How To Solve Coordinate Geometry Quickly

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 – y₁ = m(x – x₁)

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

Therefore, y – y₁ = \frac{(y_{1}- y_{2})}{(x_{1}- x_{2})} × (x – x₁)

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 (x₁,y₁), (x₂ ,y₂), (x₃,y₃)

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(x₁, y₁) and B(x₂, y₂)

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