines

1. How to find a slope of the line?

Firstly we will write the equation in a slope-intercept form.

Use inverse operations to solve for y so that it is written as y=mx+b.

2. Find the equation of a straight line that has y-intercept 4 and is perpendicular to straight line joining (2, -3) and (4, 2).

We will assume “m” to be the slope of the required line.

Now accordingly we see that the required line is perpendicular to the joining line,

Therefore,

m × Slope of PQ = -1

⇒ m ×  (2+3)/(4−2) = -1

⇒ m × ( 5/2)= -1

⇒ m = -2/5

3. How can we define the equation of straight lines?

When we talk about the relation between the coordinate points of lying in any given line under the equation derived Mathematically can be defined as a straight line. The most commonly used forms of the equation of straight line are y = mx + c and ax + by = c.

4. Find the equation of a straight line which makes an angle of tan-√3 with the x-axis, and cuts off an intercept of √3/2 with the y-axis.

Here, m = tan Ө = tan (tan-√3) = √3

The equation in  y = mx +c form is:

y  = √3x + √3/2

5. The ⏊PERPENDICULAR from the origin (0,0) to a line meets at the point (- 3, 7). Find the equation of the line.

We have a slope, m = (y2 - y1)/(x2 - x1)  = (7 - 0)/(- 3  - 0)= -7/3

Slope of a line perpendicular to this line = - 1/m = 3/7

(y - 7) =  3/7(x + 3)

Required equation=3x - 7y +52 = 0

6. If the pairs of lines  2x2 - 3xy + qy2=0,and 4x2 - xy -3y2=0 have one line in common, then q =?

We have L1 = 4x2 - xy -3y2=0 , and L2 = 2x2 - 3xy + qy2=0

Now, let’s look at L1,

4x2 - 4xy + 3xy -3y2=0

4x (x - y) + 3y (x - y) = 0

= (x-y)(4x + 3y) = 0

So, x - y = 0 ⇒ y/x = 1..(1), and

4x + 3y = 0 ⇒ y/x =  - 4/3…(2)

For L2,

2x2 - 3xy + qy2=0 dividing by x2 , we get,

2 - 3(y/x) + q(y/x)2= 0…(3)

Case1: Substituting the value of y/x from eq(1) in (3)

2 - 3(1) + q(1)2 = 0, we get,

q = 1

Case2: Substituting the value of y/x from eq(2) in (3)

2 - 3(4/3) + q(4/3)2 = 0, we get,

q  = 9/8

 So, the value of q are 1, 9/8

7. The pair of lines represented by 2c2+ 5xy + (c2 - 3)y2 = 0 are perpendicular to each other for how many values of c?

We know the 2nd-degree equation, i.e., Ax2 + 2Hxy + By2=0

For a perpendicular condition, make A + B = 0,

2c + (c2 - 3) = 0 = 0 ⇒ c2+ 2c -3= 0