Gujarat Board Solutions Class 10 Maths Chapter 7 Coordinate Geometry
Gujarat Board Textbook Solutions Class 10 Maths 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:
(i) Let the given points be A(2, 3) and B(4, 1). We know that the distance between two points A(x1, y1) and B(x2, y2) is given bý
Question 6.
Name the type of quadrilateral formed, if any, by the following points and give reasons for your answer:
(j) (-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:
(i) Let the given points be A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0).
Then,
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:
We have, P(2, -3), Q( 10, y) and PQ = 10 units.
Now, PQ2 = (10)2 = 100
(10 – 2)2 + [(y – (-3)]2= 100
(8)2 + (y + 3)2 = 100
64+y2 + 6y + 9 = 100
y2 + 6y – 27 = 0
y2 + 9y – 3y – 27 = o
y(y + 9) -3 (y + 9) = 0
(y + 9)(y – 3) = 0
y + 9 = 0
or y – 3 = 0
y = -9
or y = 3
y = -9, 3
Hence, the required value of y is -9 or 3.
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:
We have, P(5, 3) and R(x, 6) and Q(0, 1). It is also given that,
PQ = RQ
PQ2 = RQ2
(0 – 5)2 + [1 – (3)]2
= (0 – x)2 + (1 – 6)2
25 + 16 = x2 + 25
x2 = 16
x = ±4
Therefore, co-ordinates of R are R(±4, 6).
Question 10.
Find the relation between x andy such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Solution:
Let P(x, y), A(3, 6) and B(-3, 4) be the given points.
It is given that,
AP = BP
Ex 7.2
Question 1.
Find the coordinates of the point which divides the line segment joining the points (-1, 7) and (4, -3) in the ratio 2: 3.
Solution:
Let the given points be A(-1, 7) and B(4, -3).
Here, we have
x1 = -1, y1 = 7
x2 = 4, y2 = -3
and m1 = 2, m2 = 3
Let the required point be P(x, y).
Question 2.
Find the coordinates of the points of tri-section of the line segment joining (4, -1) and (-2, -3).
Solution:
Let P and Q be the points of the tri-section of AB.
Then,
AP = PQ = QB = 1
Case I: Here P divides AB in the ratio 1 : 2.
So, we have
x1 = 4, y1 = -1
x2 = -2, y1 = -3
and m1 = 1, m2 = 2
∴ The coordinates of P are given by
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 1 m from each other along AD, as shown in Figure. Niharika runs 1/4th the distance AD on the 2 line and posts a green flag. Preet runs 1/5th distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Question 4.
Find the ratio in which the segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Let P(-1, 6) divides the line segment joining the points A(-3, 10) and B(6, -8) in the ratio k: 1.
Then the coordinates of P are given by –
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 coordinates of the point of divisions.
Solution:
Let the given points be A(1, -5) and B(-4, 5).
Let the x-axis cuts AB at the point P in the ratio k :1.
Then, the coordinates of P are given as
Here, we have
x1 = 1, y1 = -5
x2 = -4 y2 = 5
and m1 = k, m1 = 1
So, coordinates of P are
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 A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram.
Question 7.
Find the coordinates of point A, where AB is the diameter of a circle whose center is (2, -3) and B is (1, 4).
Solution:
Let the given point be A(x, y). Since C is the mid-point of AB.
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 on the line segment AB.
Solution:
Question 9.
Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Solution:
Let P, Q and R be the three points that divide the line segment joi fling the points A(-2, 2) and (2, 8) in four equal parts.
Case I: For point P, we have
Question 10.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.
Solution:
Let the given points are A(3, 0), B(4, 5), C(-1, 4)
and D(-2, -1).
We have,
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 the given point be A(2, 3), B(-1, 0) and C(2, -4).
Here, we have
x2 = 2 ,y1= 3
x2 = -1, y2 = 0
and x3 = 2, y3 = -4
Now, area of ΔABC
= 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= 1/2 [2[(0) – (-4)1 +(-1)(-4 -3) + 2(3 – 0)1
= 1/2(8 + 7 + 6) = 1/2 x 21
= 21/2 sq. units
(ii) Let the given points be A(-5, -1), B(3, -5) and C(5, 2).
Here, we have
x1 = -5, y1 = -1
x2 = 3, y2 = -5
and x3 = 5, y3 = 2
Now, area of ΔABC
= 1/2[x1(y1 – y1) + x2(y3 – y1) + x3(y1 – y2)]
= 1/2(-5) [(-5-2)] +(3) [2-(-1)] + (5)[(-1) – (-5)]
= 1/2 [35 + 9 + 20]
= 32 sq.units
Question 2.
In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, -2), (5, 1), (3, k)
(ii) (8, 1), (k, -4), (2, -5)
Solution:
(i) Let the given points be A(7, -2), B(5, 1) and C(3, k).
Here, we have,
x1 = 7 y1 = -2
x2 = 5, y2 = 1
and x3 = 3, y3 = k
Now, area of ΔABC
= 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= 1/2 7[(1 -k)] + 5[k -(-2)] + 31(-2 – 1)]
= 1/2 [7 – 7k + 5k + 10 – 9]
= 1/2 [8 – 2k] = 4 – k
If the points are collinear, then area of the triangle = 0
⇒ 4 – k = 0
⇒ k = 4.
(ii) Let the given point be A(8, 1), B(k, -4) and C(2, -5).
Here, we have
x1 = 8, y1 = 1
x2 = k, y2 = -4
and x3 = 2, y3 = -5
Now, area of ΔABC
= 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= 1/2 [8 – 6k + 10]
= [18 – 6k] = 9 – 3k
If the points are collinear, then area of the triangle = 0
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 area of the given triangle.
Solution:
Let A(0, -1), B(2, 1) and C(0, 3) be the vertices of ABC. Let D, E, F be the mid-points of sides BC, CA and AB respectively.
Then, the coordinates of D, E and F are (1, 2), (0, 1) and (1, 0) respectively.
Now, Area of ΔABC
= 1/2 [x1(y1 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= 1/2 [0(1 – 3) +2[3 -(-1)] + 0(-1-1)]
= 1/2 [0 + 8 + 0] = 4 sq. units
Area of ΔDEF
= 1/2 [x1(y2 -y3) + x2(y3 – y1) + x3(y1 – y2)]
Area of ΔDEF
= [1(1 – 0) + 0(0 – 2) + [(2 – 1)]
= Area of ADEF = 1/2 [1 + 1]
= 1 sq. unit
∴ Area of DEF : Area of ABC = 1 : 4.
Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, 5), (3, -2) and (2, 3). [National Olympiad]
Solution:
Let the vertices of quadrilateral be A(-4, -2), B(-3, -5), C(3, -2) and D(2, 3).
Area of ΔABC
= 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Here, we have
x1 = -4, y1 = -2
x2= -3, y2 = -5
x3 = 3, y3 = -2
Now, area of ΔABC
= 1/2 [(-4(-5 + 2) + (-3)(-2 + 2) + 3(-2 + 5)]
= 1/2 [(-4 x -3) + (-3 x 0) + (3 x 3)]
= 1/2 [12 + 0 + 9] = 1/2 x 21
= 21/2 sq. units,
We know that, area of ΔACD
= [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Here, we have
x1 = -4, y2 = -2
x2 = 3, y2 = -2
and x3 = 2, y3= 3
Question 5.
You have studied in Class IX, that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A(4, -6), B(3, -2) and C(5, 2).
Solution:
According to the question, AD is the median of ΔABC, therefore D is the mid-point of BC.
For ΔABD,
[Let, x1 = 4, y1 = -6, x2 = 3, y2 = -2, x3 = 4, y3 = 0]
= [-8 + 18 – 16] = (-6) = -3
= 3 sq. units [∴ Area of triangle is positive.]
Area of ADC = Area of AED
Hence, the median of the triangle divides it into two triangles of equal areas.
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,
