PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

 PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

PSEB 10th Class Maths Solutions Chapter 7 Coordinate Geometry

Ex 7.1

Question 1.
Find the distance between the following pairs of points:
(i) (2, 3); (4, 1)
(ii)(-5, 7); (-1, 3)
(iii) (a, b); (-a, -b).
Solution:

Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B
discussed in section 7.2.
Solution:

  

Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in fig. Champa and Charnel walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square”? Chameli disagrees. Using distance formula, find which of them is correct, and why?

Solution:
In the given diagram, the vertices of given points are : A (3, 4); B (6, 7); C (9, 4) and D (6, 1).
Now,

Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) ( 1,- 2), (1, 0),(- 1, 2), (- 3, 0)
(ii) ( 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2).
Solution:

From above discussion, it is clear that
AB = BC = CD = DA = √8 and AC = BD = 4.
Hence, given quadrilateral ABCD is a square.

 

From above discussion, it is clear that AB = CD and BC = DA. and AC ≠ BD.
i.e., opposite sides are equal but their diagonals are not equal.
Hence, given quadrilateral ABCD is a parallelogram.

Question 7.
Find the points on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Solution:
Let required point be P (x, 0) and given points be A (2, – 5) and B (- 2, 9).
According to question,
PA = PB
(PA)2 = (PB)2
or (2 – x)2 + (- 5- 0)2 = (- 2 – x)2 + (9 – 0)2
or 4 + x2 – 4x + 25 = 4 + x2+ 4x + 81
-8x = 56
x = 4/4 = – 7
Hence, required point be (- 7, 0).

Question 8.
Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.
Solution:

Question 9.
If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distances QR and PR.
Solution:

 

Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Solution:

Ex 7.2

Question 1.
Find the co-ordinates of the point which divides the join (- 1, 7) and (4, – 3) in the ratio 2 : 3.
Solution:
Let required point be P (x, y) which divides the join of given points A (- 1, 7)
and B (4, – 3) in the ratio of 2 : 3.
(-1, 7) (x, y) (4, – 3)

Question 2.
Find the co-ordinates of the points of trisection of the line segment joining (4, – 1) and (2, – 3).
Solution:
Let P (x1, y1) and Q (x2, y2) be the required points which trisect the line segment joining A (4, – 1)and B (- 2, – 3) i.e., P(x1, y1) divides AB in ratio 1: 2 and Q divides AB in ratio 2 : 1.

Question 3.
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in fig. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag.

Preet runs 1/5 th the distance AD on the eighth line and posts a red flag. What is the distance betweenboth the flags? If Rashmi has to post a blue flag exactly half way between the line (segment) joining the two flags, where should she post her flag?

Question 4.
Find the ratio in which (he segment joining the points (- 3, 10) and (6, – 8) is divided by (- 1, 6).
Solution:
Let point P (- 1, 6) divides the line segment joining the points A (- 3, 10) and B (6, – 8) the ratio K : 1.

Question 5.
Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also find the co
ordinates of the point of division.
Solution:
Let required point on x-axis is P (x, 0) which divides the line segment joining the points A (1, – 5) and B (- 4, 5) in the
ratio K : 1.

Question 6.
If (1, 2); (4, y); (x, 6) and (3, 5)are the vertices of a parallelogram taken in order, find x and y.
Solution:
Let points of parallelogram ABCD are A (1, 2) (4, y) ; C (x, 6) and D (3, 5)
But diagonals of a || gm bisect each other.
Case I. When E is the mid point of A (1, 2) and C (x, 6)
∴ Co-ordinates of E are:

and 4 = 5+y/2
or 8 = 5 + y
or y = 3
Hence, values of x and y are 6 and 3.

Question 7.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).
Solution:
Let, coordinates of A be (x, y). But, centre is the’ niij ioint of the vertices of the diameter.

Question 8.
If A and B are (- 2, – 2) and (2, – 4) respectively, find the coordinates of P such that AP = AB and P lies ¡n the line segment AB.
Solution:

Question 9.
Find the coordinates of the points which divides the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.
Solution:
Let required points are C, D and E which divide the line segment joming the points A (- 2, 2) and B (2, 8) into four equal parts. Then D is mid point of A and B ; C is the mid point of A and D ; E is the mid point of D and B such that
AC = CD = DE = EB

Question 10.
Find the area of a rhombus if the vertices are (3, 0); (4, 5); (- 1, 4) and(- 2, – 1) taken in order.
[Hint: Areas of a rhombus = 1/2 (Product of its diagonals)]
Solution:

Ex 7.3

Question 1.
Find the area of the triangle whose vertices are:
(i) (2, 3); (- 1, 0); (2, – 4)
(ii) (- 5, – 1); (3, – 5); (5, 2)
Solution:
(i) Let vertices of the ∆ABC are A (2, 3); B(- 1, 0) and C (2, – 4)
Here x1 = 2, x2 = – 1 x3 = 2
y1 = 3, y2 = 0, y3 = – 4 .
∴ Area of ∆ABC = 1/2 [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)
= 1/2 [2 × (0 + 4) – 1 × (- 4 – 3) + 2 × (3 – 0)]
= 1/2 [8 + 7 + 6] = 21/2
= 10.5 sq units.

(ii) Let vertices of the ∆ABC are A (- 5, – 1); B (3, – 5) and C (5, 2)
Here x1 = – 5, x2 = 3, x3 = 5
y1 = – 1, y2 = – 5, y3 = 2
∴ Area of ∆ABC = 1/2 [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
= 1/2 [- 5 (- 5 – 2) + 3 (2 + 1) + 5 (- 1 + 5)]
= 1/2 [35 + 9 + 20]
= 1/2 × 64 = 32 sq units.

Question 2.
In each of the following find the value of ‘k’ for which the points are coimear.
(i) (7, – 2); (5, 1); (3, k)
(ii) (8, 1); (k, – 4); (2, – 5)
Solution:
(i) Let given points be A (7, – 2); B (5, 1) and C (5, k)
Here x1 = 7, x2 = 5, x3 = 3
y1 = – 2, y2 = 1 y3 = k
Three points are collinear iff
1/2 [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)] = 0
or 1/2 [7 (1 – k) + 5(k + 2) + 3(- 2 – 1)] = 0
or 7 – 7k + 5k +10 – 9 = 0
or – 2k + 8 = 0
or – 2k = – 8
or – k = −8/−2 = 4 .
Hence k = 4.

ii) Let given points be A (8, 1); B (k, – 4) and C(2, – 5)
Here x1 = 8 x2 = k, x3 = 2
y1 = 1, y = – 4, y = – 5
Three points are collinear iff
1/2 [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)] = 0
or 1/2 [8 (- 4 + 5) + k (- 5 – 1) + 2 (1 + 4) = 0]
or 8 – 6k + 10 = 0
or – 6k = – 18 .
or k = −18/−6 = 3.
Hence k = 3.

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, – 1), (2, 1) and (0, 3). FInd the ratio of the area of the triangle formed to the area of the given triangle.
Solution:
Let vertices of given triangle ABC are A(0, – 1); B (2, 1) and C (0, 3).
Also, D, E, F be the mid points of AB, BC, CA respectively.
Using mid point formula,

 

Question 4.
Find the area of the quadrilateral whose vertices taken in order, are (- 4, – 2); (- 3, – 5); (3, – 2); (2, 3).
Solution:
Let co-ordinates of the given quadrilateral ABCD are A(- 4, – 2); B(-3, – 5); C(3, – 2) and D (2, 3).
Join AC then Quad. ABCD divides in two triangles
i.e. ∆ABC and ∆CDA

 

Question 5.
You have studied in Class IX, (Chapter 9, Q. 3) that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A(4, – 6), B(3, – 2) and C(5, 2).
Solution:
Given that coordinates of the vertices of ∆ABC are A(4, – 6); B (3, – 2) and C (5, 2)
Let CD is the median i.e. D is the mid point of AB which divides AABC into two pails i.e.

In ∆CDB
x = 5, x = 35, x = 3
y = 2, y = – 4, y = – 2
Area of ∆CDB = 1/2 [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
= 1/2 [5 (- 4 + 2) + 3.5 (- 2 – 2) + 3 (2 + 4)]
= 1/2 [- 10 – 14 + 18]
= 1/2 × – 6 = – 3
= 3 sq. units(∵ area cannot be negalive)
From above discussion it is clear that area of ∆ADC = area of ∆CDB = 3 sq. units
Hence, a median of a triangle divides it into two triangles of equal areas.

Ex 7.4

Question 1.
Determine the ratio in which the line it + y – 4 = 0 divides the line segment joining the points A (2, – 2) and B (3, 7).
Solution:
Let line 2x + y – 4 = 0 divides the line segment joining the points A (2,- 2) and B(3, 7) at C (x, y) in the ratio k : 1

Question 2.
Find a relation between x and y if (x, y) ; (1, 2) and (7, 0) are collinear.
Solution:
Let given points are A (x, y); B (1, 2) and C (7, 0).
Here x1 = x, x2 = 1, x3 = 7
y1 = y, y2 = 2, y3 = 0
∵ Three points are collinear
iff 1/2 [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)] = 0
or 1/2 x (2 – 0) + 1 (0 – y) + 7 (y – 2)] = 0
or 2x – y + 7y – 14 = 0
or 2x + 6y – 14 = 0
or x + 3y – 7 = 0 is the required relation.

Question 3.
Find the centre of a cirçle passing through the points (6, —6); (3, —7) and (3,3).
Solution:
Let O (x, y) be the required centre of the circle which passes through points P(6, – 6); Q(3, – 7) and R (3, 3).
∴ radii of circle are equal.

∴ OP = OQ = OR
or (OP)2 = (OQ)2 = (OR)2
Now, (OP)2 = (OQ)2
(x – 6)2 + (y + 6)2 = (x – 3)2 + (y + 7)2
or x2 + 36 – 12x + y2 + 36 + 12y = x2 + 9 – 6x + y2 + 49 + 14y
or – 12x + 12y + 72 = – 6x + 14y + 58
or – 6x – 2y + 14 = 0
or 3x + y – 7 = 0 ………………(1)
Also, (OQ)2 = (OR)2
or (x – 3)2 + (y + 7)2 = (x – 3)2 + (y – 3)2
or (y + 7)2 = (y – 3)2
or y2 + 49 + 14y = y2 + 9 – 6y
or 20y = – 40
y = −40/20 = – 2
Substitute this value of)’ in (1), we get
3x – 2 – 7 = 0
or 3x – 9 = 0
or 3x = 9
or x = 9/3 = 3
∴ Required centre is (3, – 2).

Question 4.
The two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of other two vertices.
Solution:
Let two opposite vertices of a square ACBD are A (- 1, 2) and B (3, 2) and coordinates of C are (x, y)
∵ Length of each sides of square are equal.
∴ AC = BC
or (AC)2 = (BC)2
or (x + 1)2 + (y – 2)2 = (x – 3)2 + (y – 2)2

or (x + 1)2 = (x – 3)2
or x2 + 1 + 2x = x2 + 9 – 6x
or 8x = 8
or x = 8/8 = 1
Now, in rt ∠d ∆ACB,
Using Pythagoras Theorem,
(AC)2 + (BC)2 = (AB)2
(x + 1)2 + (y – 2)2 + (x – 3)2 + (y – 2)2 = (3 + 1)2 + (2 – 2)2
or x2 + 1 + 2x + y2 + 4 – 4y + x2 + 9 – 6x + y2 + 4 – 4y = 16
or 2x2 + 2y2 – 4x – 8y + 2 = 0
or x2 + y2 – 2x – 4y + 1 = 0
Putting the value of x = 1 in (1), we get
(1)2 + y2 – 2 (1) – 4y + 1 = 0
or y2 – 4y = 0
or y (y – 4) = 0
Either y = 0 or y – 4 = 0
Either y = 0 or y = 4
∴ y = 0, 4
∴ Required points are (1. 0) and (1.4).

Question 5.
The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1m from each other. There ¡s a triangular grassy lawn in the plot as shown in the Fig. The students are to sow seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of A PQR if C is the origin? Also calculate the areas of the triangles In these cases. What do you observe?
Solution:
Case I:
When taking A as origin then AD is X-axis and AB is Y-axis.
∴ Coordinates of triangular grassy Lawn
PQR are P (4, 6); Q (3, 2) and R(6, 5).
Here x1 = 4, x2 = 3, x3 = 6
y1 = 6, y2 = 2, y3 = 50
Now, area of ∆PQR = 1/2 [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
= 1/2 [4 (2 – 5) + 3 (5 – 6) + 6 (6 – 2)]
= 1/2 [- 12 – 3 + 24] = 9/2
= 4.5 sq. units.

Case II: When taking C as origin then CB is X – axis and CD is Y – axis.
∴ Coordinates of triangular grassy lawn PQR
are P(12, 2); Q (13,6) and R (10, 3)
Here x1 = 12, x2 = 13, x3 = 10
y1 = 2, y2 = 6, y3 = 3
Now, area of ∆PQR = 1/2 [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
= 1/2 [12 (6 – 3) + 13 (3 – 2) + 10 (2 – 6)]
= 1/2 [36 + 13 – 40]
= 9/2 = 4.5 sq. units.
From above two cases, it is clear that area of triangular grassy lawn is same.

  

Question 7.
Let (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the potnt 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 do you observe?
[Note : The point which is common to all the three medians ¡s called centroid and this point divides each median in the ratio 2: 1]
(v) if A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle.
Solution:
Given that vertices of ∆ABC are A (4, 2); B (6, 5) and C (1, 4).
(i) AD is the median from the vertex A.
∴ D is the mid point of BC.

 

(iii) Le BE and CF are the medians of ∆ABC to AC and AB respectively.
∴ E and F are mid points of AC and AB respectively.

 

(iv) From above discussion, it is clear that coordinates of P, Q and R are same and coincide at a point, is known as centroid of triangle, which divides each median in the ratio 2: 1.

(v) The vertices of given ∆ABC are
A (x1, y1); B (x2, y2) and C (x3, y3).

Question 8.
ABCD is a rectangle formed by the points A (- 1, – 1), B (- 1, 4), C (5, 4) and D (5, – 1). P, Q R and S are the mid points
of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square ? a rectangle? or a rhombus ? Justify your answer.
Solution:
Given: The vertices ot’ given rectangle ABCD are
A(- 1, – 1); B(- 1, 4); C(5, 4) and D (5, – 1).

  

Form above discussion it is clear that PQ = QR = RS = SP.
Also, PR ≠ QS.
⇒ All sides of quad. PQRS are equal but their diagonals are not equal.
Quad. PQRS is a rhombus.