### Type 1: How To Solve Quickly Coordinate Geometry -Distance and Equation of Straight Line

**Question 1: Find the distance between points A(–2,-5) and B(6,1)?**

**Options:**

**A. 100**

**B. 10**

**C. 20**

**D. 5**

**Solution: **AB = \sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}

AB =\sqrt{(6- (-2))^2 + (1- (-5))^2}

AB = \sqrt{(6+2)^2 + (1+5)^2}

AB = \sqrt{(8)^2 + (6)^2}

AB = \sqrt{64 + 36}

AB =\sqrt{100}

AB = 10

**Correct option: B**

**Question 2: Find the equation of line whose end points are (4, 5) and (2, 8). **

**Options:**

**A. 3x – 2y -2 = 0**

**B. 3x + 2y -2 = 0**

**C. 3x + 2y + 2 = 0**

**D. 3x – 2y + 2 = 0**

**Solution: ** We know that,

y – y_{1} = m(x – x_{1})

m = \frac{(y_{1}- y_{2})}{(x_{1}- x_{2}}

Therefore, y – y_{1} = \frac{(y_{1}- y_{2})}{(x_{1}- x_{2})} × (x – x_{1})

y – 5 = \frac{(8- 5)}{(4 – 2)} × x – 4

y – 5 =\frac{3}{2} ×( x – 4)

2y – 10 = 3x – 12

3x – 2y -2 = 0

**Correct option: A**

**Question 3: Two points (a + 3, b + k) and (a, b) are on the line x – 2y + 9 = 0. Find the value of k.**

**Options**:

**A. \frac{1}{2}**

**B. \frac{2}{3}**

**C.\frac{3}{4}**

**D.\frac{3}{2}**

**Solution: **x – 2y + 9 = 0

y = **\frac{x}{2}**+ **\frac{9}{2}**

Slope of the line = **\frac{1}{2}**

Slope of the line using two points = b + k – **\frac{b}{a}** + 3 –a

= **\frac{k}{3}**

**\frac{k}{3}** = **\frac{1}{2}**

k = **\frac{3}{2}**

**Correct option: D**

### Type 2: How To Solve Coordinate Geometry Quickly. In which quadrant does the point lie?

**Question 1: In which quadrant does the point (-7, 3) lie?**

**Options:**

**A. I quadrant**

**B. II quadrant**

**C. III quadrant**

**D. IV quadrant**

**Solution: **The point is negative in the x axis and positive for the y axis, thus the point must lie in the 2nd quadrant.

**Correct option: B**

**Question 2: In which quadrant does the point (2, 3) lie?**

**Options:**

**A. I quadrant**

**B. II quadrant**

**C. III quadrant**

**D. IV quadrant**

**Solution: **Both points are positive. Therefore they will lie in 1st quadrant

**Correct option: A**

**Question 3: In which quadrant does the point (-10, -3) lie?**

**Options:**

**A. I quadrant**

**B. II quadrant**

**C. III quadrant**

**D. IV quadrant**

**Solution: **Both points are negative. Therefore they will lie in 3th quadrant.

**Correct option: C**

##

### Type 3: Solve Quickly Coordinate Geometry Questions.

Find the Coordinates

**Question 1: Find the co-ordinates of the centroid of a triangle whose vertices are (0, 4), (6, 10) and (9, 4).**

**Options: **

**A. 6, 5**

**B. 7, 5**

**C. 5, 6**

**D. 5, 7**

**Solution**: We know that, Centroid of a triangle with its vertices (x_{1},y_{1}), (x_{2} ,y_{2}), (x_{3},y_{3})

C =\frac{x_{1}+ x_{2} + x_{3}}{3}, \frac{y_{1}+ y_{2} + y_{3}}{3}

C =\frac{0+ 6 + 9}{3}, \frac{4+ 10 + 4}{3}

C =\frac{15}{3}, \frac{18}{3}

C = 5, 6

**Correct option: C**

**Question 2: If the distance between two points A(a, –3) and B(3, a) is 6 unit, then a = ?**

**Options: **

**A. ± 3**

**B. – 5**

**C. 4**

**D. 0**

**Solution: **We know that, Distance between two points A(x_{1}, y_{1}) and B(x_{2}, y_{2})

AB = \sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}

AB = \sqrt{(3- a)^2 + (a+ 3)^2}

AB = √2a² + 18

According to question,

\sqrt{2}a²+18 = 6

2a² + 18 = 36

a² = 9

a = ±3

**Correct option: A**

**Question 3: The line passing through (4,3) and (y,0) is parallel to the line passing through (–1,–2) and (3,0). Find the value of y?**

**Options: **

**A. 2**

**B. -2**

**C. -1**

**D. 1**

**Solution: **Slope of line passing through (4,3) (y,0)

m = \frac{(y_{1}- y_{2})}{(x_{1}- x_{2})}

m_{1}= \frac{(3- 0)}{(4- y)}

Slope of line passing through (–1,–2) (3,0)

m_{2}= \frac{(-2 – 0)}{(-1- 3)}

If two lines are parallel then, there slopes are equal

m_{1}= m_{2}

\frac{3}{4- y } = \frac{-2}{-4}

\frac{3}{4- y } = \frac{2}{4}

8 – 2y = 12

– 2y = 12 – 8

– 2y = 4

y = \frac{4}{-2}

y = – 2

**Correct option: B**

### Type 4: Coordinate Geometry Solve Questions Quickly.

Find the area

**Question 1: A(a,b), B(c,d) and C(e,f) are the vertices of a triangle. Given that:**

**Option 1: AB + BC > AC**

**Option 2: Area of the triangle = \frac{1}{2} ×[ (a(c-e) + (c(f-b) ) + (e(b-d) )]**

**Which of the following are true?**

**Options: **

**A. Option I is true**

**B. Option II is true**

**C. Both are true**

**D. None of the above**

**Solution: **The formula given in option II is wrong.

The correct formula is

A = **\frac{1}{2} ×[ (a(d-f) + (c(f-b) ) + (e(b-d) )]**

**Correct option: A**

**Question 2: Find the area of the triangle formed by the vertices (31, 25), (23, 30) and (33, 28)**

**Options: **

**A. 25**

**B. 15**

**C. 19**

**D. 17**

**Solution: ** Area of triangle = \frac{1}{2} ×[ (x_{1}(y_{2}-y_{3}) + (x_{2}(y_{3}-y_{1}) ) + (x_{3}(y_{1}-y_{2}) )]

A=\frac{1}{2} ×[ (31(30-28) + (23(28-25) ) + (33(25-30) )]

A = 17

**Correct option: D**

**Question 3: Find the area of the triangle formed by the vertices (-8, 19), (-7, 31) and (13, 25)**

**Options:**

**A. 125**

**B. 135**

**C. 119**

**D. 123**

**Solution: **Area of triangle = \frac{1}{2} ×[ (x_{1}(y_{2}-y_{3}) + (x_{2}(y_{3}-y_{1}) ) + (x_{3}(y_{1}-y_{2}) )]

A= \frac{1}{2} ×[ (- 8 (31-25) + (7(25 – 19) ) + (13(19-31) )]

A = 123

**Correct option: D**

**Read Also – ****Formulas for Coordinate geometry**

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