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(x_{1}, y_{1}) and B(x_{2}, y_{2}) 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

x^{2} – 4x + 29 = x^{2} + 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, PQ^{2} = (10)^{2} = 100

(10 – 2)^{2} + [(y – (-3)]^{2}= 100

(8)^{2} + (y + 3)^{2} = 100

64+y^{2} + 6y + 9 = 100

y^{2} + 6y – 27 = 0

y^{2} + 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

PQ^{2} = RQ^{2}

(0 – 5)^{2} + [1 – (3)]^{2}

= (0 – x)^{2} + (1 – 6)^{2}

25 + 16 = x^{2} + 25

x^{2} = 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

= \(\sqrt{(x-3)^{2}+(y-6)^{2}}\)

= \(\sqrt{[x-(-3)]^{2}+(y-4)^{2}}\)

= \(\sqrt{(x-3)^{2}+(y-6)^{2}}\)

= \(\sqrt{(x+3)^{2}+(y-4)^{2}}\)

x^{2} – 6x + 9 + y^{2} – 12y + 36

= x^{2} + 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