CBSE Class 10 Maths Chapter 7 Coordinate Geometry Objective Questions have become an integral part of the CBSE Board exams. Also, with the current modification of the exam pattern, more objectivetype questions will be asked in the Class 10 Board exams for all subjects, including Mathematics. Here, we have compiled only the Multiple Choice Questions (MCQs) framed from various important topics in CBSE Maths Chapter 7 – Coordinate Geometry as per the Maths textbooks for CBSE Class 10. These CBSE Class 10 Maths Chapter 7 Coordinate Geometry MCQs are very useful to score good marks in Maths.
One of the main topics discussed in this chapter includes finding the area between two points whose coordinate values are provided. For instance, the area of a triangle. This chapter covers some basic concepts like the area of a triangle, rhombus, and the distance between sides and intersections. This chapter also deals with the relationship between Numerical and Geometry and their application in our daily lives. Students can download the topicwise CBSE Class 10 Maths objective questions from the link given here.
List of Subtopics Covered in Chapter 7
The NCERT Solutions provided at BYJUâ€™S are created in such a way that every step in a problem is explained clearly and in detail. The CBSE Class 10 examination often asks questions, either directly or indirectly, from the NCERT textbooks. These CBSE Class 10 Maths Chapter 7 – Coordinate Geometry Objective Questions with Solutions is also very helpful for the students to prepare for the exams. Following a change in the exam pattern, more MCQs are expected to be included in the exam paper in this academic year.
7.1 Basics Revisited (5 MCQs from the Topic)
7.2 Distance Formula (5 MCQs from the Topic)
7.3 Section Formula (5 MCQs from the Topic)
7.4 Area from Coordinates (5 MCQs from the Topic)
Download CBSE Class 10 Maths Chapter 7 – Coordinate Geometry Objective Questions Free PDF
Basics Revisited

 What will be the reflection of the point (4, 5) about the xaxis in the fourth quadrant?
 (4, 5)
 (4, 5)
 (4, 5)
 (4, 5)
 What will be the reflection of the point (4, 5) about the xaxis in the fourth quadrant?
Answer: (B) (4,5)
Solution: Xaxis will act as a plane mirror, and this point will form an image, following the sign convention, at (4, 5) in the fourth quadrant.

 Point P lies on the line 3x + 4y – 12 = 0. If the xcoordinate of P is a, then its ycoordinate is ______.
 (123a) / 4
 (124a)/ 3
 (12+3a)/4
 (3a12)/4
 Point P lies on the line 3x + 4y – 12 = 0. If the xcoordinate of P is a, then its ycoordinate is ______.
Answer: (A) (123a) / 4
Solution:
The points on the line should satisfy the equation of the line.
So, the point P (a, y) satisfies the equation.
Hence,
3x+ 4y – 12 = 0
3(a) + 4(y) 12 = 0
4y = 123a
Y = (123a) / 4

 If point P lies on the line y = 1, find the following a) Its ycoordinate, b) Its xcoordinate
 y = 1, x can be any real number
 y = 1, x = 1
 x = 2, y = 1
 x = 1, y = 1
 If point P lies on the line y = 1, find the following a) Its ycoordinate, b) Its xcoordinate
Answer: (A) y = 1, x can be any real number
Solution: Since it is given that the point P lies on the line y = 1, its ycoordinate will be 1, and the xcoordinate can be any real number.

 Find the values of k if the points A (k+1, 2k), B (3k, 2k+3), and C (5k1, 5k) are collinear.
 k = 5,Â 1/5
 k = 4,Â 1/4
 k = 3,Â 1/3
 k = 2,Â 1/2
 Find the values of k if the points A (k+1, 2k), B (3k, 2k+3), and C (5k1, 5k) are collinear.
Answer: (D) k = 2,Â 1/2
Solution: We know that if three points are collinear, then the area of the triangle formed by these points is zero.
Since the points A (k+1,2k), B (3k, 2k+3) and C (5k1,5k) are collinear,
The area of Î” ABC = 0.
â‡’ Â½ [x_{1}(y_{2}âˆ’y_{3}) +x_{2}(y_{3}âˆ’y_{1}) +x_{3}(y_{1}âˆ’y_{2})] =0
Multiplying the above expression by 2, we get
[x_{1}(y_{2}âˆ’y_{3}) +x_{2}(y_{3}âˆ’y_{1}) +x_{3}(y_{1}âˆ’y_{2})] =0Here,Â x_{1}=k+1,Â x_{2}=3k,Â x_{3}=5kâˆ’1
And y_{1}=2k,Â y_{2}=2k+3,Â y_{3}=5k
â‡’[(k+1)(2k+3âˆ’5k)+3k(5kâˆ’2k)+(5kâˆ’1)(2kâˆ’(2k+3))]=0â‡’[(k+1)(3âˆ’3k)+3k(3k)+(5kâˆ’1)(2kâˆ’2kâˆ’3)]=0
â‡’ [âˆ’3k^{2}+3kâˆ’3k+3+9k^{2}âˆ’15k+3] =0
â‡’ (6k^{2}âˆ’15k+6) =0
â‡’6k^{2}âˆ’15k+6=0
Dividing the equation by 3, we get
â‡’2k^{2}âˆ’5k+2=0
â‡’2k^{2}âˆ’4kâˆ’k+2=0 [by factorisation method]
â‡’2k(kâˆ’2)âˆ’1(kâˆ’2)=0
â‡’ (kâˆ’2) (2kâˆ’1) =0
If k â€“ 2 = 0, then k = 2
If 2k â€“ 1, then k =Â 1/2
âˆ´ k=2, 1/2
Hence, the required values of k are 2 andÂ 1/2.

 Name the type of triangle formed by the points A (5, 6), B (4,2), and C (7, 5).
 Equilateral triangle
 Scalene triangle
 Isosceles triangle
 Rightangled triangle
 Name the type of triangle formed by the points A (5, 6), B (4,2), and C (7, 5).
Answer: (B) Scalene triangle
Solution: To find the type of triangle, first, we determine the length of all three sides and see if the condition of the triangle is satisfied by these sides.
Now, using the distance formula between two points,
We see that ABâ‰ BCâ‰ CA
And Î”ABC does not satisfy the Pythagoras theorem
Hence, the required triangle is scalene because all sides have different lengths.
Distance Formula

 The distance of the point (â€“2, â€“2) from the origin is
Solution: Let the origin be O and the point A be (2, 2)
Using the distance formula,
OA^{2}Â = (2^{2}Â +Â 2^{2})
OA^{2}Â = 8

 The points on the Xaxis at a distance of 10 units from (11, â€“8) are:
 (5, 2) (17, 0)
 (5, 0) (17, 0)
 (6, 0) (17, 0)
 (5, 0) (16, 0)
 The points on the Xaxis at a distance of 10 units from (11, â€“8) are:
Answer: (B) (5, 0) (17, 0)
Solution: Any point on the xaxis is of the form (x, 0)
Let the point beÂ (x,0)
Using the distance formula,
(xâˆ’11)^{2}+8^{2}=10^{2}
x^{2}âˆ’22x+121+64=100
x^{2}âˆ’22x+85=0,
By factorising, we get,
x^{2}âˆ’17xâˆ’5x+85=0
(xâˆ’17)(xâˆ’5)=0
x=17, 5
Hence, the points are (17,0) and (5,0).

 The point on the xaxis, which is equidistant from (2, â€“5) and (â€“2, 9), is:
 (7, 0)
 (7, 0)
 (2, 0)
 (2, 0)
 The point on the xaxis, which is equidistant from (2, â€“5) and (â€“2, 9), is:
Answer: (B) (7, 0)
Solution: We know that a point on the xaxis is of the form (x, 0). Let the point on the xaxis be P(x, 0), and the given points are A (2, 5) and B (2, 9).
Now,
Since PA = PB,
â‡’Â (2âˆ’x)^{2}Â +Â (âˆ’5âˆ’0)^{2}Â =Â (âˆ’2âˆ’x)^{2}Â +Â (9âˆ’0)^{2}
â‡’Â 4 – 4x +Â x^{2}Â + 25 = 4 + 4x +Â x^{2}Â + 81
â‡’Â – 8x = 56
â‡’Â x = 7
Hence, the required point is (7, 0)

 The points (6. 6), (0, 6), and (6, 0) are the vertices of a right triangle, as shown in the figure. Find the distance between its centroid and circumcentre.
Solution: Circumcentre = midpoint of AB
= [(X_{1 + }X_{2})/2, (Y_{1} + Y_{2})/2]
= [(6+0)/2, (6+0)/2]
= (3, 3)
Centroid (G)Â = [(X_{1 }+ X_{2}+ X_{3})/3, (Y_{1}+Y_{2}+ Y_{3})/3]
= [(6+0+6)/3, (0+6+6)/3]
= (4, 4)
âˆµ The distance between two points (x_{1}, y_{1})Â andÂ (x_{2}, y_{2})Â is
âˆ´ The distance between the centroid and circumcentre is

 The distance between the points (a, b) and (â€“a, â€“b) is:
Solution: Let A (a, b) and B (a,b) be the two points and â€™dâ€™ be the distance between them.
By using the distance formula, we get
Section Formula

 Midpoint of the line segment joining the points (â€“5, 4) and (9, â€“8) is:
 (2,2)
 (7,6)
 (2,2)
 (7,6)
 Midpoint of the line segment joining the points (â€“5, 4) and (9, â€“8) is:
Answer: (C) (2,2)
Solution: Midpoint of a line segment joining (x_{1}, y_{1})Â andÂ (x_{2}, y_{2})Â is [(x_{1}+ x_{2})/2, (y_{1}+y_{2})/2]
âˆ´ Midpoint of the linesegment joining the points (â€“5, 4) and (9, â€“8) = [(95)/2, (48)/2] = (2, 2)

 The line x + y = 10 divides line segment AB in the ratio a:1. Find the value of a.


 Â½
 1
 2
 3

Answer: (B) 1
Solution:
The point, say P(x, y), divides the line AB into the ratio a:1.
The equation for the point that divides a line in the ratio m:n is
where (x_{1}, y_{1}) and (x_{2}, y_{2}) are the coordinates of the endpoints of the line segment.
Applying the formula, we get
This point lies on the line x + y = 10, so substitute the points in the equation for the line
6a + 2 + 8a + 4 = 10(a+1)
4a = 4
a = 1

 Points A (1, 2) and B (3, 4) are two ends of a line segment. Find the point which divides AB in the ratio 3:4.


 (4,3)
 (2,3)
 15/7, 22/7
 13/7, 20/7

Answer: (D) 13/7, 20/7
Solution: The coordinates for the point that divides a line in the ratio m:n is
Substituting in the equation, we get
= 13/7, 20/7
Therefore, the point which divides the line segment in the ratio 3:4 is 13/7, 20/7.

 Find the point (x,y) that divides the join of A(3,6) and B(7,10) in the ratio 3:1
 (6, 9)
 (4, 5)
 (8, 9)
 None of these
 Find the point (x,y) that divides the join of A(3,6) and B(7,10) in the ratio 3:1
Answer: (A) (6, 9)
Solution: If (x, y) divides the join of A(x_{1}, y_{1})Â andÂ (x_{2}, y_{2}) in the ratio m:n
Then,
and
Here,Â x_{1}=3,Â x_{2}=7,Â y_{1}=6,Â y_{2}=10, m = 3 and n = 1
and
x = 6 and y = 9
Therefore, the point is (6, 9).

 If (1, 2), (4, y), (x, 6), and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
 (2, 3)
 (4, 3)
 (6, 3)
 (2, 5)
 If (1, 2), (4, y), (x, 6), and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Answer: (C) (6, 3)
Solution:
The given points are: A(1,2), B(4, y), C(x, 6), and D(3,5)
Since the diagonals of a parallelogram bisect each other,
The coordinates of P are:
X= (x+1)/ 2 = (3+4)/2
â‡’x+1=7â‡’x=6
Y = (5+Y)/2= (6+2)/2
â‡’5+y=8â‡’y=3
âˆ´Â The required values of x and y are:
x=6, y=3.
Area from Coordinates

 The area of a quadrilateral whose vertices taken in order are (â€“4, â€“2), (â€“3, â€“5), (3, â€“2), and (2, 3) is _______.
 26 sq. units
 28 sq. units
 30 sq. units
 27 sq. units
 The area of a quadrilateral whose vertices taken in order are (â€“4, â€“2), (â€“3, â€“5), (3, â€“2), and (2, 3) is _______.
Answer: (B) 28 sq. units
Solution:
Consider the pointsÂ A(âˆ’4,âˆ’2),B(âˆ’3,âˆ’5),C(3,âˆ’2)Â andÂ D(2,3).
Area of a triangle having coordinatesÂ (x_{1}, y_{1}),Â (x_{2}, y_{2}),Â (x_{3}, y_{3})Â is given by
Â½ Ã— x_{1} (y_{2}âˆ’y_{3}) +x_{2}(y_{3}âˆ’y_{1}) +x_{3}(y_{1}âˆ’y_{2})
âˆ´ The area of Î”ABC
=Â 1/2  (4) (5 + 2) 3(2 + 2) + 3(2 + 5)
=Â Â 1/220âˆ’8âˆ’6+15=21/2Â = 10.5 sq. units
Similarly, the area ofÂ Î”ACD
=Â 1/2 (4) (2Â – 3) +3(3Â + 2) + 2(2 + 2)
=Â 1/220+15=35/2Â = 17.5Â sq. units
Now, area of quadrilateral ABCD = AreaÂ (Î”ABC)Â +Â Area (Î”ACD)Â = (10.5 + 17.5) sq. units = 28 sq. units.

 If the points (a, 0), (0, b) and (1, 1) are collinear, then which of the following is true?
 (1/a) + (1/b)= 2
 (1/a) + (1/b)= 1
 (1/a) + (1/b)= 0
 (1/a) + (1/b)= 4
 If the points (a, 0), (0, b) and (1, 1) are collinear, then which of the following is true?
Answer: (B) (1/a) + (1/b) = 1
Solution: Area of a triangle formed byÂ (x_{1}, y_{1}),Â (x_{2}, y_{2}),Â (x_{3}, y_{3})Â =Â 1/2Â [x_{1}(y_{2}Â –Â y_{3}) +Â x_{2}(y_{3}Â –Â y_{1})
+Â x_{3}(y_{1}Â –Â y_{2})]
For 3 points to be collinear, the area of the triangle should be zero:
â‡’1/2[a (bâˆ’1)Â +Â 0(1âˆ’0)Â +Â 1(0âˆ’b)] =0
â‡’1/2[a (bâˆ’1) +1(0âˆ’b)] =0
â‡’ab=a+b
â‡’ (1/a) + (1/b) = 1

 If 2 triangles have the same height, the ratio of their areas is equal to
 The ratio of any 2 sides
 The ratio of their corresponding bases
 The ratio of their heights
 1
 If 2 triangles have the same height, the ratio of their areas is equal to
Answer: (B) The ratio of their corresponding bases
Solution: Area of triangle 1 / Area of triangle 2 = (Â½ Ã— Base 1Ã— height) / (Â½ Ã— Base 2Ã— height) = base 1/ base 2

 The area of triangle with verticesÂ A(x_{1},y_{1}),B(x_{2},y_{2})Â andÂ C(x_{3},y_{3})Â is:
(A) 1/2[x_{2}(y_{2}âˆ’y_{3}) +x_{3}(y_{3}âˆ’y_{1}) +x_{2}(y_{1}âˆ’y_{2})]
(B) 1/2[x_{1}(y_{2}âˆ’y_{3}) +x_{3}(y_{3}âˆ’y_{1}) +x_{1}(y_{1}âˆ’y_{2})]
(C) 1/2[x_{1}(y_{2}âˆ’y_{3}) +x_{2}(y_{3}âˆ’y_{1}) +x_{3}(y_{1}âˆ’y_{2})]
(D)Â All of the above
Answer: (C) 1/2[x_{1}(y_{2}âˆ’y_{3}) +x_{2}(y_{3}âˆ’y_{1}) +x_{3}(y_{1}âˆ’y_{2})]
Solution:
Let ABC be any triangle whose vertices areÂ A(x_{1}, y_{1}), B(x_{2}, y_{2})Â andÂ C(x_{3}, y_{3}). Draw AP, QR and CR perpendiculars from A, B and C, respectively, to the xaxis. Clearly, ABQP, APRC and BQRC are trapezia.
Now from the figure, it is clear that
Area of â–³ABC = Area of trapezium ABQP + Area of trapezium APRC – Area of trapezium BQRC.
Area of a trapeziumÂ =1/2Â (Sum of parallel sides)(Distance between the parallel sides)
Therefore,
The area of Î”ABC=1/2(BQ+AP)QP+1/2(AP+CR)PRâˆ’1/2(BQ+CR)QR
=1/2(y_{2}+y_{1})(x_{1}âˆ’x_{2})+1/2(y_{3}+y_{1})(x_{3}âˆ’x_{1})âˆ’1/2(y_{2}+y_{3})(x_{3}âˆ’x_{2})
=1/2[x_{1}(y_{2}âˆ’y_{3})+x_{2}(y_{3}âˆ’y_{1})+x_{3}(y_{1}âˆ’y_{2})]
Thus, the area ofÂ Î”ABCÂ is the numerical value of the expressionÂ 1/2[x_{1}(y_{2}âˆ’y_{3})+x_{2}(y_{3}âˆ’y_{1})+x_{3}(y_{1}âˆ’y_{2})].

 P(x, y), A (3, 4) and B (5, 2) are the vertices of triangle PAB, such that PA=PB and area of Î”PAB=10 sq. units, and PA = kâˆš5 units.Â Find the value of k.
 2
 1
 âˆš5
 3
 P(x, y), A (3, 4) and B (5, 2) are the vertices of triangle PAB, such that PA=PB and area of Î”PAB=10 sq. units, and PA = kâˆš5 units.Â Find the value of k.
Answer: (A) 2
Solution:
Given, PA=PB
âˆ´ PA^{2}=PB^{2}
â‡’ (xâˆ’3)^{2}+ (yâˆ’4)^{2 }= (xâˆ’5)^{2}+(y+2)^{2}
â‡’x^{2}âˆ’6x+9+y^{2}âˆ’8y+16=x^{2}âˆ’10x+25+y^{2}+4y+4
â‡’4xâˆ’12y=4â‡’xâˆ’3y=1â€¦(i)
The area of Î”PAB=1/2x (4+2) +3(âˆ’2âˆ’y) +5(yâˆ’4)
â‡’10=1/26xâˆ’6âˆ’3y+5yâˆ’20
â‡’20=6x+2yâˆ’26â‡’6x+2yâˆ’26=20Â orÂ 6x+2yâˆ’26=âˆ’20
â‡’6x+2y=46Â orÂ 6x+2y=63x+y=23â€¦ (2)
orÂ 3x+y=3â€¦ (3)
By solving (1) and (3), we get
x=1, y=0
Solving (1) and (2), we get
x=7, y=2
âˆ´Â Coordinates of P are (7, 2) and (1, 0)
Length of
Also, the Length of PA
â‡’k = 2
Each question is provided with the answer and a detailed solution. Students are, however, advised to solve these questions on their own before referring to the answers in order to selfassess their performance. Also, find here some CBSE Class 10 Maths Chapter 7 Extra MCQs.
CBSE Class 10 Maths Chapter 7 Extra MCQs
1. The points (5, 1), (1, p) and (4, 2) are collinear if the value of p is _____.
(a) 1
(b) 2
(c) 1
(d) 3
2. What are the points (1,1), (2, 7) and (3, 3)?Â
(a) vertices of an isosceles triangle
(b) collinear
(c) vertices of an equilateral triangle
(d) none of the above
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