. Exercise 7.1 Page: 105

 

 

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)

Answer:

(i) Distance between the points is given by

Therefore the distance between (2,3) and (4,1) is given by

l =

=

= √4+4 = √8 = 2√2

(ii)Applying Distance Formula to find distance between points (–5, 7) and (–1, 3), we get

l =

 =

 = √16+16 = √32 = 4√2

(iii)Applying Distance Formula to find distance between points (a, b) and (–a, –b), we get

l =

=

=

 

2. Find the distance between the points (0, 0) and (36, 15). Also, find the distance between towns A and B if town B is located at 36 km east and15 km north of town A.

Answer:

Applying Distance Formula to find distance between points (0, 0) and (36, 15), we get

=

= √1296 + 225 = √1521 = 39

Yes, we can find the distance between the given towns A and B.

Assume town A at origin point (0, 0).

Therefore, town B will be at point (36, 15) with respect to town A.

And hence, as calculated above, the distance between town A and B will be 39km.

 

3. Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.

Answer:

Let A = (1, 5), B = (2, 3) and C = (–2, –11)

Using Distance Formula to find distance AB, BC and CA.

BC =

CA =

Since AB+BC ≠ CA

Therefore, the points (1, 5), (2, 3), and (−2, −11) are not collinear.

 

4. Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.

Answer:

Let A = (5, –2), B = (6, 4) and C = (7, –2)

Using Distance Formula to find distances AB, BC and CA.

AB =

BC =

CA =

Since AB = BC.

Therefore, A, B and C are vertices of an isosceles triangle.

 

5. In a classroom, 4 friends are seated at the points A (3, 4), B (6, 7), C (9, 4) and D (6, 1). 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.

Answer:

We have A = (3, 4), B = (6, 7), C = (9, 4) and D = (6, 1)

Using Distance Formula to find distances AB, BC, CD and DA, we get

AB =

BC =

CD =

AD =

Therefore, All the sides of ABCD are equal here

Now, we will check the length of its diagonals.

AC =

BD =

So, Diagonals of ABCD are also equal.

we can definitely say that ABCD is a square.

Therefore, Champa is correct.

 

6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.

(i) (–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)

Answer:

(i)Let A = (–1, –2), B = (1, 0), C= (–1, 2) and D = (–3, 0)

Using Distance Formula to find distances AB, BC, CD and DA, we get

AB =

BC =

CD =

AD =

Therefore, all four sides of quadrilateral are equal.

Now, we will check the length of diagonals.

AC =

BD =

Therefore, diagonals of quadrilateral ABCD are also equal.

we can say that ABCD is a square.

(ii)Let A = (–3, 5), B= (3, 1), C= (0, 3) and D= (–1, –4)

Using Distance Formula to find distances AB, BC, CD and DA, we get

AB =

BC =

CD =

DA =

We cannot find any relation between the lengths of different sides.

Therefore, we cannot give any name to the quadrilateral ABCD.

(iii)Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)

Using Distance Formula to find distances AB, BC, CD and DA, we get

AB =

BC =

CD =

DA =

Here opposite sides of quadrilateral ABCD are equal.

We can now find out the lengths of diagonals.

AC =

BD =

Here diagonals of ABCD are not equal.

 we can say that ABCD is not a rectangle therefore it is a parallelogram.

 

7. Find the point on the x–axis which is equidistant from (2, –5) and (–2, 9).

Answer:

Let the point be (x, 0) on x–axis which is equidistant from (2, –5) and (–2, 9).

Using Distance Formula and according to given conditions we have:

⇒

Squaring both sides, we get

⇒

(x-2)² + 25 = (x+2)² + 81

x² + 4 - 4x + 25 = x² + 4 + 4x + 81

8x = - 25 - 81

8x = -56

x = - 7

Therefore, point on the x–axis which is equidistant from (2, –5) and (–2, 9) is (–7, 0)

 

8. Find the values of y for which the distance between the points P (2, –3) and Q (10, y) is 10 units.

Answer:

Using Distance formula, we have

⇒

⇒ 64 + (y +3)² = 100

⇒ (y+3)² = 100-64 = 36

⇒ y+3 = ± 6

⇒ y+3=6 or y+3 = - 6

Therefore y = 3 or -9

 

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.

Answer:

It is given that Q is equidistant from P and R. Using Distance Formula, we get

PQ = RQ

⇒√25+16 = √x² + 25

⇒41 = x² + 25

 16 = x²

 x = ± 4

Thus, R is (4, 6) or (–4, 6).

When point R is (4,6)

 PR =

 QR =

When point R is (- 4,6)

PR = 

 QR =

 

10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).

Answer:

It is given that (x, y) is equidistant from (3, 6) and (–3, 4).

Using Distance formula, we can write

⇒

⇒ (x-3)² + (y-6)² = (x+3)² + (y-4)²

⇒ x² + 9 -6x + y² + 36 - 12y = x² + 9 + 6x + y² + 16 - 8y

⇒36- 16 = 6x + 6x + 12y  - 8y

⇒20 = 12x + 4y

⇒3x + y = 5

⇒3x + y - 5 = 0

. Exercise 7.2 Page: 111

 

Solve the followings Questions.

 

1. Find the coordinates of the point which divides the join of (- 1, 7) and (4, - 3) in the ratio 2:3.

Answer:

Let P(x, y) be the required point. Using the section formula

Therefore the point is (1,3).

 

2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Answer:

Let P (x₁,y₁) and Q (x₂,y₂) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

Therefore P(x₁,y₁) = (2, -5/3)

Point Q divides AB internally in the ratio 2:1.

Q (x₂ ,y₂) = (0, -7/3)

 

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 the following figure. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the 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 flagexactly halfway between the line segment joining the two flags, where should she post her flag?

Answer:

4. Find the ratio in which the line segment joining the points (-3, 10) and (6, - 8) is divided by (-1, 6).

Answer:
Let the ratio in which the line segment joining ( -3, 10) and (6, -8) is divided by point ( -1, 6) be k:1.
Therefore, -1 = 6k-3/k+1
-k - 1 = 6k -3
7k = 2
k = 2/7
Therefore, the required ratio is 2:7.

 

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 division.

Answer:
Let the ratio in which the line segment joining A (1, - 5) and B ( - 4, 5) is divided by x-axis be k:1.
Therefore, the coordinates of the point of division is (-4k+1/k+1, 5k-5/k+1).

We know that y-coordinate of any point on x-axis is 0.

∴ 5k-5/k+1 = 0

Therefore, x-axis divides it in the ratio 1:1.

To find the coordinates let's substitute the value of k in equation(1)

 

Required point = [(- 4(1) + 1) / (1 + 1), (5(1) - 5) / (1 + 1)]

 

= [(- 4 + 1) / 2, (5 - 5) / 2]

 

= [- 3/2, 0]

6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Answer:

Let A,B,C and D be the points (1,2) (4,y), (x,6) and (3,5) respectively.

7. Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, - 3) and B is (1, 4).

Answer:

 

 

Let (x,y) be the coordinate of A.

Since AB is the diameter of the circle, the centre will be the mid-point of AB.

now, as centre is the mid-point of AB.

x-coordinate of centre = (2x+1)/2​

y-coordinate of centre = (2y+4)/2​

But given that centre of circle is (2,−3).

Therefore,

(2x+1)/2​=2⇒x=3

(2y+4​)/2=−3⇒y=−10

Thus the coordinate of A is (3,−10).

8. If A and B are (–2, –2) and (2, –4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.

Answer:

As given the coordinates of A(−2,−2) and B(2,−4) and P is a point lies on AB.

And AP = 3/7 ​AB

∴BP = 4/7

 

Then, ratio  of AP and PB = m_1​:m_2 ​= 3:4

 

Let the coordinates of P be (x,y).

 

∴ x = (m_1​x_2​ + m₂​x₁) / (m₁​ + m₂​)​​

 

⇒ x = (3 × 2 + 4 × (−2)) / (3 + 4) ​ = (6 − 8) / 7 ​= −2 / 7​

 

And y = (m_1​y_2​ + m₂​y1) / (m₁​ + m₂​)​​​​

 

⇒ y = ((3 × (−4) + 4 × (−2)) / (3 + 4) ​= (−12−8) / 7 ​= −20 / 7​

 

∴ Coordinates of P = −2 / 7​, −20 / 7​

9. Find the coordinates of the points which divide the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.

Answer:

From the figure, it can be observed that points X,Y,Z are dividing the line segment in a ratio 1:3,1:1,3:1 respectively.

Using Sectional Formula, we get,

Coordinates of X = ((1 × 2 + 3 × (−2)) / (1 + 3), (1 × 8 + 3 × 2) / (1 + 3))


= (−1, 7/2)

Coordinates of Y = (2 − 2) / 2, (2 + 8) / 2 = (0,5)


Coordinates of Z = ((3 × 2 + 1 × (−2)) / (1 + 3), (3 × 8 + 1 × 2) / (1 + 3)


= (1, 13/2)

10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order. [Hint: Area of a rhombus = 1/2(product of its diagonals)]

Answer:
Let (3, 0), (4, 5), ( - 1, 4) and ( - 2, - 1) are the vertices A, B, C, D of a rhombus ABCD.

Length of the diagonal AC=


Length of the diagonal BD=

Area of rhombus ABCD = 1/2 X 4√2 X 6√2= 24 square units.
Therefore, the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order, is 24 square units.