











How To Solve Coordinate Geometry Questions Quickly
Ways to solve Coordinate Geometry Problems Quickly
In coordinate geometry , points are placed on the coordinate plane . There are two scales , one is running across the plane called the x-axis and another scale is a right angles to it called the y – axis


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
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