NCERT Exemplar Problems Class 10 Maths Chapter 7 Coordinate Geometry

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NCERT Exemplar Problems Class 10 Maths Chapter 7 Coordinate Geometry

Exercise 7.1 Multiple Choice Questions (MCQs)

Question 1:
The distance of the point P(2, 3) from the X-axis is
(a) 2                           (b) 3                           (c) 1                           (d) 5
Solution:
(b) We know that, if (x, y) is any point on the cartesian plane in first quadrant.
Then, x = Perpendicular distance from Y-axis
and y = Perpendicular distance from X-axis

Distance of the point P(2, 3) from the X-axis = Ordinate of a point P(2, 3)= 3.

Question 2:
The distance between the points A(0, 6) and 5(0,- 2) is
(a) 6                        (b) 8                             (c) 4                           (d) 2
Solution:
(b) v Distance between the points (x₁, y₁) and (x₂, y₂),

Question 3:
The distance of the point P(- 6, 8) from the origin is
(a) 8                          (b) 2√7                      (c) 10                         (d) 6
Solution:
(c) ∴Distance between the points (x_1,y₂)and (x₂, y₂)

Question 4:
The distance between the points (0, 5) and (- 5, 0) is
(a) 5                          (b) 5√2                      (c)2√5                       (d) 10
Solution:
(b) ∴ Distance between the points (x₁,y₁) and (x₂, y₂),

Question 5:
If AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0), then the length of its diagonal is
(a) 5                          (b) 3                           (c) √34                      (d) 4
Solution:
(c)

Now, length of the diagonal AB = Distance between the points A(0, 3) and B(5, 0).
∴ Distance between the points (x,, y,) and (x₂, y₂),

Hence, the required length of its diagonal is √34.

Question 6:
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(a) 5                                        (b) 12                                       (c)11                                          (d)7+√5
Solution:
(b) we Further, adding all the distance of a triangle to get the perimeter of a triangle.We plot the vertices of a triangle i.e., (0, 4), (0,0) and (3,0) on the paper shown as given below

Now,perimeter of ΔAOB=Sum of the length of all its sides = d(AO) + d(OB) + d(AB)
∴ Distance between the points (x₁,y₁) and (x₂, y₂),

Hence, the required perimeter of triangle is 12.

Question 7:
The area of a triangle with vertices A(3,0), B(7, 0) and C(8, 4) is
(a) 14                     (b) 28                         (c) 8                           (d) 6
Solution:
(c) Area of Δ ABC whose Vertices A≡(x₁,y₁),B≡(x₂,y₂) and C≡(x₃, y₃) are given by

Hence, the required area of AABC is 8.

Question 8:
The points (- 4, 0), (4, 0) and (0, 3) are the vertices of a
(a) right angled triangle                            (b) isosceles triangle
(c) equilateral triangle                               (d) scalene triangle
Solution:
(b) Let A(- 4, 0), B(4, 0), C(0, 3) are the given vertices.
Now, distance between A (-4, 0) and B (4, 0),

Hence, ΔABC is an isosceles triangle because an isosceles triangle has two sides equal.

Question 9:
The point which divides the line segment joining the points (7, – 6) and (3, 4) in ratio 1: 2 internally lies in the
(a) I quadrant                    (b) II quadrant                    (c) III quadrant                           (d) IV quadrant
Solution:
(d) If P(x, y) divides the line segment joining A(x₁,y₂) and B(x₂, y₂) internally in the ratio

Question 10:
The point which lies on the perpendicular bisector of the line segment joining the points A(-2, – 5) and B(2, 5) is
(a) (0,0)                     (b) (0, 2)                    (c) (2, 0)                    (d)(-2,0)
Solution:
(a) We know that, the perpendicular bisector of the any line segment divides the^jjpe segment into two equal parts i.e., the perpendicular bisector of the line segment always passes through the mid-point of the line segment.
Mid-point of the line segment joining the points A (-2, -5) and S(2, 5)

Hence, (0, 0) is the required point lies on the perpendicular bisector of the lines segment.

Question 11:
The fourth vertex D of a parallelogram ABCD whose three vertices are A(- 2, 3), B(6, 7) and C(8, 3) is
(a) (0,1)                           (b) (0,-1)                         (c) (-1,0)                                (d) (1,0)
Solution:
(b) Let the fourth vertex of parallelogram, D≡(x_{4 ,}y₄) and L, M be the middle points of AC and BD, respectively,

Since, ABCD is a parallelogram, therefore diagonals AC and BD will bisect each other. Hence, L and M are the same points.

Hence, the fourth vertex of parallelogram is D s (x₄, y₄) s (0,-1).

Question 12:
If the point P(2,1) lies on the line segment joining points A(4, 2) and 6(8, 4), then
(a)AP =  AB                              (b) AP = PB                            (c)PB =  AB                      (d)AP = AB
Solution:
(d) Given that, the point P(2,1) lies on the line segment joining the points A(4,2) and 8(8, 4), which shows in the figure below:

Hence, required condition is AP =

Question 13:
If P(,4) is the mid-point of the line segment joining the points Q(- 6,    5) and fl(- 2, 3), then the value of a is
(a)-4                           (b) -12                        (c) 12                        (d) -6
Solution:
(b) Given that, P(,4) is the mid-point of the line segment joining the points Q(-6, 5) and
R (-2, 3), which shows in the figure given below


Hence, the required value of a is -12.

Question 14:
The perpendicular bisector of the line segment joining the points A(1,5) and 8(4,6) cuts the y-axis at
(a) (0,13)                  (b) (0,-13)                 (c) (0,12)                  (d) (13,0)
Solution:
(a) Firstly, we plot the points of the line segment on the paper and join them.

We know that, the perpendicular bisector of the line segment AB bisect the segment AB, i.e., perpendicular bisector of line segment AB passes through the mid-point of AB.

Now, we draw a straight line on paper passes through the mid-point P. We see that the perpendicular bisector cuts the Y-axis at the point (0,13).
Hence, the required point is (0,13).
Alternate Method
We know that, the equation of line which passes through the points (x₁, y₁) and (x₂, y₂) is

Also, we know that the perpendicular bisector of the line segment is perpendicular on the line segment.
Let slope of line segment is m₂.
From Eq. (iii),

So, the required point is (0,13).

Question 15:
The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the figure is

Solution:
(a) Let the coordinate of the point which is equidistant from the three vertices 0(0, 0), A(0,2y) and B(2x, 0) is P(h,k).
Then,                                     PO = PA = PB
⇒                                           (PO)² = (PA)²= (PB)²                                                        … (i)
By distance formula,

Question 16:
If a circle drawn with origin as the centre passes through (,0), then the point which does not lie in the interior of the circle is

Solution:

Question 17:
A line intersects the y-axis and X-axis at the points P and Q, respectively. If (2, – 5) is the mid-point of PQ, then the coordinates of P and Q are, respectively                                  ’
(a) (0,-5) and (2, 0)                                    (b) (0, 10) and (- 4, 0)
(c) (0, 4) and (- 10, 0)                                (d) (0, – 10) and (4, 0)
Solution:
(d) Let the coordinates of P and 0 (0, y) and (x, 0), respectively.

So, the coordinates of P and Q are (0, -10) and (4, 0).

Question 18:
The area of a triangle with vertices (a, b + c) , (b, c + a) and (c, a + b) is
(a) (a + b + c)²                       (b) 0                              (c) (a + b + c)                           (d) abc
Solution:
(b) Let the vertices of a triangle are, A ≡ (x₁, y₁) ≡ (a, b + c)
B ≡ (x₂, y₂) ≡ (b,c + a) and C = (x₃, y₃) ≡ (c, a + b)

Hence, the required area of triangle is. 0.

Question 19:
If the distance between the points (4, p) and (1, 0) is 5, then the value of pis
(a) 4 only                   (b) ±4                        (c) – 4 only                (d) 0
Solution:
(b) According to the question, the distance between the points (4, p) and (1, 0) = 5

Hence, the required value of p is ± 4,

Question 20:
If the points A(1, 2), B(0, 0) and C(a, b) are collinear, then
(a) a = b                      (b) a = 2b                  (c) 2a = b                   (d) a = – b
Solution:
(c) Let the given points are B = (x₁,y₁) =  (1,2),
B = (x₂,y₂) = (0,0) and C3 = (x₃,y₃)= (a, b).

Hence, the required relation is 2a = b.

Exercise 7.2 Very Short Answer Type Questions

Question 1:
Δ ABC with vertices A(0 ,- 2,0), B(2, 0) and C(0,2) is similar to ΔDEF with vertices D (- 4, 0), E(4, 0) and F(0, 4).
Solution:
True

Hence, both the triangles are similar.                                                                                                                                [by SSS rule]

Question 2:
The point P(- 4, 2) lies on the line segment joining the points A(- 4, 6) and B(- 4, – 6).
Solution:
True
We plot all the points P(-4,2),  A(-4, 6) and B(-4, – 6) on the graph paper,

From the figure, point P(- 4,2) lies on the line segment joining the points A(- 4,6) and B(- 4, – 6),

Question 3:
The points (0, 5), (0, -9) and (3, 6) are collinear.
Solution:
False

If the area of triangle formed by the points (0, 5), (0 – 9) and (3, 6) is zero, then the points are collinear.
Hence, the points are non-collinear.

Question 4:
Point P(0, 2) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points A(-l, 1) and B(3, 3).
Solution:
False
We know that, the points lies on perpendicular bisector of the line segment joining the two
points is equidistant from these two points.


Hence, the required intersection point is (0, 4).

Question 5:
The points A(3,1), B(12, – 2) and C(0, 2) cannot be vertices of a triangle.
Solution:
True

Hence, the points A(3,1), B(12, -2) and C(0,2) are collinear.
So, the points A(3,1), B(12, – 2) and C(0,2) cannot be the vertices of a triangle.

Question 6:
The points A(4, 3), B(6, 4), C(5, – 6) and D(- 3, 5) are vertices of a parallelogram.
Solution:
False

In parallelogram, opposite sides are equal. Here, we see that all sides AB, BC, CD and DA are different.
Hence, given vertices are not the vertices of a parallelogram.

Question 7:
A circle has its centre at the origin and a point P (5, 0) lies on it. The point Q (6, 8) lies outside the circle.
Solution:
True
First,we draw a circle and a point from the given information

We know that, if the distance of any point from the centre is less than/equal to/ more than the radius, then the point is inside/on/outside the circle, respectively.
Here, we see that, OQ > OP
Hence, it is true that point Q (6, 8), lies outside the circle.

Question 8:
The point A (2, 7) lies on the perpendicular bisector of the line segment joining the points P (5, – 3) and Q (0, – 4).
Solution:
False
If A (2,7) lies on perpendicular bisector of P(6,5) and Q (0, – 4), then AP=AQ

So, A does not lies on the perpendicular bisector of PQ.
Alternate Method
If the point A (2,7) lies on the perpendicular bisector of the line segment, then the point A satisfy the equation of perpendicular bisector.
Now, we find the equation of perpendicular bisector. For this, we find the slope of perpendicular bisector.

Hence, the point A (2,7) does not lie on the perpendicular bisector of the line segment.

Question 9:
The point P(5, – 3) is one of the two points of trisection of line segment joining the points A(7, – 2) and B(1, – 5).
Solution:
True
Let P (5,-3) divides the line segment joining the points A (7,-2) and B (1 ,-5) in the ratio k: 1 internally.
By section formula, the coordinate of point P will be

So the point P divides the line segment AB in ratio 1: 2. Hence, point P in the point of trisection of AB.

Question 10:
The points A (-6,10), B(- 4, 6) and C(3, -8) are collinear such that
AB = AC.
Solution:
True
If the area of triangle formed by the points (x₁,y₂), (x₂, y₂) and (x₃, y₃) is zero, then the points are collinear,

which is the required relation.

Question 11:
The point P(- 2, 4) lies on a circle of radius 6 and centre (3, 5).
Solution:
False
If the distance between the centre and any point is equal to the radius, then we say that point lie on the circle.
Now, distance between P (-2,4) and centre (3, 5)

which is not equal to the radius of the circle.
Hence, the point P(-2, 4) does not lies on the circle.

Question 12:
The points A(- 1, – 2), B (4, 3), C (2, 5) and D (- 3, 0) in that order form a rectangle.
Solution:
True

Since, diagonals AC and BD are equal.
Hence, the points A (-1, – 2), B (4, 3), C (2, 5)and D (- 3 0)form a rectangle.

Exercise 7.3 Short Answer Type Questions

Question 1:
Name the type of triangle formed by the points A (-5, 6), B(- 4, – 2) and C(7, 5).
Solution:
To find the type of triangle, first we determine the length of all three sides and see whatever condition of triangle is satisfy by these sides.
Now, using distance formula between two points,

We see that,                                    AB ≠ BC ≠ CA
and not hold the condition of Pythagoras in a ΔABC.
i.e.,                        (Hypotenuse)² =( Base)² + (Perpendicular)²
Hence, the required triangle is scalene because all of its sides are not equal i.e., different to each other.

Question 2:
Find the points on the X-axis which are at a distance of 2√5 from the point (7, -4). How many such points are there?
Solution:
We know that, every point on the X-axis in the form (x, 0). Let P(x, 0) the point on the X-axis
have 2√5 distance from the point Q (7, – 4).

Hence, there are two points lies on the axis, which are (5, 0) and (9, 0), have 2V5 distance from the point (7, – 4).

Question 3:
What type of quadrilateral do the points A (2, -2), B (7, 3) C (11, – 1) and D (6, – 6) taken in that order form?
Solution:
To find the type of quadrilateral, we find the length of all four sides as well as two diagonals and see whatever condition of quadrilateral is satisfy by these sides as well as diagonals. Now, using distance formula between two points,

Here, we see that the sides AB = CD and BC = DA
Also, diagonals are equal i.e., AC = BD
which shows the quadrilateral is a rectangle.

Question 4:
Find the value of a, if the distance between the points A(- 3, – 14) and B (a, – 5) is 9 units.
Solution:
According to the question,
Distance between A (- 3, -14) and 8 (a, – 5), AB = 9

Hence, the required value of a is – 3.

Question 5:
Find a point which is equidistant from the points A (- 5, 4) and B (- 1, 6). How many such points are there?
Solution:
Let P (h, k) be the point which is equidistant from the points A (- 5, 4) and B (-1, 6).

So, the mid-point of AB satisfy the Eq. (i). Hence, infinite number of points, in fact all points which are solution of the equation 2h + k +1 = 0, are equidistant from the points A and B.
Replacing h, k by x, y in above equation, we have 2 x +y+1= 0

Question 6:
Find the coordinates of the point Q on the x-axis which lies on the perpendicular bisector of the line segment joining the points A (- 5, – 2) and B (4, – 2). Name the type of triangle formed by the point Q, A and B.
Solution:
Firstly, we plot the points of the line segment on the paper and join them.

We know that, the perpendicular bisector of the line segment AB bisect the segment AB, i.e.,perpendicular bisector of the line segment AB passes through the mid-point of AB.

Alternate Method                                                                                              •
(i) To find the coordinates of the point of 0 on the X-axis. We find the equation of perpendicular bisector of the line segment AS.
Now, slope of line segment AB,

To know the type of triangle formed by the points Q, A and B. We find the length of all three sides and see whatever condition of triangle is satisfy by these sides.
Now, using distance formula between two points,

which shows that the triangle formed by the points Q, A and 6 is an isosceles.

Question 7:
Find the value of m, if the points (5,1), (- 2, – 3) and (8, 2m) are collinear.
Solution:
Let A ≡ (x₁,y₁) s (5,1), B = (x₂, y₂) = (- 2, – 3), C s (x₃, y₃) = (8,2m)
Since, the points A ≡ (5,1), B ≡ (- 2, – 3) and C ≡ (8,2m) are collinear.

Question 8:
If the point A(2, – 4) is equidistant from P(3, 8) and Q(- 10, y), then find the value of y. Also, find distance PQ.
Solution:
According to the question,
A (2, – 4) is equidistant from P (3, 8) = 0 (-10, y) is equidistant from A (2, – 4)

Hence, the values of y are -3, – 5 and corresponding values of PQ are √290 and √338 = 1342, respectively.

Question 9:
Find the area of the triangle whose vertices are (-8,4), (-6,6) and (- 3, 9).
Solution:
Given that, the vertices of triangles

Hence, the required area of triangle is 0.

Question 10:
In what ratio does the X-axis divide the line segment joining the points (- 4, – 6) and (- 1, 7)? Find the coordinates of the points of division.
Solution:
Let the required ratio be λ: 1. So, the coordinates of the point M of division A (- 4, – 6) and B(-1,7) are

Question 11:
Find the ratio in which the point P(,) divides the line segment joinnig the points A(,) and B(2,5).
Solution:
Let P(,) divide AB internally in the ratio m:n
using the section formula ,we get


Hence, the required ratio is 1 : 5.

Question 12:
If P(9a -2, – b) divides line segment joining A(3a + 1,-3) and B(8a, 5) in the ratio 3 : 1, then find the values of a and b.
Solution:
Let P(9a – 2, – b) divides AS internally in the ratio 3:1.
By section formula,

Hence, the required values of a and b are 1 and – 3.

Question 13:
If (a, b) is the mid-point of the line segment joining the points A(10, – 6), B(k, 4) and a – 2b= 18, then find the value of k and the distance AB.
Solution:
Since, (a, b) is the mid-point of line segment AB.

Hence, the required distance of AB is 2√61.

Question 14:
If the centre of a circle is (2a, a – 7), then Find the values of a, if the  circle passes through the point (11, – 9) and has diameter 10√2 units.
Solution:
By given condition,

Distance between the centre C(2a, a – 7) and the point P( 11, – 9), which lie on the circle = Radius of circle

Hence, the required values of a are 5 and 3.

Question 15:
The line segment joining the points A(3, 2) and B(5,1) is divided at the point P in the ratio 1 : 2 and it lies on the line
3x – 18y + k = 0. Find the value of k.
Solution:
Given that, the line segment joining the points 4(3,2) and 6(5,1) is divided at the point P in the ratio 1 : 2.

Hence, the required value of k is 19.

Question 16:
If D(,) E(7,3) and  F(,) are the mid-points of sides of ΔABC,then find the area of the ΔABC.
Solution:

Question 17:
If the points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a ABC right angled at B, then find the values of a and hence the area of ΔABC.
Solution:
Given that, the points A (2, 9), B(a, 5) and C(5, 5) are the vertices of a ΔABC right angled at B.
By Pythagoras theorem, AC² = AB² + BC²                                                                                                                                          ,,.(i)

Hence, the required area of ΔABC is 6 sq units.

Question 18:
Find the coordinates of the point R on the line segment joining the points P(- 1, 3) and Q(2, 5) such that
PR = PQ.
Solution:
According to the question,

Question 19:
Find the values of k, if the points A(k + 1, 2k), B(3k, 2k + 3) and C (5k – 1, 5k) are colli near.
Solution:
We know that, if three points are collinear, then the area of triangle formed by these points is zero.
Since, the points A(k + 1,2k), B(3k, 2k + 3) and C(5k -1, 5k) are collinear.
Then, area of ΔABC = 0


Hence, the required values of k are 2 and

Question 20:
Find the ratio in which the line 2x+ 3y – 5 = 0 divides the line segment joining the points (8, – 9) and (2,1). Also, find the coordinates of the point of division.
Solution:
Let the line 2x + 3y – 5 = 0 divides the line segment joining the points A (8, – 9) and B (2,1) in the ratio λ: 1 at point P.

Exercise 7.4 Long Answer Type Questions

Question 1:
If (- 4, 3) and (4, 3) are two vertices of an equilateral triangle, then find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.
Solution:
Let the third vertex of an equilateral triangle be (x, y). Let A (- 4, 3), B(4 3) and C (x, y).
We know that, in equilateral triangle the angle between two adjacent side is 60 and all three sides are equal.

Question 2:
A(6,1), B (8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the mid-point of DC, then find the area of ΔADE.
Solution:
Given that, A (6,1), B (8,2) and C (9, 4) are three vertices of a parallelogram ABCD.
Let the fourth vertex of parallelogram be (x, y).
We know that, the diagonals of a parallelogram bisect each other.


Hence, the required area of ΔADE is sq units.

Question 3:
The points A(x₁, y₁), B(x₂ y₂) and C(x₃, y₃) are the vertices of ΔABC.
(i)   The median from A meets BC at Find the coordinates of the point D.
(ii)  Find the coordinates of the point P on AD such that AP : PD = 2:1
(iii) Find the coordinates of points Q and R on medians BE and CF, respectively such that BQ:QE = 2:1 and CR:RF = 2:1
(iv ) What are the coordinates of the centroid of the ΔABC?
Solution:
Given that, the points A (x_1, y₁), B(x₂, y₂)andC (x₃, y₃)are the vertices of ΔABC.
(i) We know that, the median bisect the line segment into two equal parts i.e., here D is the mid-point of BC.

Question 4:
If the points A (1, – 2), B (2, 3), C(a, 2) and D(- 4, – 3) form a parallelogram, then find the value of a and height of the parallelogram taking AB as base.
Solution:
In parallelogram, we know that, diagonals are bisects each other i.e., mid-point of AC = mid-point of BD

So, the required value of a is – 3.
Given that, AS as base of a parallelogram and drawn a perpendicular from D to AS which meet AS at P. So, DP is a height of a parallelogram.

Question 5:
Students of a school are standing in rows and columns in their playground for a drill practice. A, B, C and D are the positions of four students as shown in figure. Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students A, B, C and D? If so, what should be his position?

Solution:
Yes, from the figure we observe that the positions of four students A, B, C and D are (3, 5), (7, 9), (11, 5) and (7,1) respectively i.e., these are four vertices of a quadrilateral. Now, we will find the type of this quadrilateral. For this, we will find all its sides.
We see that, AB = BC = CD = DA i.e., all sides are equal.

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Hence, the required position of Jaspal is (7, 5).

Question 6:
Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distance covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13,14) and office at (13, 26) and coordinates are in km.
Solution:

So, extra distance travelled by Ayush in reaching his office = 27 – 24.6 = 2.4 km Hence, the required extra distance travelled by Ayush is 2.4 km.

All Chapter NCERT Exemplar Problems Solutions For Class 10 maths

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All Subject NCERT Exemplar Problems Solutions For Class 10

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