Gujarat Board Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4
Gujarat Board Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4
Gujarat Board Textbook Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4
Question 1.
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, -2) and B(3, 7).
Solution:
Let the line 2x + y – 4 = 0 divide the line segment joining the points A(2, -2) and B(3, 7) in the ratio λ : 1. Let the point of intersection be P. Then,
Question 2.
Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
If the given points are collinear, then the area of the triangle with these points as vertices will be zero.
1/2[x(2 – 0) + 1(0 – y) + 7(y – 2)]= 0
= 1/2 [2x – y + 7y – 14] = 0
= [2x + 6y – 14] = 0
2x + 6y – 14 = 0
x + 3y – 7 = 0
Question 3.
Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).
Solution:
Let the given points be A(6, -6), B(3, -7) and C(3, 3).
Let the centre of the circle be O(x, y).
Then, OA = OB = OC [radii of circle]
(OA)2 = (OB)2 = (OC)2 ……….(1)
From (1), we have .
OA2 = OB2
⇒ (x-6)2 +(y+6)2 = (x-3)2 + (y+ 7)2
⇒ x2 – 12x + 36 + y2 + 12y + 36
⇒ x2 – Gx + 9 + y2 + 14y + 49
⇒ 6x + 2y = 14
⇒ 3x + y = 7
Again, we have
(OB)2 = (OC)2
(y + 7)2 = (y – 3)2
y2 + 49 + 14y = y2 + 9 – 6y
20 y = 9 – 49
20 y = -40
y = -2
Putting the value ofy in (2), we get x = 3.
Hence, centre of a circle = 0(3, -2).
Question 4.
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
Solution:
Let A(-1, 2) and C(3, 2) be the two opposite vertices of a square ABCD. Let B(x, y) be the unknown vertex.
Then, AB = BC
AB2 = BC2
⇒ (x + 1)2 + (y – 2)2 = (x – 3)2 + (y – 2)2
⇒ x2 + 2x + 1 + y2 – 4y + 4
⇒ x2 – 6x + 9 + y2 – 4y + 4
⇒ 8x = 8
x = 1
Also, AB2 + BC2 = AC2
(∴ ∠B = 90º and therefore using Pythagoras Theorem]
(x + 1)2 + (y – 2)2 + (x – 3)2 + (y – 2)2
= (3 + 1)2 + (2 – 2)2
=x2 + 2x + 1 + y2 – 4y + 4 + x2 – 6x + 9 + y2 – 4y + 4 = 16
= 2x2 + 2y2 – 4x – 8y + 2 = 0
= x2 + y2 – 2x – 4y + 1 = 0
[Dividing throughout by 2]
Putting x = 1, we get
1 + y2 – 2 – 4y + 1 = 0
y(y – 4) = 0
y = 0, 4
Hence, the other vertices are (1, 0) and (1, 4).
Question 5.
The Class X students of a secondary school in Krishna Nagar have been allotted a rectangular plot of a land for their gardening
activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 metre from each other. There is triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of the 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 ∠PQR if C is the origin?
Also, calculate the area of the triangle in these cases. What do you observe?
Question 6.
The vertices of a tABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/AC = 1/4 calculate the area of the
ΔADE and compare it with the area of ΔABC.
Solution:
Question 7.
Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ΔABC.
(i) The median from A meets BC at D. 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 do you observe?
(v) If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ΔABC, find the coordinates of the centroid of the triangle.
Solution:
Question 8.
ABCD is a rectangle formed by the points A(- 1, -1), B(-i, 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? JustifSr your answer.
Solution:
We have, following given points
A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1).
Therefore, coordinates of P are
Now,
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