Quantitative Aptitude > Coordinate Geometry

Refresher Test - Coordinate Geometry Launch Test Discuss

Practice Tests
Practice Test - Coordinate Geometry Launch Test Discuss

Add Comment Bookmark share + Refresher Material


Free Online Preparation for CAT with Minglebox e-CAT Prep. Cover basic concepts of Co-ordinate Geometry under Quantitative Aptitude for MBA Entrance Exam Preparation with Study material, solved examples and tests prepared by CAT coaching experts.

 

In two dimensional Coordinate Geometry, location of any point lying in the plane, is given by specifying the perpendicular distances of the point, from a set of fixed mutually perpendicular lines. The fixed mutually perpendicular lines are known as X-axes and Y-axes respectively. The point of intersection is known as the origin 'O'.

 

 

 

These 2 lines divide the given plane in 4 parts known as quadrants. Distances measured to the right hand side of origin O are treated as positive and distances measured towards the left of origin are treated as negative. In a similar way distances measured along the Y-axis and above the X-axis are treated as positive distance measured below the X-axis are treated as positive distances measured below the X-axis are treated as negative.

 

The Coordinates of a point are specified as an ordered pair, Comprising of its distance measured along the X and Y axis . Distance measured along one X-axis from the origin is known as abscissa, and usually denoted X, the G distance measured along the Y-axis From the origin is called the ordinate of the point and is denoted by y. Thus coordinate is of a point are specified as (x,y) I.e as (abscissa, ordinate).

 

The 4 quadrants are known as 1 st ,2 nd 3 rd & 4 th quadrant. The below diagram summarizes the sign of the abscissa and the ordinate of x points lying in 1 st , 2 nd , 3 rd , 4 th quadrant.

 

 

 

Distance Formula

 

The distance d between 2 points and having coordinates

( ) and is given by

D =

 

Section formula and mid point formula

 

 

The coordinates P of a point which divides the join of points and in the

 

ratio m:n is given by X = ; Y =

 

 

Mid point Formula

 

The Coordinates of the mid point of the line joining and is given by X =

 

; Y =

 

Coordinates of the centroid of a triangle

 

Let , and be the coordinates

of the vertices of triangle , then coordinates of it's centroid (x,y) is given by X =

  ; Y =

 

 

The coordinates of the incentre I (x,y) of the triangle with vertices ;

and side length a,b,c is given by X =

 

Area of a triangle

 

The area of triangle having vertices ; and is

given by = [ + + ]

 

Note – If area of is Zero points are collinear.

 

Condition for 4 points , no three of which are collinear to be a parallelogram . Let

, , be 4 points lying in a plane then , , , are the verticals of

a parallelogram if and

 

are the vertices of a square if in addition to above or

 

Equation of a Line

 

Slope of a line

The slope of a line is defined as tangent of the angle which the line makes in the positive direction of the x-axis in the anti – clockwise direction.

The slope is generally represented by 'm' thus in the notion used above

The equation of a line is given by y = mx+c where 'c' is the intercept mode on the y -axis by the line.

Equation of x-axis is y = 0

Equation of y-axis is x = 0

 

2 lines are said to be parallel if they have the same slope that is if where where m1 and m2 are the slopes of the 2lines.

2 lines said to be perpendicular if they product of there slope is -1. that is if then lines are perpendicular to each other. Different forms of the equation of a line.

Line passing through point and having slope 'm' .  

Line passing through points and .

If a line makes intercept a,b on x and y-axis respectively then equation of line is given by

Angle between 2lines having slopes and is given by where is the actual

angle between the 2 lines.

Perpendicular distance between parallel lines , Let the equation of 2 lines be and line is given by

Perpendicular distance of a point from a line . Let be any point and let y = mx +c be any line, then the perpendicular distance of point p from line y = mx + c is given by

 

Equation of a Circle

 

The equation of a circle having center at point (h,x) and radius 'r' given by

 

Solved Examples

Question

 

Find the equation of the line with slope 2 and intercept on the y-axis as -7 ?

 

Solution

 

we know that equation of the line having slope 'm' and intercept on y-axis is 'c' the equation is y = mx +c , hence equation of line is y = 2x – 7

 

Question

 

The equation of a line which makes an intercept of 3 on x-axis and -3 on y-axis is

  1. x – 2y = 5

  2. x – 2y = 3

  3. 4x + 13y = 7

  4. none of these

 

Solution

 

The line with the x-axis at point (3,0) and y-axis at the point (0,-3) hence equation of line is using

y = x-3 or x – y = 3

hence answer is option (2)

 

Question

 

Find the equation of the line perpendicular to the line x + y = 2 and passing through (1,2)

  1. y = x + 3

  2. y = x – 1

  3. y = x + 1

  4. y – x = 2

 

Solution

 

Let slope of required line be 'm' then slope of line x + y = 2 is '-1' for lines to be perpendicular the product of their slope should be '-1' hence we get

m = 1

using one point slope from the equation of we get required equation as

(y – 2) = 1. (x-1)

y = x + 1

Hence (3) is the answer.

 

Question

 

The equation of the line through the intersection of the lines 2x + y = 3 and 3y – x + 2 = 0 and having slope -1/2 is

  1. 14y +5x +10 = 0

  2. 14y+7x-9 = 0

  3. 14y+7x-6 = 0

  4. 14y+7x+11 = 0

 

Solution

 

Equation of any line through the point of intersection of the given lines is of the form (2x+y-3) + (3y-x+2) = 0 where k is the constant to be determined , remaining we get

(3k+1)y + (2-k)x+2k-3 = 0,but slope of this line (given) solving we get k = 3/5

hence putting the value of k = 3/5 we get equation required line as 14y+7x-9 = 0

 

Question

 

The coordinate of the verities of a triangle are

         

  1. (3,5)
  2. (2,6)
  3. (4,4)
  4. (2,4)

 

Solution

 

The Coordinate of the centroid of the triangle with vertical is given by X

= : Y =

Hence X = Y =

Therefore option (4) is the correct answer.

 

 

 

 

Refresher Test - Coordinate Geometry Launch Test Discuss

Practice Test - Coordinate Geometry Launch Test Discuss

Comments Add Comment

arthur5534 : thanks minglebox you are gr8 to help us so much and your material and tests are perfect with less errors.

: the last question seems to carry no real meaning....its incomplete!!

nikhilnk : Helpful material. Can you please explain the solution for the last question again?