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 |
Other Exercises |
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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 x-axis is |y|.
Here, the point is (-3, 4). So, its distance from x-axis is |4| = 4
3. The point on x-axis 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 x-axis, then
AP = BP ⇒ AP2 = BP2
⇒ (x + 1)2 + (0 + 0)2 = (x – 5)2 + (0 – 0)2
⇒ x2 + 2x + 1 = x2 – 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(x1 = x, y2 = 2), B(x2 = -3, y2 = -4) and C(x3 = 7, y3 = -5) be collinear points. Then
x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 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 (x1 = 5, y1 = 0), B(x2 = 8, y2 = 0) and C(x3 = 8, y3 = 4) be the vertices of the triangle.
Then
Area(△ABC) = ½[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= 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.a2b2
(d) 1/2b2
Solution
(b) 1/2ab
Let A(x1 = a, y1 = 0), O(x2 = 0, y2 = 0) and B(x3 = 0, y3 = b) be the given vertices. So
Area (△ABO) = ½|x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)|
= ½|a(0 – b) + 0(b – 0) + 0(0 – 0)|
= ½|-ab|
= 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
(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 x-axis. Then as per the question
AP = BP ⇒ AP2 = BP2
⇒ (x – 7)2 + (0 – 6)2 = (x – 3)2 + (0 – 4)2
⇒ x2 – 14x + 49 + 36 = x2 + 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 x-axis is
(a) 3 units
(b) 4 units
(c) 5 units
(d) 1 unit
Solution
(b) 4 units
The y-coordinate the distance of the point from the x-axis
Here, the y-coordinate is 4.
21. In what ratio does the x-axis 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 x-axis 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 x-axes 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 y-axis 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 y-axis 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 y-axis, 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, (x1 = -3, y1 = b) and (x2 = 1, y2 = 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)
⇒ (x1 = 6, y1 = 5) and (x2 = 1, y2 = 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, (x1 = -1, y = 0), (x2 = 5, y = -2) and (x3 = 8, y3 = 2)
Let G(x, y) be the centroid of △ABC. Then,
x = 1/3(x1 + x2 + x3)
= 1/3(-1 + 5 + 8)
= 4
and
y = 1/3(y1 + y2 + y3)
= 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) right-angled
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) right-angled
Solution
(d) right-angled
Let P(0, 6), Q(-5, 3) and R(3, 1) be the given points. Then,
Thus, PQ2 + PR2 = QR2
Therefore, △PQR is right-angled.
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, (x1 = 2, y1 = 3), (x2 = 5, y2 = k) and (x3 = 6, y3 = 7).
Point A, B and C are collinear. Then,
x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 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, (x1 = 1, y1 = 2), (x2 = 0, y2 = 0) and (x3 = a, y3 = b)
Point A, O and C are collinear
⇒ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 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, (x1 = 3, y1 = 0), (x2 = 7, y2 = 0) and (x3 = 8, y3 = 4)
Therefore,
Area of △ABC = ½[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= 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, (x1 = 4, y1 = p) and (x2 = 1, y2 = 0)
Therefore,
AB = 5
