Multiplechoice questions have become an integral part of the CBSE Exams. Also, with the current modification to the exam pattern, the objective type questions are going to be asked more 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 chapter 7 Coordinate Geometry from the Maths textbooks for CBSE Class 10.
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, 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 topic wise CBSE Class 10 Maths objective questions from the clickable link given below.
List of SubTopics Covered In Chapter 7
The solutions provided at BYJUâ€™S are in such a way that every step while solving 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 the change in exam pattern, more MCQs are expected to be included in the exam paper from this academic year onwards.
7.1 Basics revisited (5 MCQs From The Topic)
7.2 Distance formula (5 MCQs From The Given Topic)
7.3 Section formula (5 MCQs Listed From The Topic)
7.4 Area from coordinates (5MCQs Listed From The Given 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: X – axis 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. IfÂ X coordinate of PÂ is a, thenÂ its ycoordinate is ______.
 (123a) / 4
 (124a)/ 3
 (12+3a)/4
 (3a12)/4
 Point P lies on the line 3x+ 4y – 12 = 0. IfÂ X coordinate of PÂ is a, thenÂ 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 followinga) 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 followinga) Its Ycoordinate
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 y coordinate will be 1.
Its x coordinate 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 theÂ triangle formed by these points is zero.
Since, the points A (k+1,2k), B (3k, 2k+3) and C (5k1,5k)Â are collinear.
Then, area ofÂ Î”Â 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 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
andÂ 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=0Â Â Â Â [by factorization 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 the three sides and see if the condition of the triangle is satisfied by these sides.
Now, using distance formula between two points,
We see that,Â ABâ‰ BCâ‰ CA
AndÂ Î”Â ABCÂ does not satisfy Pythagoras theorem
Hence, the required triangle is scalene because all sides are of different length.
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 distance formula,
OA^{2}Â = (2^{2}Â +Â 2^{2})
OA^{2}Â = 8

 The points on 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 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 formÂ (x,0)
Let the point beÂ (x,0)
Using distance formula
(xâˆ’11)^{2}+8^{2}=10^{2}
x^{2}âˆ’22x+121+64=100
x^{2}âˆ’22x+85=0,
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 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 pointsÂ (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)
âˆµÂ Distance between two pointsÂ (x_{1}, y_{1})Â andÂ (x_{2}, y_{2})Â is
âˆ´Â Distance between centroid and circumcentre

 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 andâ€™dâ€™ be the distance between them.
By using distance formula, we get
Section formula

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

 Point 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
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
 None of these
 (6,9)
 (4,5)
 (8,9)
 Find the point (x,y) that divides the join of A(3,6) and B(7,10) in the ratio 3:1
Answer: (B) (6, 9)
Solution: If (x, y) divides the join of A(x_{1},y_{1})Â andÂ (x_{2},y_{2})Â 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.
âˆ´Â 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 areÂ (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 areÂ (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´Â Area ofÂ 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, area ofÂ Â Â 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
 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 the
Answer: (B) Ratio of their corresponding bases
Solution: Area of triangle 1 / Area of triangle 2= (Â½ Ã—Â Base 1Ã—Â height) / (Â½ Ã—Â Base 2Ã—Â 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) AllÂ ofÂ these
(D) 1/2[x_{1}(y_{2}âˆ’y_{3}) +x_{2}(y_{3}âˆ’y_{1}) +x_{3}(y_{1}âˆ’y_{2})]
Answer: (D) 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 all trapezia.
Now from 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,
Area ofÂ 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 ofÂ Î”PAB=10Â sq. units, then 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 ofÂ Î”PAB=10Â sq. units, then 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)
Area ofÂ 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)
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, Length ofÂ PA
â‡’k=2
Each question is provided with an answer and a detailed solution. Students are, however, advised to attempt these questions on their own before referring to the answers in order to selfassess their performance.
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