Coordinate Geometry: Concepts, Formulas & Examples

# Important Formulas for CAT Coordinate Geometry

Coordinate Geometry holds significant importance in the CAT Quantitative Ability (QA) section. Typically, problems in Coordinate Geometry are not overly challenging, making it crucial not to overlook them in the CAT exam. Practicing a sufficient number of CAT Coordinate Geometry questions enhances your ability to tackle such problems effortlessly during the actual exam. In this article, we will delve into essential Coordinate Geometry questions for CAT QA, offering valuable practice.

## Introduction

• A coordinate graph is a rectangular grid with two number lines called axes. The x-axis is the horizontal number line and the y-axis is the vertical number line.
• The axes intersect at the origin which is the point (0,0).
• Coordinate geometry is used to describe various curves such as circles, parabolas, etc.

## Coordinate Geometry Formulas

(1) Distance Formula:
To Calculate Distance Between Two Points:
Let the two points be A and B, having coordinates to be (x_1,y_1) and (x_2,y_2) respectively.
Thus, the distance between two points is-
distance =

(2) Midpoint Theorem:
To Find Mid-point of a Line Connecting Two Points:
Consider the same points A and B, having coordinates to be (x1,y1) and (x2,y2) respectively. Let M(x,y) be the midpoint of lying on the line connecting these two points A and B. The coordinates of this point are –

(3) Angle Formula:
To Find The Angle Between Two Lines:
Consider two straight lines and , with given slopes as  m_1 and m_2 respectively.
Let “θ” be the angle between these two lines, we can then represent the angle between them as-

(4) Section Formula:
To Find a Point which divides a line into m:n Ratio using:
Consider a two straight lines having coordinates (x1,y1) & (x2,y2) respectively. Let a point which divides the line in some ratio as m:n, then the coordinates of this point are-
When the ratio m:n is internal:

When the ratio m:n is external:

(5) Area of a Triangle in Cartesian Plane:
We can compute the area of a triangle in Cartesian Geometry if we know all the coordinates of all three vertices. If coordinats are (x1,y1),(x2,y2) and (x3,y3) then area will be:

Now, Let us have a look at some more formulas for coordinate geometry. We will use the below picture as a reference for the formulas.

• Slope of PQ = m =
• Equation of PQ is as below:
or y = mx + c
• The product of the slopes of two perpendicular lines is –1.
• The slopes of two parallel lines are always equal.
If m1 and m2 are slopes of two parallel lines, then m1=m2.
• The distance between the points (x1, y1) and (x2, y2) is
• If point P(x, y) divides the segment AB, where A  (x1, y1) and B   (x2, y2), internally in the ratio m: n, then,
x= (mx2 + nx1)/(m+n)  and  y= (my2 + ny1)/(m+n)
• If G (x, y) is the centroid of triangle ABC, A  (x1, y1), B   (x2, y2), C   (x3, y3), then,
x = (x1 + x+ x3)/3 and y = (y1 + y2 + y3)/3
• If I (x, y) is the in-center of triangle ABC, A   (x1, y1), B   (x2, y2), C   (x3, y3), then,
• where a, b and c are the lengths of the BC, AC and AB respectively.
• The equation of a straight line is y = mx + c, where m is the slope and c is the y-intercept (tan   = m, where   is the angle that the line makes with the positive X-axis).
• If two intersecting lines have slopes m1 and m2 then the angle between two lines θ will be tan θ = (m1−m2) / (1+m1m2)
• The length of perpendicular from a point (x1 ,y1 ) on the line AX+BY+C = 0 is
P = (Ax1+By1+C) / (A2+B2)

## Equations of a Line

• General equation of a line Ax + By = C
• Slope intercept form y = mx + c (c is y intercept)
• Point-slope form y – y1 = m (x – x1) (m is the slope of the line)
• Intercept form x / a + y / b = 1 (where a and b are x and y intercepts respectively)
• Two point form:  (y−y1) / (y2−y1) = (x−x1) / (x2−x1)

• Quadrant I ⇒ X is Positive Y is Positive
• Quadrant II ⇒ X is Negative Y is Positive
• Quadrant III ⇒ X is Negative Y is Negative
• Quadrant IV ⇒ X is Positive Y is Negative

Question for Coordinate Geometry: Concepts, Formulas & Examples
Try yourself:Angles between 180 ° and 270 ° lies in:

## Solved Examples

Example 1: What is the distance between the points A (3,8) and B(-2,-7) ?
a) 5√2
b) 5
c) 5√10
d) 10√2

• The distance between 2 points (x1, y1) and (x2, y2) is given as
• Sqrt ((x2-x1)2 + (y2-y1)2)
• Hence, required distance = sqrt((-2-3)2 + (-7-8)2) = 5√10

Example 2: The points of intersection of three lines 2X + 3Y – 5=0 and 5X – 7Y + 2=0 and 9X – 5Y – 4 = 0
a) Form a triangle
b) Are on lines perpendicular to each other
c) Are on lines parallel to each other
d) Are coincident

To solve the question above, we should remember the properties of the lines for being parallel, perpendicular or intersecting:

• Two lines are parallel to each other if their slopes are equal
• Two lines are perpendicular if the product of slopes is -1
• Lines are coincident if they at least have one point which satisfies all the equation.
• The three lines can be expressed in the y=mx + c format as:
Y = (5/3) – (2X/3),   Y = (5X/7) + (2/7)   , Y = (9X/5) – (4/5)
• Therefore, the slopes of the three lines are -2/3, 5/7, 9/5 and their Y intercepts are 5/3, 2/7 and 4/5 respectively.
• We see above that the product of slopes of none of the lines is -1. Thus, lines are not perpendicular to each other.
• Also, slopes of the no two lines is same. Thus, lines are not parallel to each other.
• Solving the first two equations we get X=1 and Y = 1. If we substitute (1,1) in the third equation Y=(9X/5 – 4/5), we find that it also satisfies the equation. This suggests that the three lines intersect at a common point and hence coincident.

Example 3: The area of the triangle whose vertices are (a + 1, a + 1), (a, a) and (a+2, a) is
a) a3
b) 1
c) 2a
d) 21/2

• Let a = 0, Thus the three vertices of the triangle becomes (1, 1) (0, 0) and (2, 0)
• If we look at the below figure, Area = ½ * base * height = ½ * 2 * 1 = 1
• Imp: The main point to note here is that area will be independent of a.

Example 4: Consider a triangle drawn on the X – Y plane with its three vertices of (41,0) , (0,41) and (0,0), each vertex being represented by its (X,Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is:
a) 12
b) 8
c) 6
d) 4

• Equation of the line will be of the form => x + y = 41.
• Now, we know that if the x,y coordinates of a point are integer, the sum will also be an integer
X + y = k (k, a variable)
• As per the question we need to exclude all the values lying on the boundary of triangle, k can take all values from 1 to 40 only. K = 0 is rejected as at k =0 will give the point at A which is also not allowed.
• With K = 40, x + y = 40; this will be satisfied by points (1, 39), (2, 38), (3,37) …… (38, 2), (39, 1). That is a total of 39 points
• Similarly x + y = 38, will be satisfied by 37 points.
X + Y = 37, will be satisfied by 36 points
X + Y = 3 will be satisfied by 2 points
X + Y = 2 will be satisfied by 1 point
X + Y = 1 will not be satisfied by any point
• So, the total number of all such points is: 39 + 38 + 37 + 36 + ……………………. + 3 + 2 + 1 = n(n+1)/2 points =  (39*40) / 2 = 780 points

Example 5: Two lines P and Q intersect at point (3, 2) in the x-y plane. The slope of line P is 45 degrees and line Q is parallel to the X axis. What is the area (in sq. units) of the triangle formed by P, Q and a line perpendicular to P passing through point (5, 4) ?
a) 12
b) 8
c) 6
d) 4

• Let us look at the image below:
• As slope of line P is 45 degree. Therefore, ∠ABC = 45 degree
• In triangle ABC, length of AB = SQRT [(5-3)+ (4-2)2] = 22 units
• Therefore, length of line AC = 22 units (Since ABC is an isosceles triangle. Thus AB = AC)
• Thus, required area = ½ * 22 * 22 = 4 sq. units

Example 6: The line 3 Y = x is the radius of the circle. It meets the circle o=centered at origin O at point M (3, 1). If PQ is the tangent to the circle at M as shown, find the length of the PQ.

a) (5/2)3 units
b) 3 3 units
c) 23 units
d) 8/3 units

• PQ is perpendicular to line Y = X / 3 (Since, radius of a circle is perpendicular to the tangent of the circle)
• Therefore, slope of PQ = -1 / (1/ 3) = – 3 (Since, product of slopes of line perpendicular to each other is -1)
• Therefore, Let equation of the line PQ be y = – 3x + c
• Now at the point M, when x = 3, y = 1
• Putting the above values of x and y in the above equation, we get c = 4 The equation of the line becomes, Y = – 3x + 4
• Thus, by using the above equation, we get:
• Coordinates of point P = (0, 4) and coordinates of point Q = (4/3, 0) (Putting x = 0 in above equation, we find value of P and putting Y = 0 in above equation, we find value of Q)
• Hence PQ = sqrt [(4/3) + 42] = 8/3 units.

## CAT Coordinate Geometry Questions

Q1: What is the equation of a circle with centre of origin and radius is 6 cm?

Ans: (c)
Sol:
Given,
Center of the circle = (0,0)
Radius of the circle (r) = 6 cm

∴ Equation of the circle is

Hence, the correct answer is Option C

Q2: The equation of circle with centre (1, -2) and radius 4 cm is:
(a) x2 + y+2x - 4y = 11
(b) x2 + y+ 2x - 4y = 16
(c) x2 + y2 - 2x + 4y = 16
(d) x2 + y- 2x + 4y = 11
Ans:
(d)
Sol: Given,
Centre of the circle (a, b) = (1, -2)
Radius of the circle (r) = 4 cm
... Equation of the circle is (xa)2 + (y - b)= r2
=> (x-1)2 +(y-(-2))2 = 42
=> (x-1)2 + (y+2)2 = 42
=> x2+1- 2.x.1 + y2 + 2+2.y.2 = 16
=> x+ 1 - 2x + y+ 4 + 4y = 16
=> x2 - 2x + y2 + 4y = 16 - 1 - 4
=> x+ y- 2x + 4y = 11
Hence, the correct answer is Option (d)

Q3: In ΔABC, AB = AC. A circle drawn through B touches AC at D and intersect AB at P. If D is the mid point of AC and AP 2.5 cm, then AB is equal to:
(a) 9 cm
(b) 10 cm
(c) 7.5 cm
(d) 12.5 cm

Ans: (b)
Sol:

Given D is midpoint of AC so,

But also given AC = AB

AD is a tangent and APB is a secant. So the tangent secant theorem can be applied,

(AB/4)2 = 2.5 × AB

AB2/4 = 2.5 × AB

AB = 10 cm

The document Important Formulas for CAT Coordinate Geometry is a part of the CAT Course Quantitative Aptitude (Quant).
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## FAQs on Important Formulas for CAT Coordinate Geometry

 1. What are the basic concepts of coordinate geometry?
Ans. Coordinate geometry is a branch of mathematics that deals with the study of geometric shapes using coordinates. It involves the use of algebraic equations and graphs to describe geometric objects. The basic concepts of coordinate geometry include points, lines, and planes in a coordinate system.
 2. What are the formulas used in coordinate geometry?
Ans. Coordinate geometry involves various formulas that are used to calculate distances, slopes, and equations of lines. Some important formulas include the distance formula, midpoint formula, slope formula, and the equation of a line in different forms such as slope-intercept form, point-slope form, and general form.
 3. How do you find the equation of a line in coordinate geometry?
Ans. To find the equation of a line in coordinate geometry, you need either two points or one point and the slope of the line. If you have two points, you can use the slope formula to find the slope and then use the point-slope form or slope-intercept form to write the equation. If you have one point and the slope, you can directly use the point-slope form to write the equation.
 4. What is the significance of knowledge of quadrants in coordinate geometry?
Ans. Knowledge of quadrants is essential in coordinate geometry as it helps in determining the position of a point in the coordinate plane. The coordinate plane is divided into four quadrants (I, II, III, IV), and each quadrant has its own unique set of coordinates. By understanding quadrants, you can easily locate and identify points in the coordinate plane.
 5. Can you provide a solved example of coordinate geometry?
Ans. Certainly! Here's a solved example: Example: Find the equation of the line passing through the points (2, 3) and (5, 7). Solution: Step 1: Calculate the slope using the slope formula: m = (y2 - y1) / (x2 - x1) m = (7 - 3) / (5 - 2) m = 4 / 3 Step 2: Use the point-slope form to write the equation: y - y1 = m(x - x1) y - 3 = (4/3)(x - 2) Step 3: Simplify the equation: 3y - 9 = 4x - 8 Step 4: Rewrite the equation in the slope-intercept form: 3y = 4x + 1 So, the equation of the line passing through the points (2, 3) and (5, 7) is 3y = 4x + 1.

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