# Straight Lines: Everything You Need to Know

**Straight Lines:**

As kids we all have been taught different types of lines. The standing, sleeping, slanting etc. But, what exactly is a line? A **line** is simply an object that is characterized as a straight, thin, one-dimensional, zero width object that extends on both sides to infinity. In this article, find everything you need to know about straight lines.

A **straight line** is essentially just a line with no curves. Most of the time, when we speak about lines, we are talking about straight lines! Here are some examples of straight lines.

- Curved lines are usually called a curve or arc and are not straight lines.
- Horizontal straight lines go from left to right or vice versa whereas vertical straight lines go up and down.
- Parallel straight lines have the same slope and they will never intersect.
- Perpendicular straight lines cross each other and form four perfect right angles in the process.
- Slanted or oblique straight lines are just as they sound: they are straight lines at an angle.

### Angles on one side of a straight line always add to 180 degrees

**On the Cartesian plane**

Under Cartesian plane lines more generally, in affine coordinates, can be described algebraically by linear equations.

In two dimensions, the equation for non-vertical lines is given in the slope-intercept form:

Y = mc + C

**Where:**

*m* is the slope or gradient of the line.

*c* is the y-intercept of the line.

*x* is the independent variable of the function *y* = *f*(*x*).

**Intercepts**

The line crosses the x – axis when x = 0

The line crosses the y – axis when y = 0

**Points to be remembered: **

- Distance between two points

( x1, y1) &(x2,y2) = √ (x2 – x1)2 + (y2 – y1)2

- Centroid of a triangle having vertices

(x1, y1) ,( x2,y2) &( x3,y3) is (x1+x2+x3) / 3 , (y1+y2y+y3) / 3

- Incentre of a triangle is given by

- Equation of a line passing through a given point (x1, y1) & having slope ‘m’ is given by

y – y1 = m( x – x1)

- Equation of a line passing through two given points (x1, y1) & ( x2,y2) is given by

y – y1 = ( x – x1), y2—y1 x2—x1

- Equation of a line in normal form is given by x cosα + y sinα = p

- Angle between two lines having slopes

‘m1’ & ‘m2’ is given by Tanθ = (m1 –m2)/ (1 + m1.m2)

**Let us solve some examples:**

- Find the equation of a line which passes through the point (2, 3) and makes an angle of 30° with the positive direction of X-axis.

**Solution **

Here the slope of the line is m = tan θ = tan 30° = 1/√3

The given point is (2, 3). Therefore, using point slope formula of the equation of a line, we have y – 3 = 1/√3 (x – 2) or x- √3y + (3√3 – 2) = 0

** **

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).

**Solution:**

Let m be the slope of the required straight line.

Since the required straight line is perpendicular to the line joining P (2, -3) and Q (4, 2).

Therefore,

m × Slope of PQ = -1

⇒ m × 2+3 / 4-2 = -1

⇒ m × 5/2 = -1

⇒ m = – 2/5

The base of an equilateral triangle with side 2a lies along they y-axis such that the mid point of the base is at the origin. Find vertices of the triangle.

** Solution:**

Let ABC be the given equilateral triangle with side 2a. Accordingly, AB = BC = CA = 2a

Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin. i.e., BO = OC = a, where O is the origin.

Now, the coordinates of point C are (0, a), while the coordinates of point B are (0, –a). The line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular. Hence, vertex A lies on the y-axis.

On applying Pythagoras theorem to ∆AOC, we obtain (AC)2 = (OA)2 + (OC)2

⇒ (2a)2 = (OA)2 + a 2

⇒ 4a 2 – a 2 = (OA)2

⇒ (OA)2 = 3 a 2

⇒ OA = √3a

∴Coordinates of point A = ± (√3a,0)

Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), – (√3a, 0) and or (0, a), (0, –a), and + (√3a, 0)

3. Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).

**Solution:**

The slope of the line joining the points (3, –1) and (4, –2) is m = [(-2) – (-1)] ÷ (4 – 3) = -2 + 1 = -1 x

Now, the inclination (θ) of the line joining the points (3, –1) and (4, – 2) is given by

Tan θ = –1 ⇒ θ = (90° + 45°) = 135°

Thus, the angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°

** **

**Questions for practice**

**Short Question Types:**

- Find the in Centre of the triangle whose vertices are (1,2),(3,4) &(–5,4)
- Find the equation of the line having slope ‘2’ & passing through the point (3,–5)
- Find the equation of the line passing through two given points (–4, 3) & (4,6)

** **

**Long Question Types:**

- Find the co-ordinates of the foot of the perpendicular from the point (–1,3) to the line 3x –5y = 4
- Find the distance of the point (2,5) from the line x/2 + y/3 = 1 measured parallel to the line y = 2x + 3
- In triangle ABC with vertices A (2, 6), B (3,-1) & C (1, 2) find the equation & length of altitude from the vertex A.
- Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1, through the point where it meets the y- axis.
- Find the co-ordinates of the in center of the triangle formed by the lines y = 15, 5x – 12y = 0 & 3x+ 4y =

This was our article on straight lines. For more such articles, keep following us here!

## Are you slow and accurate?

Or quick but careless?

Practice from a bank of 300,000+ questions , analyse your strengths & improvement opportunities.

No thanks.