Gujarat Board GSEB Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Textbook Questions and Answers.
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
AB = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
AB = \(\sqrt{(4-2)^{2}+(1-3)^{2}}\)
= \(\sqrt{4+4}\) = \(\sqrt{8}\) = 2\(\sqrt{4}\)
(ii) Let the given points be A(-5, 7) and B(-1,3) We know that the distance between two points A(-5, 7) and B(-1, 3) is given by
AB = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
AB = \(\sqrt{(-1-(-5))^{2}+(3-(-1))^{2}}\)
= \(\sqrt{16 + 16}\) = \(\sqrt{32}\) = 4\(\sqrt{2}\)
(iii) Let the given points be A(a, b) and B(-a, -b). We know that the distance between two points A(a, b) and B(-a, -b) is given by
AB = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
AB = \(\sqrt{(-a-a)^{2}+(-b-b)^{2}}\)
= \(\sqrt{(-2 a)^{2}+(-2 b)^{2}}\)
= \(\sqrt{(-2 a)^{2}+(-2 b)^{2}}\)
= \(\sqrt{4 a^{2}+4 b^{2}}\)
= \(\sqrt{4\left(a^{2}+b^{2}\right)}\)
= \(2 \sqrt{a^{2}+b^{2}}\)
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:
Let the given points be A(0, 0) and B(36, 15), then
AB = \(\sqrt{(36-0)^{2}+(15-0)^{2}}\)
= \(\sqrt{(36)^{2}+(15)^{2}}\)
= \(\sqrt{1296+225}\)
= \(\sqrt{1521}\) = 39
Yes, we can find the distance between the two towns A and B discussed in section 7.2 and this distance = 39 km.
Question 3.
Determine if the points (1, 5), (2, 3) and (-2, -11) are coimear.
Solution:
Let the given points be A(1, 5), B(2, 3) and C(- 2, -11), then
AB = \(\sqrt{(2-1)^{2}+(3-5)^{2}}\)
⇒ AB = \(\sqrt{(1)^{2}+(-2)^{2}}\)
⇒ AB = \(\sqrt{1+4}\) = \(\sqrt{5}\)
AC = \(\sqrt{(-2-1)^{2}+(-11-5)^{2}}\)
⇒ AC = \(\sqrt{(-3)^{2}+(-16)^{2}}\)
⇒ AC = \(\sqrt{9+256}\)
⇒ AC = \(\sqrt{265}\)
and BC = \(\sqrt{(-2-2)^{2}+(-11-3)^{2}}\)
⇒ BC = \(\sqrt{(-4)^{2}+(-14)^{2}}\)
⇒ BC = \(\sqrt{16+196}\)
⇒ BC = \(\sqrt{212}\)
Here, we see that AB + BC AC, BC + AC AB and AB + AC BC.
Hence, the points A, B and C are not coimear.
Question 4.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:
Let the given points be A(5, -2), B(6, 4) and C(7, -2), then
AB = \(\sqrt{(6-5)^{2}+[4-(-2)]^{2}}\)
= \(\sqrt{(1)^{2}+(6)^{2}}\) = \(\sqrt{1+36}\)
= \(\sqrt{37}\)
BC = \(\sqrt{(7-6)^{2}+(-2-4)^{2}}\)
= \(\sqrt{(1)^{2}+(-6)^{2}}\)
= \(\sqrt{1+36}\) = \(\sqrt{37}\)
Since AB = BC
Therefore, ABC is an isosceles triangle.
Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in figure. Champa and Chameli 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.
Solution:
Let A(3, 4), B(6, 7), C(9, 4) and D(6, 1) be the given points. Then,
AB = \(\sqrt{(6-3)^{2}+(7-4)^{2}}\)
= \(\sqrt{(3)^{2}+(3)^{3}}\) = \(\sqrt{9+9}\)
= \(\sqrt{18}\) = 3\(\sqrt{2}\)
BC = \(\sqrt{(9-6)^{2}+(4-7)^{2}}\)
= .\(\sqrt{(3)^{2}+(-3)^{2}}\) + \(\sqrt{9+9}\)
=\(\sqrt{18}\) = 3\(\sqrt{2}\)
CD = \(\sqrt{(6-9)^{2}+(1-4)^{2}}\)
= \(\sqrt{(-3)^{2}+(-3)^{2}}\)
= \(\sqrt{9+9}\) = \(\sqrt{18}\) = 3\(\sqrt{2}\)
DA = \(\sqrt{(3-6)^{2}+(4-1)^{2}}\)
= \(\sqrt{(-3)^{2}+(3)^{2}}\) \(\sqrt{9+9}\)
= \(\sqrt{18}\) = 3\(\sqrt{2}\)
AC = \(\sqrt{(9-3)^{2}+(4-4)^{2}}\) = 6
BD = \(\sqrt{(6-6)^{2}+(1-7)^{2}}\) = 6
We see that,
AB = BC = CD = DA
and AC = BD = 6
Therefore, ABCD is a square.
Hence, Champa is correct.
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,
AB = \(\sqrt{[1-(-1)]^{2}-[0-(-2)]^{2}}\)
= \(\sqrt{(1+1)^{2}+(-2)^{2}}\)
= \(\sqrt{(2)^{2}+(-2)^{2}}\)
= \(\sqrt{4+4}\) = \(\sqrt{8}\)
BC = \(\sqrt{(-1-1)^{2}+(2-0)^{2}}\)
= \(\sqrt{(-2)^{2}+(2)^{2}}\)
= \(\sqrt{4+4}\) = \(\sqrt{8}\)
CD = \(\sqrt{[3-(-1)]^{2}+(0-2)^{2}}\)
= \(\sqrt{(-3+1)^{2}+(-2)^{2}}\)
= \(\sqrt{(-2)^{2}+(-2)^{2}}\)
= .\(\sqrt{4+4}\) = \(\sqrt{8}\)
and DA = \(\sqrt{[-1-(-3)]^{2}+(-2-0)^{2}}\)
= \(\sqrt{(-1+3)^{2}+(-2)^{2}}\)
= \(\sqrt{(2)^{2}+(-2)^{2}}\)
= \(\sqrt{4+4}\) = \(\sqrt{8}\)
Thus, we have
AB = BC = CD = DA
⇒ All sides are equal Also,
AC = \(\sqrt{[-1-(-1)]^{2}+[2-(-2)]^{2}}\)
= \(\sqrt{(-1+1)^{2}+(2+2)^{2}}\)
= \(\sqrt{0+(4)^{2}}\) = \(\sqrt{16}\) = 4
and BD = \(\sqrt{(-3-1)^{2}+0}\)
= \(\sqrt{(-4)^{2}}\) = \(\sqrt{16}\) = 4
Since, all four sides of the quadrilateral are equal and diagonals are also equal, sothe given points form a square.
(ii) Let the given points be A(-3, 5), B(3, 1), C(0, 3) and D(-1, -4).
Then, AB = \(\sqrt{[3-(-3)]^{2}+(1-5)^{2}}\)
= \(\sqrt{(3+3)^{2}+(1-5)^{2}}\)
= \(\sqrt{(6)^{2}+(-4)^{2}}\)
= \(\sqrt{36+16}\) = \(\sqrt{52}\) = 2\(\sqrt{13}\)
BC= \(\sqrt{(0-3)^{2}+(3-1)^{2}}\)
= \(\sqrt{(-3)^{2}+(2)^{2}}\)
= \(\sqrt{9+4}\) = \(\sqrt{13}\)
AB = \(\sqrt{(-3)^{2}+4}\) = \(\sqrt{13}\)
∴ AB = BC + AC
∴ A, B and C are collin ear.
So, quadrilateral ABCD is not formed.
(iii) Let the given points be A(4, 5), B(7, 6), C(4, 3) and D(1, 2)
Then, AB = \(\sqrt{(7-4)^{2}+(6-5)^{2}}\)
⇒ AB = \(\sqrt{(3)^{2}+(1)^{2}}\)
= \(\sqrt{9+1}\) = \(\sqrt{10}\)
BC = \(\sqrt{(4-7)^{2}+(3-6)^{2}}\)
= \(\sqrt{(-3)^{2}+(-3)^{2}}\)
= \(\sqrt{9+9}\) = \(\sqrt{18}\)
CD = \(\sqrt{[-1-(-3)]^{2}+(-2-0)^{2}}\)
= \(\sqrt{(-3)^{2}+(-1)^{2}}\)
= \(\sqrt{9+1}\) = \(\sqrt{10}\)
and DA = \(\sqrt{(4-1)^{2}+(5-2)^{2}}\)
= \(\sqrt{(3)^{2}+(3)^{2}}\)
= \(\sqrt{9+9}\) = \(\sqrt{18}\)
Here, we have
AB = CD = \(\sqrt{10}\)
And, BC = DA = \(\sqrt{18}\)
⇒ Opposite sides of quadrilateral are equaL
Also, AC = \(\sqrt{(4-4)^{2}+(3-5)^{2}}\)
= \(\sqrt{0+(-2)^{2}}\)
= \(\sqrt{0+4}\) = \(\sqrt{4}\) = 2
and BD = \(\sqrt{(1-7)^{2}+(2-6)^{2}}\)
= \(\sqrt{(-6)^{2}+(-4)^{2}}\)
= \(\sqrt{36 + 16}\) = \(\sqrt{52}\)
= AC ≠ BD
Here, sides AB = DC and BC = AD.
Therefore, the quadrilateral formed by given points is a parallelogram.
Question 7.
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Solution:
Let the required point be ‘P’ which is on the x-axis, so its ordinate = 0 and abscissa (say) = x.
Therefore, coordinates of the point P are (x, 0).
Let the given points be A(2, -5) and B(-2, 9).
It is given that:
AP = BP
\(\sqrt{(2-x)^{2}+(-5-0)^{2}}\) = \(\sqrt{[x-(-2)]^{2}+(0-9)^{2}}\)
= \(\sqrt{(2-x)^{2}+(-5)^{2}}\) = \(\sqrt{(x+2)^{2}+(-9)^{2}}\)
= \(\sqrt{(2)^{2}+(x)^{2}-2(2)(x)+25}\) = \(\sqrt{(x)^{2}+(2)^{2}+2(x)(2)+81}\)
= \(\sqrt{4+x^{2}-4 x+25}\) = \(\sqrt{x^{2}+4+4 x+81}\)
= \(\sqrt{x^{2}-4 x+29}\) = \(\sqrt{x^{2}+4 x+85}\)
Squaring both sides, we get
x2 – 4x + 29 = x2 + 4x + 85
-4x -4x = 85 – 29
-8x = 56
= x = -7
Therefore, the required point is (-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:
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).
Now, QR = \(\sqrt{(0 \pm 4)^{2}+(1-6)^{2}}\)
= \(\sqrt{41}\)
PR = \(\sqrt{(\pm 4-5)^{2}+[6-(-3)]^{2}}\)
= \(\sqrt{(4-5)^{2}+81}\) or \(\sqrt{(-4-5)^{2}+81}\)
= \(\sqrt{82}\) or \(\sqrt{162}\) = \(\sqrt{82}\) or 9\(\sqrt{2}\)
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
= \(\sqrt{(x-3)^{2}+(y-6)^{2}}\)
= \(\sqrt{[x-(-3)]^{2}+(y-4)^{2}}\)
= \(\sqrt{(x-3)^{2}+(y-6)^{2}}\)
= \(\sqrt{(x+3)^{2}+(y-4)^{2}}\)
x2 – 6x + 9 + y2 – 12y + 36
= x2 + 6x + 9 + y2 – 8y + 16 (Squaring both sides)
– 6x -6x -12y + 8y + 36 – 16 = 0
-12x – 4y + 20 = 0
-4(3x + y – 5) = 0
3x + y – 5 = 0