SSC CGL Exam > SSC CGL Notes > SSC CGL Mathematics Previous Year Paper (Topic-wise) > SSC CGL Previous Year Questions (2023 - 18): Coordinate Geometry
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Pinnacle Day: 89th Coordinate Geometry
Coordinate Geometry
Coordinate geometry is a system of
geometry where the position of points on
the plane is described by using an
ordered pair of numbers. The x-
coordinate and the y- coordinate of a
point taken together is known as
coordinates of the point
Key points:
Coordinate Axes : The perpendicular •
lines in the cartesian system are x axis
and y axis . Both of them together are
known as coordinate axes.
Origin : The point of intersection of •
coordinate axes is origin. The
coordinates of origin are (0 , 0).
Abscissa : The x - coordinate is called •
abscissa. It is the perpendicular distance
of a point from y axis.
Ordinate : The y-coordinate is called •
ordinate . It is the perpendicular distance
of a point from x-axis.
If y = a, where a is constant then y = a •
denotes a straight line parallel to x-axis.
If x = a, where a is constant then x = a •
denotes a straight line parallel to y-axis.
For passing a line through the origin, •
one of the coordinates of the given
equation must be (0, 0).
Important Formulae :
Distance Formula : If points P(x
1
, y
1
), •
Q(x
2 ,
y
2
) lie on the plane xy then,
PQ = ( ??
2
- ??
1
)
2
+ ( ??
2
- ??
1
)
2
Section Formula : •
The coordinates of a point which divides
the line segment joined by two distinct
points (x
1
, y
1
) and (x
2
, y
2
) in the ratio
m : n are :
(x , y) = ( ,
?? ??
2
+ ?? ??
1
?? + ??
?? ??
2
+ ?? ??
1
?? + ??
)
Note : When line segment is divided
externally in the ratio m : n then ratio
becomes ( ,
?? ??
2
- ?? ??
1
?? - ??
?? ??
2
- ?? ??
1
?? - ??
)
Midpoint Formula : •
It is derived from the Section Formula.
The coordinates of a point which divides
the line segments in the ratio 1 : 1. Then,
(x, y) = ( , )
??
1
+ ??
2
2
??
1
+ ??
2
2
Centroid Formula : •
If (x
1
, y
1
) , (x
2
, y
2
) and (x
3
, y
3
) are the
vertices of a triangle then coordinates of
the centroid are :
( , )
??
1
+ ??
2
+ ??
3
3
??
1
+ ??
2
+ ??
3
3
Incentre Formula : •
If (x
1
, y
1
) , (x
2
, y
2
) and (x
3
, y
3
) are the
vertices of a triangle then coordinates of
the incenter are :
( , )
?? ??
1
+ ?? ??
2
+ ?? ??
3
?? + ?? + ??
?? ??
1
+ ?? ??
2
+ ?? ??
3
?? + ?? + ??
Slope of a line : •
If (x
1 ,
y
1
) , (x
2
, y
2
) are any two points on
line L, then the ratio is called the
??
2
- ??
1
??
2
- ??
1
slope of the line L.
Angle between two lines of slope m
1
•
and m
2,
respectively then
tan = ?
??
2
- ??
1
1 + ??
1
??
2
|
|
|
|
|
|
Intercept form of equation of a line : •
+ = 1,where ‘a’ is x intercept of line
??
??
??
??
and ‘b’ is y intercept of line .
Perpendicular distance from a point (x
1
, •
y
1
) to line ax + by + c = 0, then
D =
?? ??
1
+ ?? ??
1
+ ??
| |
??
2
+ ??
2
Distance between lines : • ?
a
1
x + b
1
y + c
1
= 0 and a
2
x + b
2
y + c
2
= 0 is
D =
??
1
- ??
2
??
2
+ ??
2
|
|
|
|
|
|
Equation of a circle •
+ + 2gx + 2fy + c = 0, where center ??
2
??
2
of the circle is (- g , - f) and radius =
??
2
+ ??
2
- ??
If point A (x
1
, y
1
) , B(x
2
, y
2
) and C (x
3
, y
3
) •
are the vertices of a triangle then area of
a triangle is :
1
2
??
1
( ??
2
- ??
3
) + ??
2
( ??
3
- ??
1
) + ??
3
( ??
1
- ??
2
)
| |
Equation of a circle - + - = ? ( ?? h )
2
( ?? ?? )
2
??
2
Where
( ) x, y coordinate of center point h , ?? ?
x x - coordinate of circle point ?
y y - coordinate of circle point ?
Variety Questions
Q.1. What is the area (in unit squares) of
the triangle enclosed by the graphs of
the equations 2x + 5y = 12, x + y = 3 and
the x-axis?
SSC CGL Tier II (03/02/2022)
(a) 2.5 (b) 3.5 (c) 3 (d) 4
Q.2. The equation of circle with centre (1
, – 2) and radius 4 cm is :
SSC CHSL 17/03/2020 (Afternoon)
(a) ??
2
+ ??
2
+ 2 ?? - 4 ?? = 16
(b) ??
2
+ ??
2
- 2 ?? + 4 ?? = 16
(c) ??
2
+ ??
2
+ 2 ?? - 4 ?? = 11
(d) ??
2
+ ??
2
- 2 ?? + 4 ?? = 11
Q.3. What is the area (in square units) of
the triangular region enclosed by the
graphs of the equations x + y = 3,
2x + 5y = 12 and the x axis?
SSC CGL Tier II (13/09/2019)
(a) 2 (b) 3 (c) 4 (d) 6
Practice Questions
SSC CHSL 2023 Tier - 1
Q.4. Find the radius of the circle
. ?? ² + ?? ² = 25
SSC CHSL 08/08/2023 (4th Shift)
(a) 25 units (b) 5 units
(c) 2 units (d) 12 units
SSC CGL 2022 Tier - 2
Q.5. Find the coordinates of the points
where the graph 57x 19y = 399 cuts -
the coordinate axis.
SSC CGL Tier II (07/03/2023)
(a) x-axis at (-7,0) and y-axis at (0,-21)
(b) x-axis at (-7,0) and y-axis at (0,21)
(c) x-axis at (7,0) and y-axis at (0,-21)
(d) x-axis at (7,0) and y-axis at (0,21)
SSC CGL 2021 Tier - 2
Q.6. The graph of the equation x = a
(a 0) is a ______. ?
SSC CGL Tier II (08/08/2022)
(a) line parallel to x axis
(b) line parallel to y axis
(c) line at an angle of 45 degree to y axis
(d) line at an angle of 45 degree to x axis
SSC CGL 2020 Tier - 2
Q.7. The graphs of the equations
4x + y = and x + y + = 0
1
3
8
3
1
2
3
4
5
2
and intersect at a point P . The point P
also lies on the graph of the equation:
SSC CGL Tier II (29/01/2022)
(a) 4x y 7 0 (b) x 3y 12 0 - + = - - =
(c) 3x y 7 0 (d) x 2y 0 - - = + - 5 =
Q.8. What is the area (in unit squares) of
the region enclosed by the graphs of the
equations 2x – 3y + 6 = 0, 4x + y = 16 and
y = 0 ?
SSC CGL Tier II (29/01/2022)
(a) 11.5 (b) 14 (c) 10.5 (d) 12
Q.9. The graphs of the equations
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Page 2
Pinnacle Day: 89th Coordinate Geometry
Coordinate Geometry
Coordinate geometry is a system of
geometry where the position of points on
the plane is described by using an
ordered pair of numbers. The x-
coordinate and the y- coordinate of a
point taken together is known as
coordinates of the point
Key points:
Coordinate Axes : The perpendicular •
lines in the cartesian system are x axis
and y axis . Both of them together are
known as coordinate axes.
Origin : The point of intersection of •
coordinate axes is origin. The
coordinates of origin are (0 , 0).
Abscissa : The x - coordinate is called •
abscissa. It is the perpendicular distance
of a point from y axis.
Ordinate : The y-coordinate is called •
ordinate . It is the perpendicular distance
of a point from x-axis.
If y = a, where a is constant then y = a •
denotes a straight line parallel to x-axis.
If x = a, where a is constant then x = a •
denotes a straight line parallel to y-axis.
For passing a line through the origin, •
one of the coordinates of the given
equation must be (0, 0).
Important Formulae :
Distance Formula : If points P(x
1
, y
1
), •
Q(x
2 ,
y
2
) lie on the plane xy then,
PQ = ( ??
2
- ??
1
)
2
+ ( ??
2
- ??
1
)
2
Section Formula : •
The coordinates of a point which divides
the line segment joined by two distinct
points (x
1
, y
1
) and (x
2
, y
2
) in the ratio
m : n are :
(x , y) = ( ,
?? ??
2
+ ?? ??
1
?? + ??
?? ??
2
+ ?? ??
1
?? + ??
)
Note : When line segment is divided
externally in the ratio m : n then ratio
becomes ( ,
?? ??
2
- ?? ??
1
?? - ??
?? ??
2
- ?? ??
1
?? - ??
)
Midpoint Formula : •
It is derived from the Section Formula.
The coordinates of a point which divides
the line segments in the ratio 1 : 1. Then,
(x, y) = ( , )
??
1
+ ??
2
2
??
1
+ ??
2
2
Centroid Formula : •
If (x
1
, y
1
) , (x
2
, y
2
) and (x
3
, y
3
) are the
vertices of a triangle then coordinates of
the centroid are :
( , )
??
1
+ ??
2
+ ??
3
3
??
1
+ ??
2
+ ??
3
3
Incentre Formula : •
If (x
1
, y
1
) , (x
2
, y
2
) and (x
3
, y
3
) are the
vertices of a triangle then coordinates of
the incenter are :
( , )
?? ??
1
+ ?? ??
2
+ ?? ??
3
?? + ?? + ??
?? ??
1
+ ?? ??
2
+ ?? ??
3
?? + ?? + ??
Slope of a line : •
If (x
1 ,
y
1
) , (x
2
, y
2
) are any two points on
line L, then the ratio is called the
??
2
- ??
1
??
2
- ??
1
slope of the line L.
Angle between two lines of slope m
1
•
and m
2,
respectively then
tan = ?
??
2
- ??
1
1 + ??
1
??
2
|
|
|
|
|
|
Intercept form of equation of a line : •
+ = 1,where ‘a’ is x intercept of line
??
??
??
??
and ‘b’ is y intercept of line .
Perpendicular distance from a point (x
1
, •
y
1
) to line ax + by + c = 0, then
D =
?? ??
1
+ ?? ??
1
+ ??
| |
??
2
+ ??
2
Distance between lines : • ?
a
1
x + b
1
y + c
1
= 0 and a
2
x + b
2
y + c
2
= 0 is
D =
??
1
- ??
2
??
2
+ ??
2
|
|
|
|
|
|
Equation of a circle •
+ + 2gx + 2fy + c = 0, where center ??
2
??
2
of the circle is (- g , - f) and radius =
??
2
+ ??
2
- ??
If point A (x
1
, y
1
) , B(x
2
, y
2
) and C (x
3
, y
3
) •
are the vertices of a triangle then area of
a triangle is :
1
2
??
1
( ??
2
- ??
3
) + ??
2
( ??
3
- ??
1
) + ??
3
( ??
1
- ??
2
)
| |
Equation of a circle - + - = ? ( ?? h )
2
( ?? ?? )
2
??
2
Where
( ) x, y coordinate of center point h , ?? ?
x x - coordinate of circle point ?
y y - coordinate of circle point ?
Variety Questions
Q.1. What is the area (in unit squares) of
the triangle enclosed by the graphs of
the equations 2x + 5y = 12, x + y = 3 and
the x-axis?
SSC CGL Tier II (03/02/2022)
(a) 2.5 (b) 3.5 (c) 3 (d) 4
Q.2. The equation of circle with centre (1
, – 2) and radius 4 cm is :
SSC CHSL 17/03/2020 (Afternoon)
(a) ??
2
+ ??
2
+ 2 ?? - 4 ?? = 16
(b) ??
2
+ ??
2
- 2 ?? + 4 ?? = 16
(c) ??
2
+ ??
2
+ 2 ?? - 4 ?? = 11
(d) ??
2
+ ??
2
- 2 ?? + 4 ?? = 11
Q.3. What is the area (in square units) of
the triangular region enclosed by the
graphs of the equations x + y = 3,
2x + 5y = 12 and the x axis?
SSC CGL Tier II (13/09/2019)
(a) 2 (b) 3 (c) 4 (d) 6
Practice Questions
SSC CHSL 2023 Tier - 1
Q.4. Find the radius of the circle
. ?? ² + ?? ² = 25
SSC CHSL 08/08/2023 (4th Shift)
(a) 25 units (b) 5 units
(c) 2 units (d) 12 units
SSC CGL 2022 Tier - 2
Q.5. Find the coordinates of the points
where the graph 57x 19y = 399 cuts -
the coordinate axis.
SSC CGL Tier II (07/03/2023)
(a) x-axis at (-7,0) and y-axis at (0,-21)
(b) x-axis at (-7,0) and y-axis at (0,21)
(c) x-axis at (7,0) and y-axis at (0,-21)
(d) x-axis at (7,0) and y-axis at (0,21)
SSC CGL 2021 Tier - 2
Q.6. The graph of the equation x = a
(a 0) is a ______. ?
SSC CGL Tier II (08/08/2022)
(a) line parallel to x axis
(b) line parallel to y axis
(c) line at an angle of 45 degree to y axis
(d) line at an angle of 45 degree to x axis
SSC CGL 2020 Tier - 2
Q.7. The graphs of the equations
4x + y = and x + y + = 0
1
3
8
3
1
2
3
4
5
2
and intersect at a point P . The point P
also lies on the graph of the equation:
SSC CGL Tier II (29/01/2022)
(a) 4x y 7 0 (b) x 3y 12 0 - + = - - =
(c) 3x y 7 0 (d) x 2y 0 - - = + - 5 =
Q.8. What is the area (in unit squares) of
the region enclosed by the graphs of the
equations 2x – 3y + 6 = 0, 4x + y = 16 and
y = 0 ?
SSC CGL Tier II (29/01/2022)
(a) 11.5 (b) 14 (c) 10.5 (d) 12
Q.9. The graphs of the equations
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Pinnacle Day: 89th Coordinate Geometry
7x + 11y = 3 and 8x + y = 15 intersect at
the point P , which also lies on the graph
of the equation.
SSC CGL Tier II (03/02/2022)
(a) 2x + y = 2 (b) 2x – y = 1
(c) 3x + 5y = 1 (d) 3x + 2y = 3
SSC CGL 2019 Tier - 2
Q.10. The graph of the equations
3x – 20y – 2 = 0 and 11x – 5y + 61 = 0
intersect at P(a, b). What is the value of
? ( ??
2
+ ??
2
- ???? ) / ( ??
2
- ??
2
+ ???? )
SSC CGL Tier II (15/11/2020)
(a) (b) (c) (d)
37
35
31
41
5
7
41
31
Q.11. The area (in sq units) of the
triangle formed by the graphs of
equations 8x + 3y = 24 , 2x + 8 = y and
the x-axis is:
SSC CGL Tier II (15/11/2020)
(a) 28 (b) 14 (c) 15 (d)24
Q.12. The graph of the linear equation
3x – 2y = 8 and 4x + 3y = 5 intersect at
the point ( ). What is the value of a, ß
(2 ? a - ß)
SSC CGL Tier II (16/11/2020)
(a) 4 (b) 6 (c) 3 (d) 5
Q.13. What is the re?ection of the point
(5, – 3) in the line Y = 3 ?
SSC CGL Tier II (18/11/2020)
(a) (5, -6) (b) ( -5, 3) (c) (5, 9) (d) (5, 3)
Q.14. The graphs of the linear equations
4x – 2y = 10 and 4x + ky = 2 intersect at a
point (a, 4). The value of k is equal to:
SSC CGL Tier II ( 18/11/2020)
(a) 3 (b) – 3 (c) – 4 (d) 4
SSC CGL 2018 Tier - 2
Q.15. The graphs of the equations
3x + y – 5 = 0 and 2x – y – 5 = 0 intersect
at the point P( ). What is the value of a, ß
( ) ? 3 a + ß
SSC CGL Tier II (11/09/2019)
(a) 4 (b) - 4 (c)3 (d) 5
Q.16. The graph of the equation
x – 7y = – 42, intersects the y -axis at
P( ) and the graph of 6x + y – 15 = 0, a , ß
intersects the x-axis at Q( ). What is ? , d
the value of a + ß + ? + d ?
SSC CGL Tier II (11/09/2019)
(a) (b) 6 (c) (d) 5
17
2
9
2
Q.17. The point of intersection of the
graphs of the equations 3x – 5y = 19 and
3y – 7x + 1 = 0 is P( ) . What is the a , ß
value of ( ) ? 3 a - ß
SSC CGL Tier II (12/09/2019)
(a) - 2 (b) –1 (c) 1 (d) 0
Q.18. The graphs of the equations
2x + 3y = 11 and x – 2y + 12 = 0
intersects at P(
1 , 1
) and the graph of ?? ??
the equation x – 2y + 12 = 0 intersects
the x-axis at Q( ). What is the value ??
2
, ??
2
of ( ) ? ??
1
- ??
2
+ ??
1
+ ??
2
SSC CGL Tier II (12/09/2019)
(a) 13 (b) - 11 (c) 15 (d) -9
Q.19. The graph of the equations
5x – 2y + 1 = 0 and 4y – 3x + 5 = 0,
intersect at the point P( ). What is the a, ß
value of ( ) ? 2 a - 3 ß
SSC CGL Tier II (13/09/2019)
(a) 4 (b) 6 (c) - 4 (d) - 3
Answer Key :-
1.(c) 2.(d) 3.(b) 4.(b)
5.(c) 6.(b) 7.(c) 8.(b)
9.(c) 10.(b) 11.(a) 12.(d)
13.(c) 14.(c) 15.(d) 16.(a)
17.(b) 18.(c) 19.(a)
Solutions :-
Sol.1.(c) Intersection point of 2x + 5y
= 12 and x + y = 3 is (1 , 2)
Intersection point of 2x + 5y = 12 and
y = 0 is (6 , 0)
Intersection point of x + y = 3 and y = 0
is (3 , 0)
Area of triangle = ? +
1
2
[ ??
1
( ??
2
- ??
3
)
+ ] ? ??
2
( ??
3
- ??
1
) ??
3
( ??
1
- ??
2
)
? ? [1(0 – 0) + 6(0 – 2) + 3(2 – 0)] ?
1
2
? = 3
1
2
[- 12 + 6 ] | | ????????
2
Sol.2.(d)
Center = (1, –2) and radius = 4 cm
Equation of circle
(x – 1)
2
+ (y + 2)
2
= (4)
2
?
x
2
+ 1 – 2x + y
2
+ 4 + 4y = 16
x
2
– 2x + y
2
+ 4y = 16 – 5 ?
x
2
+ y
2
– 2x + 4y = 11 ?
Sol.3.(b) In x + y = 3
put x = 0 ? ?? = 3
put y = 0 ? ?? = 3
Line formed by the equation will be AB
In 2x + 5y = 12
put x = 0 ? ?? = 2 . 4
put y = 0 ? ?? = 6
Line formed by the equation will be CD
AB and CD will intersect at point E.
Given, x + y = 3 ………(1)
and 2x + 5y = 12 …….(2)
Multiply equation (1) by 2 and subtract it
from equation (2)
? 3 ?? = 6 ? ?? = 2
Put the value of y in any of the equations
x + 2 = 3 ? ?? = 1
So, E(x , y) = (1 , 2)
Area formed by the the equations
x + y = 3 , 2x + 5y = 12 and the x axis is
the shaded region or ? ??????
Now, AC = OC – OA = 6 – 3 = 3
Height of the triangle will be the y
coordinate of point E = 2
Required area = base Height
1
2
× ×
= 3 2 = 3 square units.
1
2
× ×
Sol.4.(b) Equation of circle passing
through origin :-
+ = …..( = radius) ??
2
??
2
??
2
??
Here , + = radius = 5 units ??
2
??
2
25 ?
Sol.5.(c) 57x 19y = 399 -
At x-axis y = 0 ?
57x 19y = 399 x = = 7 - ?
399
57
At y-axis x = 0 ?
57x 19y = 399 y = = 21 - ?
- 399
19
-
So , the graph 57x 19y = 399 cuts the -
coordinate axis ,
x-axis at (7,0) and y-axis at (0, 21) -
Sol.6.(b)
The graph of the equation x = a is a line
parallel to the y axis.
Sol.7.(c) 4x + y =
1
3
8
3
12x + y = 8 ………..e.q .(1)
x + y + = 0
1
2
3
4
5
2
2x + 3y = 10 ………..e.q .(2) -
Multiplying (1) by 3 and subtracting from
(2) we get,
– 34x = – 34 , x = 1, ?
Putting x = 1 in e.q .(1) , we get
y = – 4
Checking these values of x and y we ?nd
only option (c) satis?es these values.
Sol.8.(b) Intersection point of 2x – 3y +
6 = 0 and 4x + y = 16 is (3 , 4)
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Page 3
Pinnacle Day: 89th Coordinate Geometry
Coordinate Geometry
Coordinate geometry is a system of
geometry where the position of points on
the plane is described by using an
ordered pair of numbers. The x-
coordinate and the y- coordinate of a
point taken together is known as
coordinates of the point
Key points:
Coordinate Axes : The perpendicular •
lines in the cartesian system are x axis
and y axis . Both of them together are
known as coordinate axes.
Origin : The point of intersection of •
coordinate axes is origin. The
coordinates of origin are (0 , 0).
Abscissa : The x - coordinate is called •
abscissa. It is the perpendicular distance
of a point from y axis.
Ordinate : The y-coordinate is called •
ordinate . It is the perpendicular distance
of a point from x-axis.
If y = a, where a is constant then y = a •
denotes a straight line parallel to x-axis.
If x = a, where a is constant then x = a •
denotes a straight line parallel to y-axis.
For passing a line through the origin, •
one of the coordinates of the given
equation must be (0, 0).
Important Formulae :
Distance Formula : If points P(x
1
, y
1
), •
Q(x
2 ,
y
2
) lie on the plane xy then,
PQ = ( ??
2
- ??
1
)
2
+ ( ??
2
- ??
1
)
2
Section Formula : •
The coordinates of a point which divides
the line segment joined by two distinct
points (x
1
, y
1
) and (x
2
, y
2
) in the ratio
m : n are :
(x , y) = ( ,
?? ??
2
+ ?? ??
1
?? + ??
?? ??
2
+ ?? ??
1
?? + ??
)
Note : When line segment is divided
externally in the ratio m : n then ratio
becomes ( ,
?? ??
2
- ?? ??
1
?? - ??
?? ??
2
- ?? ??
1
?? - ??
)
Midpoint Formula : •
It is derived from the Section Formula.
The coordinates of a point which divides
the line segments in the ratio 1 : 1. Then,
(x, y) = ( , )
??
1
+ ??
2
2
??
1
+ ??
2
2
Centroid Formula : •
If (x
1
, y
1
) , (x
2
, y
2
) and (x
3
, y
3
) are the
vertices of a triangle then coordinates of
the centroid are :
( , )
??
1
+ ??
2
+ ??
3
3
??
1
+ ??
2
+ ??
3
3
Incentre Formula : •
If (x
1
, y
1
) , (x
2
, y
2
) and (x
3
, y
3
) are the
vertices of a triangle then coordinates of
the incenter are :
( , )
?? ??
1
+ ?? ??
2
+ ?? ??
3
?? + ?? + ??
?? ??
1
+ ?? ??
2
+ ?? ??
3
?? + ?? + ??
Slope of a line : •
If (x
1 ,
y
1
) , (x
2
, y
2
) are any two points on
line L, then the ratio is called the
??
2
- ??
1
??
2
- ??
1
slope of the line L.
Angle between two lines of slope m
1
•
and m
2,
respectively then
tan = ?
??
2
- ??
1
1 + ??
1
??
2
|
|
|
|
|
|
Intercept form of equation of a line : •
+ = 1,where ‘a’ is x intercept of line
??
??
??
??
and ‘b’ is y intercept of line .
Perpendicular distance from a point (x
1
, •
y
1
) to line ax + by + c = 0, then
D =
?? ??
1
+ ?? ??
1
+ ??
| |
??
2
+ ??
2
Distance between lines : • ?
a
1
x + b
1
y + c
1
= 0 and a
2
x + b
2
y + c
2
= 0 is
D =
??
1
- ??
2
??
2
+ ??
2
|
|
|
|
|
|
Equation of a circle •
+ + 2gx + 2fy + c = 0, where center ??
2
??
2
of the circle is (- g , - f) and radius =
??
2
+ ??
2
- ??
If point A (x
1
, y
1
) , B(x
2
, y
2
) and C (x
3
, y
3
) •
are the vertices of a triangle then area of
a triangle is :
1
2
??
1
( ??
2
- ??
3
) + ??
2
( ??
3
- ??
1
) + ??
3
( ??
1
- ??
2
)
| |
Equation of a circle - + - = ? ( ?? h )
2
( ?? ?? )
2
??
2
Where
( ) x, y coordinate of center point h , ?? ?
x x - coordinate of circle point ?
y y - coordinate of circle point ?
Variety Questions
Q.1. What is the area (in unit squares) of
the triangle enclosed by the graphs of
the equations 2x + 5y = 12, x + y = 3 and
the x-axis?
SSC CGL Tier II (03/02/2022)
(a) 2.5 (b) 3.5 (c) 3 (d) 4
Q.2. The equation of circle with centre (1
, – 2) and radius 4 cm is :
SSC CHSL 17/03/2020 (Afternoon)
(a) ??
2
+ ??
2
+ 2 ?? - 4 ?? = 16
(b) ??
2
+ ??
2
- 2 ?? + 4 ?? = 16
(c) ??
2
+ ??
2
+ 2 ?? - 4 ?? = 11
(d) ??
2
+ ??
2
- 2 ?? + 4 ?? = 11
Q.3. What is the area (in square units) of
the triangular region enclosed by the
graphs of the equations x + y = 3,
2x + 5y = 12 and the x axis?
SSC CGL Tier II (13/09/2019)
(a) 2 (b) 3 (c) 4 (d) 6
Practice Questions
SSC CHSL 2023 Tier - 1
Q.4. Find the radius of the circle
. ?? ² + ?? ² = 25
SSC CHSL 08/08/2023 (4th Shift)
(a) 25 units (b) 5 units
(c) 2 units (d) 12 units
SSC CGL 2022 Tier - 2
Q.5. Find the coordinates of the points
where the graph 57x 19y = 399 cuts -
the coordinate axis.
SSC CGL Tier II (07/03/2023)
(a) x-axis at (-7,0) and y-axis at (0,-21)
(b) x-axis at (-7,0) and y-axis at (0,21)
(c) x-axis at (7,0) and y-axis at (0,-21)
(d) x-axis at (7,0) and y-axis at (0,21)
SSC CGL 2021 Tier - 2
Q.6. The graph of the equation x = a
(a 0) is a ______. ?
SSC CGL Tier II (08/08/2022)
(a) line parallel to x axis
(b) line parallel to y axis
(c) line at an angle of 45 degree to y axis
(d) line at an angle of 45 degree to x axis
SSC CGL 2020 Tier - 2
Q.7. The graphs of the equations
4x + y = and x + y + = 0
1
3
8
3
1
2
3
4
5
2
and intersect at a point P . The point P
also lies on the graph of the equation:
SSC CGL Tier II (29/01/2022)
(a) 4x y 7 0 (b) x 3y 12 0 - + = - - =
(c) 3x y 7 0 (d) x 2y 0 - - = + - 5 =
Q.8. What is the area (in unit squares) of
the region enclosed by the graphs of the
equations 2x – 3y + 6 = 0, 4x + y = 16 and
y = 0 ?
SSC CGL Tier II (29/01/2022)
(a) 11.5 (b) 14 (c) 10.5 (d) 12
Q.9. The graphs of the equations
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Pinnacle Day: 89th Coordinate Geometry
7x + 11y = 3 and 8x + y = 15 intersect at
the point P , which also lies on the graph
of the equation.
SSC CGL Tier II (03/02/2022)
(a) 2x + y = 2 (b) 2x – y = 1
(c) 3x + 5y = 1 (d) 3x + 2y = 3
SSC CGL 2019 Tier - 2
Q.10. The graph of the equations
3x – 20y – 2 = 0 and 11x – 5y + 61 = 0
intersect at P(a, b). What is the value of
? ( ??
2
+ ??
2
- ???? ) / ( ??
2
- ??
2
+ ???? )
SSC CGL Tier II (15/11/2020)
(a) (b) (c) (d)
37
35
31
41
5
7
41
31
Q.11. The area (in sq units) of the
triangle formed by the graphs of
equations 8x + 3y = 24 , 2x + 8 = y and
the x-axis is:
SSC CGL Tier II (15/11/2020)
(a) 28 (b) 14 (c) 15 (d)24
Q.12. The graph of the linear equation
3x – 2y = 8 and 4x + 3y = 5 intersect at
the point ( ). What is the value of a, ß
(2 ? a - ß)
SSC CGL Tier II (16/11/2020)
(a) 4 (b) 6 (c) 3 (d) 5
Q.13. What is the re?ection of the point
(5, – 3) in the line Y = 3 ?
SSC CGL Tier II (18/11/2020)
(a) (5, -6) (b) ( -5, 3) (c) (5, 9) (d) (5, 3)
Q.14. The graphs of the linear equations
4x – 2y = 10 and 4x + ky = 2 intersect at a
point (a, 4). The value of k is equal to:
SSC CGL Tier II ( 18/11/2020)
(a) 3 (b) – 3 (c) – 4 (d) 4
SSC CGL 2018 Tier - 2
Q.15. The graphs of the equations
3x + y – 5 = 0 and 2x – y – 5 = 0 intersect
at the point P( ). What is the value of a, ß
( ) ? 3 a + ß
SSC CGL Tier II (11/09/2019)
(a) 4 (b) - 4 (c)3 (d) 5
Q.16. The graph of the equation
x – 7y = – 42, intersects the y -axis at
P( ) and the graph of 6x + y – 15 = 0, a , ß
intersects the x-axis at Q( ). What is ? , d
the value of a + ß + ? + d ?
SSC CGL Tier II (11/09/2019)
(a) (b) 6 (c) (d) 5
17
2
9
2
Q.17. The point of intersection of the
graphs of the equations 3x – 5y = 19 and
3y – 7x + 1 = 0 is P( ) . What is the a , ß
value of ( ) ? 3 a - ß
SSC CGL Tier II (12/09/2019)
(a) - 2 (b) –1 (c) 1 (d) 0
Q.18. The graphs of the equations
2x + 3y = 11 and x – 2y + 12 = 0
intersects at P(
1 , 1
) and the graph of ?? ??
the equation x – 2y + 12 = 0 intersects
the x-axis at Q( ). What is the value ??
2
, ??
2
of ( ) ? ??
1
- ??
2
+ ??
1
+ ??
2
SSC CGL Tier II (12/09/2019)
(a) 13 (b) - 11 (c) 15 (d) -9
Q.19. The graph of the equations
5x – 2y + 1 = 0 and 4y – 3x + 5 = 0,
intersect at the point P( ). What is the a, ß
value of ( ) ? 2 a - 3 ß
SSC CGL Tier II (13/09/2019)
(a) 4 (b) 6 (c) - 4 (d) - 3
Answer Key :-
1.(c) 2.(d) 3.(b) 4.(b)
5.(c) 6.(b) 7.(c) 8.(b)
9.(c) 10.(b) 11.(a) 12.(d)
13.(c) 14.(c) 15.(d) 16.(a)
17.(b) 18.(c) 19.(a)
Solutions :-
Sol.1.(c) Intersection point of 2x + 5y
= 12 and x + y = 3 is (1 , 2)
Intersection point of 2x + 5y = 12 and
y = 0 is (6 , 0)
Intersection point of x + y = 3 and y = 0
is (3 , 0)
Area of triangle = ? +
1
2
[ ??
1
( ??
2
- ??
3
)
+ ] ? ??
2
( ??
3
- ??
1
) ??
3
( ??
1
- ??
2
)
? ? [1(0 – 0) + 6(0 – 2) + 3(2 – 0)] ?
1
2
? = 3
1
2
[- 12 + 6 ] | | ????????
2
Sol.2.(d)
Center = (1, –2) and radius = 4 cm
Equation of circle
(x – 1)
2
+ (y + 2)
2
= (4)
2
?
x
2
+ 1 – 2x + y
2
+ 4 + 4y = 16
x
2
– 2x + y
2
+ 4y = 16 – 5 ?
x
2
+ y
2
– 2x + 4y = 11 ?
Sol.3.(b) In x + y = 3
put x = 0 ? ?? = 3
put y = 0 ? ?? = 3
Line formed by the equation will be AB
In 2x + 5y = 12
put x = 0 ? ?? = 2 . 4
put y = 0 ? ?? = 6
Line formed by the equation will be CD
AB and CD will intersect at point E.
Given, x + y = 3 ………(1)
and 2x + 5y = 12 …….(2)
Multiply equation (1) by 2 and subtract it
from equation (2)
? 3 ?? = 6 ? ?? = 2
Put the value of y in any of the equations
x + 2 = 3 ? ?? = 1
So, E(x , y) = (1 , 2)
Area formed by the the equations
x + y = 3 , 2x + 5y = 12 and the x axis is
the shaded region or ? ??????
Now, AC = OC – OA = 6 – 3 = 3
Height of the triangle will be the y
coordinate of point E = 2
Required area = base Height
1
2
× ×
= 3 2 = 3 square units.
1
2
× ×
Sol.4.(b) Equation of circle passing
through origin :-
+ = …..( = radius) ??
2
??
2
??
2
??
Here , + = radius = 5 units ??
2
??
2
25 ?
Sol.5.(c) 57x 19y = 399 -
At x-axis y = 0 ?
57x 19y = 399 x = = 7 - ?
399
57
At y-axis x = 0 ?
57x 19y = 399 y = = 21 - ?
- 399
19
-
So , the graph 57x 19y = 399 cuts the -
coordinate axis ,
x-axis at (7,0) and y-axis at (0, 21) -
Sol.6.(b)
The graph of the equation x = a is a line
parallel to the y axis.
Sol.7.(c) 4x + y =
1
3
8
3
12x + y = 8 ………..e.q .(1)
x + y + = 0
1
2
3
4
5
2
2x + 3y = 10 ………..e.q .(2) -
Multiplying (1) by 3 and subtracting from
(2) we get,
– 34x = – 34 , x = 1, ?
Putting x = 1 in e.q .(1) , we get
y = – 4
Checking these values of x and y we ?nd
only option (c) satis?es these values.
Sol.8.(b) Intersection point of 2x – 3y +
6 = 0 and 4x + y = 16 is (3 , 4)
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Pinnacle Day: 89th Coordinate Geometry
Intersection point of 2x – 3y + 6 = 0 and
y = 0 is ( – 3 , 0)
Intersection point of 4x + y = 16 and
y = 0 is (4 , 0)
Area of triangle = ? +
1
2
[ ??
1
( ??
2
- ??
3
)
+ ] ? ??
2
( ??
3
- ??
1
) ??
3
( ??
1
- ??
2
)
? ? [3(0 – 0) – 3(0 – 4) + 4(4 – 0)] ?
1
2
? ? [ ] ? = 14
1
2
12 + 16 ????????
2
Short Trick :-
Required area = × 7 units × 4 units =
1
2
14 ????????
2
Sol.9.(c) 7x + 11y = 3 …..(1)
8x + y = 15 ……(2)
On solving eqn. (1) and (2) , we get
y = – 1 and x = 2
If we put the obtained value of x and y in
options given, then only option (c) gets
satis?ed.
Sol.10.(b) The point where both the lines
meet is the only point which satis?es
both the equations.
On solving both the e.q . , we get
a = 6 and b = 1
2 2
( ??
2
+ ??
2
- ???? ) / ( ?? - ?? + ???? )
= =
36 + 1 - 6
36 - 1 + 6
31
41
Sol.11.(a) Points of intersection
By 8x + 3y = 24 and 2x + 8 = y is (0 , 8)
y = 0 and 2x + 8 = y is ( 4 ,0) -
y = 0 and 8x + 3y = 24 is (3 ,0)
Base = 7 and height = 8
Area = 8 7 = 28
1
2
× × ????????
2
Sol.12.(d) 3x – 2y = 8 and 4x + 3y = 5
On solving both the e.q . , we get
= 2 , = – 1 (2 = 5 a ß ? a - ß)
Sol.13.(c) y = 3 is parallel to x axis so on
re?ection x axis will not change
Perpendicular distance of (5 , – 3) from
y = 3 = 6 units
6 units on the other side of the line
(y = 3 ) = 3 + 6 = 9
So the point = (5 , 9)
Sol.14.(c) They intersect at y = 4
So , 4x – 2(4) = 10 x = 4.5 ?
4x + ky = 2 4(4.5) + k(4) = 2 ?
k = – 4
Sol.15.(d) 3x + y = 5 ………e.q .(1)
2x – y = 5 ……..e.q .(2)
Adding e.q . (1) and (2) , we get
x = 2 ?
Put the value of x in e.q . (1) , we get
3x + y = 5 3(2) + y = 5 y = 1 ? ? -
So, intersecting point of the given lines
{P( )} = (2 , – 1) a , ß
= 5 ? ( 3 a + ß) = { 3 ( 2 ) + (- 1 )}
Sol.16.(a) Given ,
x – 7y = – 42 ………e.q .(1)
6x + y – 15 = 0 ………e.q .(2)
On y axis, x = 0
Put the value of x in e.q . (1) , we get
0 – 7y = – 42 ? ? ?? = 6
? ?? (a , ß) = ( 0 , 6 )
On x axis , y = 0
Put the value of y in e.q . (2) , we get
6x + 0 = 15 x = ? ?
5
2
? ?? (? , d) = (
5
2
, 0 )
? ??h?? ?????????? ???? a + ß + ? + d
= 0 + 6 + =
5
2
+ 0
17
2
Sol.17.(b) Given,
3x – 5y =19 ……e.q .(1)
and 3y – 7x + 1= 0
7x – 3y = 1 ……..e.q .(2)
Multiply e.q :- (1) by 3 and e.q . (2) by 5 and
subtract e.q :- (2) from e.q .(1) , we get
? 26 ?? = - 52 ? ?? = - 2
Putting the value of x in e.q . (1) ,
3( – 2) – 5y = 19
? ?? = - 5 ? ?? (a , ß) = (- 2 , - 5 )
( ) 3 a - ß ? 3 (- 2 ) - (- 5 ) = - 1
Sol.18.(c) 2x + 3y =11 ……….e.q .(1)
and x – 2y + 12 = 0
2y – x = 12 ………..e.q . (2)
Multiply e.q .(2) by 2 and then add both
the equation ,
and x = – 2 ? 7 ?? = 35 ? ?? = 5
1
,
1
? ?? ( ?? ?? ) = (- 2 , 5 )
Now,
x – 2y + 12 = 0
At x axis y = 0 ? ?? = - 12
Q(
2 , 2
) = ( – 12 , 0) ?? ??
( ) ??
1
- ??
2
+ ??
1
+ ??
2
{ – 2 – ( – 12) + 5 + 0 } = 15 ?
Sol.19.(a)
5x – 2y + 1 = 0 or 2y – 5x = 1…e.q .(1)
4y – 3x + 5 = 0 or 3x – 4y = 5 …e.q .(2)
Multiply eq(1) by 2 and add it in eq (2)
- 7 ?? = 7 ? ?? = - 1
Put this value in any of the equation
5( – 1) – 2y + 1 = 0
? ?? = - 2
? ?? (a , ß) = (- 1 , - 2 )
= 4 ? ( 2 a - 3 ß) = { 2 (- 1 ) - 3 (- 2 )
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FAQs on SSC CGL Previous Year Questions (2023 - 18): Coordinate Geometry - SSC CGL Mathematics Previous Year Paper (Topic-wise)
| 1. What are the key concepts of Coordinate Geometry that are important for SSC CGL exams? |  |
Ans. Key concepts include the Cartesian coordinate system, distance formula, section formula, area of triangles, slopes of lines, equations of lines (slope-intercept and point-slope forms), and properties of various geometric shapes like circles, triangles, and rectangles in the coordinate plane.
| 2. How do you calculate the distance between two points in Coordinate Geometry? | |
Ans. The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
| 3. What is the section formula in Coordinate Geometry and how is it applied? | |
Ans. The section formula is used to find the coordinates of a point that divides a line segment joining two points \((x_1, y_1)\) and \((x_2, y_2)\) in the ratio \(m:n\). The coordinates of the dividing point \(P\) are given by:
\[ P\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right) \]
| 4. How do you find the area of a triangle given its vertices in Coordinate Geometry? | |
Ans. The area \(A\) of a triangle with vertices at \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be found using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
| 5. What are the various forms of the equation of a line in Coordinate Geometry, and when are they used? | |
Ans. The main forms of the equation of a line include:
1. Slope-intercept form: \(y = mx + c\) (used when the slope and y-intercept are known).
2. Point-slope form: \(y - y_1 = m(x - x_1)\) (used when a point on the line and the slope are known).
3. Two-point form: \(y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\) (used when two points on the line are known).
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SSC CGL Previous Year Questions (2023 - 18): Coordinate Geometry SSC CGL
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The content of SSC CGL Previous Year Questions (2023 - 18): Coordinate Geometry is prepared as per the latest SSC CGL syllabus.
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