# Coordinate Geometry for Competitive Exams

## Owing to its vast syllabus, the quantitative aptitude section is referred to as the most complex yet essential… The post Coordinate Geometry for Competitive Exams appeared first on Leverage Edu.

Owing to its vast syllabus, the quantitative aptitude section is referred to as the most complex yet essential one to be found in numerous **competitive exams**. While the elements of this section range from **permutation and combination** to number system, the difficulty level also varies from basic to intermediate level. One such topic included in the quant section is Coordinate Geometry which utilises the position of points on a plane to solve several mathematical equations. If you are preparing for competitive exams and are struggling to ace this topic, we are here to impart you with the **success mantra** to effectively study Coordinate Geometry through this detailed blog explaining the concept as well as helpful tricks to crack these questions.

**Also Read: ****Mastering the GMAT Quantitative Ability**

## What is Coordinate Geometry?

In simpler terms, coordinate geometry is a mathematical concept and is defined as a system of geometry in which the position of points of the plane is elucidated by using an ordered pair of numbers. Recognized as the blend of both geometry and algebra, Coordinate Geometry is also known as Analytical Geometry. The core concept lies in positioning specific points on a plane and through which the distance between two points, two lines, midpoint of a line, equation of the segment, are as well as the perimeter of a polygon can be determined.

## Understanding Coordinate Geometry

Think about a plane which has a flat surface and goes in both directions. Now, if you place a point anywhere on this plane, the concept of coordinate geometry can assist you in defining where this point is by using two numbers. Moreover, knowing the coordinates of a group of points can further help in various ways, namely,

- Finding the distance between those points
- Locating the midpoint, slope as well as the equation of a line segment
- Understanding if the lines are either parallel or perpendicular
- Determine the area and perimeter of a polygon using its coordinates
- Change a shape by rotating, moving or reflecting it
- Decipher the equation of circular figures, curves or ellipses

## Some Formulas & Examples

Now that you have learned the basics as well as useful to solve coordinate geometry questions, let’s take a look at some of the major formulas through suitable examples.

1. **Equation of line parallel to the y-axis**

X=a

**Example**: Rohan plotted four points on a graph. Find out which point represents the line parallel to the y-axis.

a. (0,6)

b. (8,0)

c. (3,5)

d. (-2,-4)

**Solution**: Part b

2. ** Equation of line parallel to the x-axis **

Y=b

**Example**: Gurleen plotted four points on a graph. Find out which point represents the line parallel to the x-axis.

a. (3,5)

b. (0,6)

c. (-3,-1)

d. (8,0)

**Solution**: b. (0,6)

3. **Equations of the line**

**Normal equation of the line **

ax+by+c=0

**Slope-intercept Form**

y=mx+c

Where c= intercept on y-axis & m= Slope of the line

**Example**– Find the slope of the line formed by the equation 5y-3x-10=0?

**Solution: **5y-3x-10=0 ,

5y=3x+10

y= 3/5x + 2

Therefore, slope of the line = ⅗

**Intercept Form**

x/A + y/B = 1

Where A & B are x-intercepts & y-intercepts respectively.

**Example: **Himanshu** **draws a triangle using the line 4x +** **3y – 12= 0, x-axis and y-axis. Find the total area of the triangle. ** **

**Solution: **For this Coordinate Geometry problem, the formula for finding the area of the triangle is:

**1/2 * x-intercept * y-intercept**

Equation of the line is 4x + 3y – 12=0

4x + 3y= 12

4x/12 + 3y/12 = 1

x/3 + y/4 = 1

Therefore, area of triangle= 1/2 * 3 * 4 = 6

**Trigonometric Form of the equation of the line, ax + by + c= 0**

X cosθ + Y sinθ = P

Where, cosθ = -a/ √(a2 + b2), sinθ= -b/ √(a2 + b2), p= c/ √(a2 + b2)

**Equation of the line passing through the point (x1, y1) & has a slope m**

** **Y – y1 = m(X – x1)

4. **Slope of the line **

(y2-y1)/ (x2-x1) = – (coefficient of x)/ (coefficient of y)

5. **Angle between two lines **

Tanθ = 土 (m2-m1)/ (1+ m1m2) where, m1, m2 = slope of the lines

*Note: **In coordinate geometry, if the lines are parallel, then tanθ= 0 and if the lines are perpendicular, then cotθ= 0*

**Example**: Anukriti has drawn to lines and the equations of the lines are 7x- 4y= 0 and 3x- 11y + 5= 0. Find the acute angles between the two lines. **Solution: **In this coordinate geometry question, begin with simplifying the equation depicting the slope of both the lines,

7x – 4y= 0

y= 7/4x

Therefore, the slope of the line 7x -4y = 0 is 7/4

Similarly, 3x- 11y + 5= 0

y= 3/11x + 5/11

Therefore, the slope of the line 3x – 11y + 5 = 3/11

The angle between the given lines 7x – 4y = 0 and 3x- 11y + 5 = 0 is θ** **

Now,

Tanθ = 土 (m2-m1)/ (1+ m1m2) = 土 [(7/4 – 3/11)]/ [1+(7/4) * (3/11)] = 土1

Since θ is acute, hence we take, tan θ= 1 = tan 45°

Therefore, θ= 45°

Therefore, the required acute angle between the lines is 45.

6. **Equation of two lines parallel to each other**

ax + by + c1= 0

bx – ay + c2= 0

*Note: The coefficients of x and y here are opposite and there is a negative sign in the other equation.*

7. In Coordinate Geometry, the formula for finding the **distance (D) between two points** **(x1,y1), (x2,y2) is:** **D= √ (x2 – x1)^2 + (y2 – y1)^2**

**Example**: Determine the total distance between (-1,1) and (3,4)

**Solution: **D= √ (x2 – x1)^2 + (y2 – y1)^2

= √ (3 – (-1))^2 + (4 – 1)^2 = √ (16 + 9) = √25 = 5

9. **The midpoint of the line formed by (x1, y1), (x2, y2)**

M= (x1 + x2)/2, (y1 + y2)/2

10. **Area of a triangle whose coordinates are (x1,y1), (x2,y2), (x3, y3)**

½ | x1 (y2 -y3) + x2 (y3 – y1) + x3 (y1 – y2) |

**Example**: Find out of the area of the triangle whose vertices are (1,1), (2,3), (4,5)

**Solution**:(x1,y1)** = ** (1,1)

(x2,y2)**= **(2,3)

(x3, y3)**= **(4,5)

Area of triangle=

½ | x1 (y2 -y3) + x2 (y3 – y1) + x3 (y1 – y2) |

½ | 1 (3 -5) + 2 (5 – 1) + 4 (1 – 3) |

½ | (-2 + 8 – 8) |

½ | (-2) | = | -1| = 1

## Coordinate Geometry Class 9 Questions

- In the following equations, verify whether the given value of the variable is a solution of the equation:

(i) x + 4 = 2x; x = 4

(ii) y – 7 = 3y + 8 ; y = 3

(iii) 3u + 2 = 2u + 7; u = 5 (iv) 2x – 3 = x/2 -2; x = 12

(v) (5/2)x+ 3 = 21/2; x = 3 (vi) 24 – 3(u – 2) = u + 8 ; U = -1

(vii) (x – 2) + (x + 3) = x + 8; x = 0 - Which of the following equations are linear equations?

(i) (3/2)x + 4 = 2x – 3

(ii) 5y – 3 = 2y + 4

(iii) u + 4 = u2 – 4

(iv) 3x + 2 = 5x – 7

(v) x2 + 2 = x + 1

(vi) y – 3 = 3y + 4

(vii) u + 1/u = 5

- In the following equations, verify whether the given value of the variable is a solution of the equation:

(i) x + 4 = 2x; x = 4

(ii) y – 7 = 3y + 8 ; y = 3

(iii) 3u + 2 = 2u + 7; u = 5 (iv) 2x – 3 = x/2 -2; x = 12

(v) (5/2)x+ 3 = 21/2; x = 3 (vi) 24 – 3(u – 2) = u + 8 ; U = -1

(vii) (x – 2) + (x + 3) = x + 8; x = 0

- Draw the graph of 2(x + 3) – 3 (y + 1) = 0. Read a few solutions from the graph and verify the same by actual substitution. In each case, find the points where the line meets the two axes.
- Which of the following equations are linear?

(i) 5×2 + 4x + 1 = 0

(ii) x3 – 1 = 0

(iii) x3 + 1 = 0

(iv) (x – 5)(x – 7) = 8

(v) 6x = 42

(vi) x/8 = 9

(vii) x + 10 = 17

- Find four solutions for the following equation: 5x-3y = 0

## Coordinate Geometry Class 10 Questions

- Find the coordinates of the points of trisection (i.e., points dividing into three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4).
- Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.
- Name the type of triangle formed by the points A (–5, 6), B (–4, –2) and C (7, 5).
- The distance of the point P (2, 3) from the x-axis is (A) 2 (B) 3 (C) 1 (D) 5
- Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).
- Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).
- Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

## Coordinate Geometry Class 12 Questions

- Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11) are collinear.
- If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.
- Find the angle between the pair of lines given below:

(i) (x + 3)/3 = (y -1)/5 = (z + 3)/4

(ii) (x + 1)/1 = (y – 4)/1 = (z – 5)/2 - Show that the lines (x – 5)/7 = (y + 2)/-5 = z/1 and x/1 = y/2 = z/3 are perpendicular to each other.
- Find the intercepts cut off by the plane 2x + y – z = 5
- Find the equations of the planes that passes through three points (1, 1, 0), (1, 2, 1), and (– 2, 2, – 1)
- Find the direction cosines of a line that makes equal angles with the coordinate axes. (Answer: 1/√3, 1/√3, 1/√3)

Thus, considered as one of the pivotal as well as enjoyable topics in the quant section, one needs to remember the copious formulas and basics of coordinate geometry to crack the different questions successfully. Further, coordinate geometry is a vital part for the quant syllabus for international-level exams like GMAT and GRE and if you find yourself stuck in the rigorous preparations, reach out to our Leverage Edu experts and we’ll guide you throughout the process of studying the different sections and acing your exam with flying colors.

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