PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4
PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4
PSEB 10th Class Maths Solutions Chapter 7 Coordinate Geometry 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.
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