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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » Equation of plane and line

Equation of plane and line

In this article, we will discuss a straight line, which is a simple line without any curves.

Table of Content
  •  

INTRODUCTION

A straight line has only one dimension without any breath. When there is no turning point in between two points then it’s called a straight line. In other words we can say that the shortest distance between two points is known as a straight line.

In this topic we learn the concepts of lines like slopes , angle b/w  two lines , equation of line.

Now let us look forward and start learning in detail about straight lines and their equation , equation of plane , point form , intercept form and also solve some questions that are frequently asked and see the graphs of point slope , intercept and much more . Now we will start studying in depth.

DEFINITION AND EQUATION OF PLANE

Plane is two dimensional surface that extends indefinitely. The general form of plane is ax+by+cz+d=0  here, a, b and c are the components of normal vector n=(a , b , c)

DEFINITION AND EQUATION OF LINE

When there is no turning point in b/w the two points are said to be straight lines. Or we can say that a straight line has only one dimension as we talk about it early.

EQUATION OF LINE

The general equation is

ax+by+c=0

here x&y are variables and a,b,c are constants.

INTERCEPT FORM OF LINE

The equation of line which cuts off intercepts a and b from x-axis and y-axis.

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POINT FORM

Let the equation of x-intercept is ‘a’ and y-intercept is ‘b’

 

TWO POINT FORM

The equation line when passes through the points (x1,y1) and (x2,y2) is

                                          y-y1 = y2-y1 (x-x1)

                                                     x2-x1

                                                    OR                           

                                           y-y2 = y2-y1 (x-x1)

                                                      x2-x1

(x1,y1) and (x2,y2)  two points on a line (x,y) are variables. 

SLOPE OF LINE

The slope of straight line is tanƟ

                       y=mx+c

  • When the line is parallel to the x-axis then the slope is 0.

  • When the line is parallel to the y-axis then the slope is undefined.

SLOPE OF LINE WHEN PASSING FROM TWO GIVEN POINTS

RELATION BETWEEN TWO LINES

Assume P1 and P2 are two lines

P1 = a1x+b1y+c1=0

P2 = a2x+b2y+c2=0

  • a1/a2 =b1/b2 ≠ c1/c2 

when two lines are parallel lines

  • a1/a2  ≠ b1/b2

when two lines intersect a point

  • a1/a2 =b1/b2 =c1/c2 

when the lines are coincident

ANGLE BETWEEN TWO LINES

THE ANGULAR BISECTOR OF STRAIGHT LINE

Let  the P1 = a1x+b1y+c1 = 0

And P2 = a2x+b2y+c2 = 0

When the angle of bisector containing the origin then the equation is:

THREE LINES CONCURRENCY

Let assume the lines be

P1 = a1x+b1y+c1 = 0

P2 = a2x+b2y+c2 = 0

P3 = a3x+b3y+c3 = 0

And concurrency of lines condition is

CONCLUSION

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faq

Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

find the equation of line when the line intersects the y-axis at a distance of 2 units above the origin and make the angle of 30 in the positive direction of the x-axis.

...Read full

find the equation of line when the line passing through the points (-1,1) and(2,-4)

...Read full

find the measure of the angle b/w the lines x+y+7=0 and x-y+1=0.

Ans: x+y+7=0 m1 = -1/1 x-y+1=0 m2 = -1/-1 = 1 Slopes of the two lines are 1 and -1 as the product...Read full

find the calculation whether the points (13,8),(26,-4) lie in the same,adjacent,opposite angle formed by the st. line 5x+6y-112=0?

Ans: First we put (13,8) in equation 5x+6y-112=0 and we will get 1 . Now we put (26,-4) in equation 5x+6y-112=0 and we will get -6. Therefor...Read full

equation

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Ans: x+y+7=0

m1 = -1/1

x-y+1=0

m2 = -1/-1 = 1

Slopes of the two lines are 1 and -1 as the product of these two slopes is -1, the lines are at  90 degree.

 

Ans: First we put (13,8) in equation 5x+6y-112=0 and we will get 1 . Now we put (26,-4) in equation 5x+6y-112=0 and we will get -6.

Therefore it says that points lie opposite to the line.

Question 5: Find the ratio in which the line joining (-2, 5) and (-5, -6) is divided by the line y = -3..

Ans: let ratio be k:1

The line diving is y = -3 , therefore point will be (x,-3)

By using section formula

-3 = -6k+5/k+1

-3k-3 = -6k +5

3k=8

K=8/3

Therefore the ratio is 8:3.

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