RS Aggarwal Solutions Chapter 16 Coordinate Geometry MCQ Class 10 Maths
Chapter Name  RS Aggarwal Chapter 16 Coordinate Geometry 
Book Name  RS Aggarwal Mathematics for Class 10 
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Related Study  NCERT Solutions for Class 10 Maths 
Coordinate Geometry MCQ Solutions
1. The distance of the point P(6, 8) from the origin is
(a) 8
(b) 2√ 7
(c) 6
(d) 10
Solution
(d) 10
(a) 3
(b) 3
(c) 4
(d) 5
Solution
The distance of a point (x, y) from xaxis is y.
Here, the point is (3, 4). So, its distance from xaxis is 4 = 4
3. The point on xaxis which is equidistant from the points A(1, 0) and B(5, 0) is
(a) (0, 2)
(b) (2, 0)
(c) (3, 0)
(d) (0, 3)
Solution
(b) (2, 0)
Let P(x, 0) the point on xaxis, then
AP = BP ⇒ AP^{2} = BP^{2}
⇒ (x + 1)^{2} + (0 + 0)^{2} = (x – 5)^{2} + (0 – 0)^{2}
⇒ x^{2 }+ 2x + 1 = x^{2} – 10x + 25
⇒ 12x = 24
⇒ x = 2
Thus, the required point is (2, 0).
4. If R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, 4) then y equals
(a) 5
(b) 7
(c) 12
(d) 6
Solution
(b) 7
Since R(5, 6) is the midpoint of the line segment AB joining the points
A(6, 5) and B(4, y), therefore,
(5 + y)/2 = 6
⇒ 5 + y = 12
⇒ y = 12 – 5 = 7
5. If the point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3 then the value of k is
(a) 16
(b) 28/5
(c) 16/5
(d) 8/5
Solution
(c) 16/5
The point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3. So,
k = (2 × 5 + 3 × 2)/(2 + 3) = (10 + 6)/5
= 16/5
6. The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is
(a) (7 + √5)
(b) 5
(c) 10
(d) 12
Solution
(d) 12
Let A(0, 4), B(0, 0) and C(3, 0) be the given vertices. So,
Therefore,
AB + BC + AC = 4 + 3 + 5 = 12.
7. If A(1, 3), B(1, 2), C(2, 5) and D(x, 4) are the vertices of a gm ABCD the value of x is
(a) 3
(b) 4
(c) 0
(d) 3/2
Solution
(b) 4
The diagonals of a parallelogram bisect each other. The vertices of the gm ABCD are A(1, 3), B(1, 2) and C(2, 5) and D(x, 4)
Here, AC and BD are the diagonals. So
(1 + 2)/2 = (1 + x)/2
⇒ x – 1 = 3
⇒ x = 1+ 3 = 4
8. If the points A(x, 2), B(3, 4) and C(7, 5) are collinear then the value of x is
(a) 63
(b) 63
(c) 60
(d) – 60
Solution
(a) – 63
Let A(x_{1} = x, y_{2 }= 2), B(x_{2} = 3, y_{2 }= 4) and C(x_{3 }= 7, y_{3} = 5) be collinear points. Then
x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1 }– y_{2}) = 0
⇒ x(  4 + 5) + (3) (5 – 2) + 7(2 + 4) = 0
⇒ x + 21 + 42 = 0
⇒ x = 63
9. The area of a triangle with vertices A(5, 0), B(8, 0) and C(8, 4) in square units is
(a) 20
(b) 12
(c) 6
(d) 16
Solution
(c) 6
Let A (x_{1 }= 5, y_{1} = 0), B(x_{2 }= 8, y_{2 }= 0) and C(x_{3} = 8, y_{3} = 4) be the vertices of the triangle.
Then
Area(△ABC) = ½[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]
= ½[5(0 – 4) + 8(4 – 0) + 8(0 – 0)]
= ½[20 + 32 + 0]
= 6 sq. units.
10. The area of △ABC with vertices A(a, 0), O(0, 0) and B(0, b) in square units is
(a) ab
(b) ½.ab
(c) ½.a^{2}b^{2}
(d) 1/2b^{2}
Solution
(b) 1/2ab
Let A(x_{1 }= a, y_{1} = 0), O(x_{2 }= 0, y_{2} = 0) and B(x_{3} = 0, y_{3} = b) be the given vertices. So
Area (△ABO) = ½x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})
= ½a(0 – b) + 0(b – 0) + 0(0 – 0)
= ½ab
= 1/2ab
11. If P(a/2, 4) is the midpoint of the line segment joining the points A(6, 5) and B(2, 3) then the value of a is
(a) 8
(b) 3
(c) 4
(d) 4
Solution
(a)  8
The point P(a/2, 4) is the midpoint of the line segment joining the points A(6, 5) and B(2, 3).
So, a/2 = (6 – 2)/2
⇒ a/2 = 4
⇒ a = 8
12. ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). The length of one of its diagonals is
(a) 5
(b) 4
(c) 3
(d) 245
Solution
(a) 5
Here, AC and BD are two diagonals of the rectangle ABCD. So
13. The coordinates of the point P dividing the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1 is
(a) (2, 4)
(b) (3, 5)
(c) (4, 2)
(d) (5, 3)
Solution
(b) (3, 5)
Here, the point P divides the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1. Then,
Coordinates of P = (2 × 4 + 1 × 1)/(2 + 1), (2 × 6 + 1 × 3)/(2 + 1)
= (8 + 1)/3, (12 + 3)/3
= (9/3, 15/3)
= (3, 5)
14. If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (2, 5), then the coordinates of the other end of the diameter are
(a) (6, 7)
(b) (6, 7)
(c) (4, 2)
(d) (5, 3)
Solution
(a) (6, 7)
Let (x, y) be the coordinates of the other end of the diameter. Then
2 = (2 + x)/2
⇒ x = 6
5 = (3 + y)2
⇒ y = 7
15. In the given figure P(5, 3) and Q(3, y) are the points of trisection of the line segment joining A(7, 2) and B(1, 5). Then, y equals
(a) 2(b) 4
(c) 4
(b) 5/2
Solution
(c) 4
Here, AQ : BQ = 2 : 1. Then,
y = {2 × (5) + 1 × (2)}/(2 + 1)
= (10 – 2)/3
= 4
16. The midpoint of segment AB is P(0, 4). If the coordinates of B are(2, 3), then the coordinates of A are
(a) (2, 5)
(b) (2, 5)
(c) (2, 9)
(d) (2, 11)
Solution
(a) (2, 5)
Let (x, y) be the coordinates of A. Then,
0 = (2 + x)/2
⇒ x = 2
4 = (3 + y)/2
⇒ y = 8 – 3 = 5
Thus, the coordinates of A are (2, 5).
17. The point P which divides the line segment joining the points A(2, 5) and B(5, 2) in the ratio 2 : 3 lies in the quadrant
(a) I
(b) II
(c) III
(d) IV
Solution
(d) IV
Let (x, y) be the coordinates of P. Then,
x = (2 × 5 + 3 × 2)/(2 + 3) = (10 + 6)/5 = 16/5
y = (2 × 2 + 3 × (5)/(2 + 3) = (4 – 15)/5 = 11/5
Thus, the coordinates of point P are (16/5, 11/5) and so it lies in the fourth quadrant.
18. If A(6, 7) and B(1, 5) are two given points then the distance 2AB is
(a) 13
(b) 26
(c) 169
(d) 238
Solution
(b) 26
The given points are A(6, 7) and B(1, 5). So,
(a) (0, 4)
(b) (4, 0)
(c) (3, 0)
(d) (0, 3)
Solution
(c) (3, 0)
Let p(x, 0) be the point on xaxis. Then as per the question
AP = BP ⇒ AP^{2} = BP^{2}
⇒ (x – 7)^{2} + (0 – 6)^{2} = (x – 3)^{2 }+ (0 – 4)^{2}
⇒ x^{2} – 14x + 49 + 36 = x^{2 }+ 6x + 9 + 16
⇒ 60 = 20x
⇒ x = 60/20 = 3
Thus, the required point is (3, 0).
20. The distance of P(3, 4) from the xaxis is
(a) 3 units
(b) 4 units
(c) 5 units
(d) 1 unit
Solution
(b) 4 units
The ycoordinate the distance of the point from the xaxis
Here, the ycoordinate is 4.
21. In what ratio does the xaxis divide the join of A(2, 3) and B(5, 6) ?
(a) 2 : 3
(b) 3 : 5
(c) 1 : 2
(d) 2 : 1
Solution
(c) 1 : 2
Let AB be divided by the xaxis in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of P are
P(5k + 2)/(k + 1), (6k – 3)/(k + 1)
But P lies on the xaxes so, its ordinate is 0.
(6k – 3)/(k + 1) = 0
⇒ 6k – 3 = 0
⇒ 6k = 3
⇒ k = 1/2
Hence, the required ratio is 1/2 : 1 which is same as 1 : 2.
22. In what ratio does the yaxis divide the join of P(4, 2) and Q(8, 3) ?
(a) 3 : 1
(b) 1 : 3
(c) 2 : 1
(d) 1 : 2
Solution
(d) 1 : 2
Let AB be divided by the yaxis in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of P are
P(8k – 4)/(k + 1), (3k + 2)/(k + 1)
But, P lies on the yaxis, so, its abscissa is 0.
⇒ (8k – 4)/(k + 1) = 0
⇒ 8k – 4 = 0
⇒ 8k = 4
⇒ k = 1/2
Hence, the required ratio is ½: 1 which is same as 1 : 2.
23. If P(1, 1) is the midpoint of the line segment joining A(3, b) and B(1, b +4) then b = ?
(a) 1
(b) 1
(c) 2
(d) 0
Solution
(b) 1
The given points are A(3, b) and B(1, b + 4)
Then, (x_{1 }= 3, y_{1} = b) and (x_{2} = 1, y_{2} = b + 4)
Therefore,
x = [3 + 1]/2
= 2/2
=  1
And
y = [b + (b + 4)]/2
= (2b + 4)/2
= b + 2
But the midpoint is P(1, 1).
Therefore,
b + 2 = 1
⇒ b = 1
24. The line 2x + y – 4 = 0 divide the line segment joining A(2, 2) and B(3, 7) in the ratio
(a) 2 : 5
(b) 2 : 9
(c) 2 : 7
(d) 2 : 3
Solution
(b) 2 : 9
Let the line 2x + y – 4 = 0 divide the line segment in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are
P(3k + 2)/(k + 1), (7k – 2)/(k + 1)
Since P lies on the line 2x + y – 4 = 0, we have
2(3k + 2)/(k + 1) + (7k – 2)/(k + 1) – 4 = 0
⇒ (6k + 4) + (7k – 2) – (4k + 4) = 0
⇒ 9k = 2
⇒ k = 2/9
Hence, the required ratio is 2/9 : 1 which is same as 2 : 9.
25. If A(4, 2), B(6, 5) and C(1, 4) be the vertices of △ABC and AD is a median, then the coordinates of D are
(a) (5/2, 3)
(b) (5, 7/2)
(c) (7/2, 9/2)
(d) none of these
Solution
(c) (7/2, 9/2)
D is the midpoint of BC
So, the coordinates of D are
D(6 + 1)/2, (5 + 4) [B(6, 5) and C(1, 4)
⇒ (x_{1} = 6, y_{1} = 5) and (x_{2} = 1, y_{2 }= 4)]
i.e., D(7/2, 9/2)
26. If A(1, 0), B(5, 2) and C(8, 2) are the vertices of △ABC then its centroid is
(a) (12, 0)
(b) (6, 0)
(c) (0, 6)
(d) (4, 0)
Solution
(d) (4, 0)
The given point are A(1, 0), B(5, 2) and C(8, 2).
Here, (x_{1} = 1, y = 0), (x_{2} = 5, y = 2) and (x_{3} = 8, y_{3} = 2)
Let G(x, y) be the centroid of △ABC. Then,
x = 1/3(x_{1} + x_{2 }+ x_{3})
= 1/3(1 + 5 + 8)
= 4
and
y = 1/3(y_{1} + y_{2} + y_{3})
= 1/3(0 – 2 + 2)
= 0
Hence, the centroid of △ABC is G(4, 0).
27. Two vertices of △ABC are A(1, 4) and B(5, 2) and its centroid is G(0, 3). Then the coordinates of C are
(a) (4, 3)
(b) (4, 15)
(c) (4, 15)
(d) (15, 4)
Solution
(c) (4, 15)
Two vertices of △ABC are A(1, 4) and B(5, 2).
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are
G(1 + 5 + a)/3, (4 + 2 + b)/3
i.e., G(4 + a)/3, (6 + b)/3
But it is given that the centroid is G(0, 3)
Therefore,
(4 + a)/3 = 0 and (6 + b)/3 =  3
⇒ 4 + a = 0 and 6 + b = 9
⇒ a = 4 and b = 15
Hence, the third vertex of △ABC is C(4, 15)
28. The points A(4, 0), B(4, 0) and C(0, 3) are the vertices of a triangle, which is
(a) isosceles
(b) equilateral
(c) scalene
(d) rightangled
Solution
(a) isosceles
Let A(4, 0), B(4, 0) and C(0, 3) be the given points. Then,
BC = AC = 5 units
Therefore, △ABC is isosceles
29. The points P(0, 6), Q(5, 3) and R(3, 1) are the vertices of a triangle, which is
(a) equilateral
(b) isosceles
(c) scalene
(d) rightangled
Solution
(d) rightangled
Let P(0, 6), Q(5, 3) and R(3, 1) be the given points. Then,
Thus, PQ^{2} + PR^{2} = QR^{2}
Therefore, △PQR is rightangled.
30. If the points A(2, 3), B(5, k) and C(6, 7) are collinear then
(a) k = 4
(b) k = 6
(c) k = 3/2
(d) k = 11/4
Solution
(b) k = 6
The given points are A(2, 3),B(5, k) and C(6, 7)
Here, (x_{1} = 2, y_{1} = 3), (x_{2 }= 5, y_{2 }= k) and (x_{3} = 6, y_{3} = 7).
Point A, B and C are collinear. Then,
x_{1}(y_{2 }– y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2}) = 0
⇒ 2(k – 7) + 5(7 – 3) + 6(3 – k) = 0
⇒ 2k – 14 + 20 + 18 – 6k = 0
⇒  4k = 24
⇒ k = 6
31. If the point A(1, 2), O(0, 0) and C(a, b) are collinear, then
(a) a = b
(b) a = 2b
(c) 2a = b
(d) a + b = 0
Solution
(c) 2a = b
The given points are A(1, 2), O(0, 0) and C(a, b)
Here, (x_{1 }= 1, y_{1} = 2), (x_{2} = 0, y_{2 }= 0) and (x_{3} = a, y_{3 }= b)
Point A, O and C are collinear
⇒ x_{1}(y_{2} – y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1} – y_{2}) = 0
⇒ 1(0 – b) + 0(b – 2) + a(2 – 0) = 0
⇒  b + 2a = 0
⇒ 2a = b
32. The area of △ABC with vertices A(3, 0), B(7, 0) and C(8, 4) is
(a) 14 sq units
(b) 28 sq units
(c) 8 sq units
(d) 6 sq units
Solution
(c) 8 sq units
The given points are A(3, 0), B(7, 0) and C(8, 4).
Here, (x_{1 }= 3, y_{1} = 0), (x_{2} = 7, y_{2} = 0) and (x_{3} = 8, y_{3} = 4)
Therefore,
Area of △ABC = ½[x_{1}(y_{2 }– y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1} – y_{2})]
= ½[3(0 – 4) + 7(4 – 0) + 8(0 – 0)]
= ½[12 + 28 + 0]
= (1/2 × 16)
= 8 sq. units
33. AOBC is rectangle whose three vertices A(0, 3), O(0, 0) and B(5, 0). The length of each of its diagonals is
(a) 5 units
(b) 3 units
(c) 4 units
(d) √34 units
Solution
(c) 4 units
A(0, 3), O(0, 0) and B(5, 0) are the three vertices of a rectangle. Let C be the fourth vertex
Then, the length of the diagonal,
(a) p = 4 only
(b) p =  4
(c) p = ± 4
(d) p = 0
Solution
(c) p = ± 4
The given points are A(4, p) and B(1, 0) and AB = 5
Then, (x_{1} = 4, y_{1} = p) and (x_{2 }= 1, y_{2} = 0)
Therefore,
AB = 5