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 2 : 3

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 = CBAC

By Section formula ,

{.text-center} 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 ADBD.

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 (2, ).

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 (8, )

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) = (102, 452245)

P(x, y) = (5 , )

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

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 .

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, then the coordinates of O are

[1+x2, 2+62]

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

If O is the mid-point of BDAB, then the coordinates of O are

[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, AB/AP = 7337

We know that AB = AP + from figure,

Thus, AB/AP = 73 can be written as,

AP+PBAP = 3+43

+ PBAP = + 43

PBAP = 4334

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)

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] = (, 72)

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 parallelopposite.

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

Also, Area of a rhombus = 12 × (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

= 2

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

Length of diagonal BD = 4−−22+5−−12

= 36+36

= 2

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

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

= square units