Exercise 7.2

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

Solution:

Let the coordinates of the point be P(x, y) which divides the line segment joining the points (-1, 7) and (4, - 3) in the ratio :

Let two points be A (x₁, y₁) and B(x₂, y₂). P (x, y) divides internally the line joining A and B in the ratio m₁: m₂. Then, coordinates of P(x, y) is given by the section formula

P (x, y) = [mx2+nx1m+n , my2+ny1m+n]

Let x1 = , y1 = , x2 = and y2 = , m = , n =

By Section formula, P (x, y) = [mx2+nx1m+n , my2+ny1m+n] --- (1)

By substituting the values in the equation (1)

x = 2×4+3×−12+3 and y = 2×−3+3×72+3

x = 8−35 and y =−6+215

x = 55 = and y = 155 =

Therefore, the coordinates of point P are (1, 3).

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

Solution:

The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2) internally in the ratio m1: m2 is given by the section formula.

Let the points be A(4, ) and B(, - 3).

Let P (x1, y1) and Q (x2, y2) be the points of trisection of the line segment joining the given points.

Then, AP = PC =

By Section formula ,

P (x, y) = [mx2+nx1m+n , my2+ny1m+n] ..... (1)

Considering A(4, - 1) and B(- 2, - 3), by observation point P(x₁, y₁) divides AB internally in the ratio 1 : .

Hence m : n = 1 : 2

By substituting the values in the Equation (1)

x1 = 1×−2+2×41+2

x1 = −2+83 =

y1 = 1×−3+2×−11+2

y1 = −3−21+2= −5353

Hence, P(x1 ,y1) = (2, −53)

Now considering A(4, - 1) and B(- 2, - 3), by observation point C(x2, y2) divides AB internally in the ratio : 1.

Hence m : n = 2 : 1

By substituting the values in the Equation (1)

x2 = 2×−2+1×42+1

= −4+43=

y2 = 2×−3+1×−12+1

= −6−13

= −7373

Therefore, C(x2 , y2) = (0, −73)

Hence, the points of trisection are P(x1 , y1) = (2, −53) and C (x₂ , y₂) = (0, −73)

3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs 14th the distance AD on the 2nd line and posts a green flag. Preet runs 15th 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 flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Solution:

Given: 100 flower pots have been placed at a distance of 1m from each other along .

Let Niharika post the green flag at a distance P, that is, (14 × 100) m = m from the starting point of the 2nd line.

Therefore, the coordinates of the point P are (, ).

Similarly, Preet posted a red flag at the distance Q, that is, (15 × 100) m = m from the starting point of the 8th line.

Therefore, the coordinates of the point Q are (, ).

We know that the distance between the two points is given by the Distance Formula,

To find the distance between these flags, we will find PQ using the distance formula,

PQ = x2−x12+y2−y12

PQ = 8−22+20−252

= 36+25

= 6160m

Let the point be A (x, y) at which Rashmi should post her blue flag exactly at the centre of the line joining the coordinates P(2, 25) and Q(8, 20).

By midpoint formula,

P(x, y) = [x1+x22,y1+y22]

P(x, y) = [2+82, 25+202]

P(x, y) = (10292, 452245)

P(x, y) = ( , )

Therefore, Rashmi should post her blue flag at a distance of 22.5 m on the 5th line.

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

Solution:

The coordinates of the point P(x, y) which divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁: m₂ is given by the Section Formula : P(x, y) = [mx2+nx1m+n , my2+ny1m+n]

Let the ratio in which the line segment joining A(-3, ) and B(, -8) be divided by point C(-1, ) be k : 1.

By Section formula, C(x, y) = [mx2+nx1m+n , my2+ny1m+n]

m = , n =

Therefore,

- = 6k−3k+1

-k - 1 = 6k -

7k =

k = 2772

Hence, the point C divides line segment AB in the ratio : .

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.

Solution:

The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1 ) and B(x2, y2), internally, in the ratio m1: m2 is given by the Section Formula: P(x, y) = [mx2+nx1m+n , my2+ny1m+n]

Let the ratio be k : 1.

Let the line segment be AB joining A (1, -) and B (-, 5)

By using the Section formula,

P (x, y) = [mx2+nx1m+n , my2+ny1m+n]

m = , n =

Therefore, the coordinates of the point of division is

(x, 0) = [−4k+1k+1,5k−5k+1] ---------- (1)

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

Therefore, 5k−5k+1 =

5k =

k =

Therefore, the x-axis divides the line segment in the ratio of : .

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

Required point = [−41+11+1, 51−51+1]

= [−4+12,5−52]

= −32, 032, 0

Therefore the coordinates of the point of division is −32, 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.

Solution:

The coordinates of the point P(x, y) which divides the line segment joining the points A (x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁: m₂ is given by the section formula: P(x, y) = [mx2+nx1m+n , my2+ny1m+n]

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

Let A (1, 2), B (4, y), C(x, ), and D (, 5) be the vertices of a parallelogram ABCDBDAC.

Since the diagonals of a parallelogram bisect each other. The intersection point O of diagonal AC and BD also divides these diagonals in the ratio : .

Therefore, O is the mid-point of ACAD and BDAD.

According to the mid point formula,

O(x, y) = [x1+x22,y1+y22]

If O is the mid-point of ACBD:

Coordinates of O = [1+x2, 2+62]

[x+12,4] ----- (1)

If O is the mid-point of BDAB:

Coordinates of O = [4+32, 5+y2]

⇒ [72, 5+y2] ------ (2)

Since both the coordinates are of the same point O, so, x+12 = 72 and 4 = 5+y2 [From equation(1) and (2)]

x + 1 = and 5 + y = (By cross multiplying & transposing)

x = and y =

Therefore, x = 6 and y = 3.

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

Solution:

The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2), internally, in the ratio m1 : m2 is given by the section formula.

Let the coordinates of point A be (x, y).

Mid-point of AB is (2, ), which is the center of the circle.

According to the mid point formula,

{.text-center}O(x, y) = [x1+x22,y1+y22]

We have A(x, y) and B(, 4) and the center is (2, )

Therefore by using midpoint formula,

(2, -3) = [x+12, y+42]

x+12 = and y+42 = (By Cross multiplying & transposing)

x + 1 = and y + 4 =

x = and y =

Therefore, the coordinates of A are (3, - 10).

8. If A and B are (–2, –2) and (2, –4), respectively, find the coordinates of P such that AP = 37 AB and P lies on the line segment AB.

Solution:

The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2) internally, in the ratio m1 : m2 is given by the section formula: P (x, y) = [mx2+nx1m+n , my2+ny1m+n]

The coordinates of point A and B are (- 2, ) and (2, ) respectively.

AP = (37) AB

Hence, ABAP = 7337

We know that AB = AP + from figure,

Thus, ABAP = 73 can be written as,

AP+PBAP = 3+43

+ PBAP = + 43

PBAP =

Therefore, AP : PB = :

Point P(x, y) divides the line segment AB joining A(-2, -2) and B(2, -4) in the ratio :4.

By using section formula,

P (x, y) = [mx2+nx1m+n , my2+ny1m+n]

P (x, y) = [3×2+4×−23+4 , 3×−4+4×−23+4]

= (6−87, −12−87)

= (−2727, −207207)

Therefore the coordinates of P −27, −207.

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:

The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2), internally, in the ratio m1: m2 is given by the section formula.

By observation, points P, Q, R divides the line segment A (- 2, ) and B (2, ) into four equal parts.

Point P divides the line segment AQ into twothree equal parts.

Therefore, AP : PB is : .

Using section formula which is given by:

P (x, y) = [mx2+nx1m+n , my2+ny1m+n]

Hence, coordinates of P = [1×2+3×−23+1, 1×8+3×23+1] = (, )

Point Q divides the line segment AB into two equal parts.

Using mid point formula,

Q = [2+−22, 2+82] = (, )

Point R divides the line segment BQ into two equal parts

Coordinates of R = [2+02, 8+52] = 1, 1321, −132

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 =12(product of its diagonals)]

Solution:

A rhombus has all sides of equal length and opposite sides are .

Let A(, 0), B(, 5), C(- 1, ) and D(- 2, ) be the vertices of a rhombus ABCD.

Also, Area of a rhombus = × (product of its diagonals)

Hence we will calculate the values of the diagonals AC and .

We know that the distance between the two points is given by the distance formula,

Distance formula = x2−x12+y2−y12

Therefore, distance between A (3, 0) and C (- 1, 4) is given by

Length of diagonal AC =3−−12+0−42

= 16+16

= 4232

The distance between B (4, ) and D (, - 1) is given by

Length of diagonal BD = 4−−22+5−−12

= 36+36

= 6252

Area of the rhombus ABCD = 12 × (Product of lengths of diagonals) = 12 × ACAD × BDAC

= 12 × 4222 × 6252

Therefore, the area of the rhombus ABCD = 12 × 42 × 62 = square units