Exercise 7.1

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 =chapter 7-Coordinate Geometry Exercise 7.1/image002.png

=chapter 7-Coordinate Geometry Exercise 7.1/image002.png

= √4+4 = √8 = 2√2

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

l =chapter 7-Coordinate Geometry Exercise 7.1/image002.png

 =chapter 7-Coordinate Geometry Exercise 7.1/image002.png

 = √16+16 = √32 = 4√2

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

l =chapter 7-Coordinate Geometry Exercise 7.1/image002.png

=chapter 7-Coordinate Geometry Exercise 7.1/image002.png

=chapter 7-Coordinate Geometry Exercise 7.1/image002.png

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

=chapter 7-Coordinate Geometry Exercise 7.1/image004.png

= √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.

chapter 7-Coordinate Geometry Exercise 7.1/image007.png

BC =chapter 7-Coordinate Geometry Exercise 7.1/image008.png

CA =chapter 7-Coordinate Geometry Exercise 7.1/image009.png

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 =chapter 7-Coordinate Geometry Exercise 7.1/image011.png

BC =chapter 7-Coordinate Geometry Exercise 7.1/image012.png

CA =chapter 7-Coordinate Geometry Exercise 7.1/image013.png

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.

NCERT solutions for class 10 maths/image014.jpg

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 =NCERT solutions for class 10 maths/image015.png

BC =NCERT solutions for class 10 maths/image016.png

CD =NCERT solutions for class 10 maths/image017.png

AD =NCERT solutions for class 10 maths/image018.png

Therefore, All the sides of ABCD are equal here

Now, we will check the length of its diagonals.

NCERT solutions for class 10 maths/image019.png

AC =NCERT solutions for class 10 maths/image019.png

BD =NCERT solutions for class 10 maths/image020.png

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 =NCERT solutions for class 10 maths/image021.png

BC =NCERT solutions for class 10 maths/image022.png

CD =NCERT solutions for class 10 maths/image023.png

AD =NCERT solutions for class 10 maths/image024.png

Therefore, all four sides of quadrilateral are equal.

Now, we will check the length of diagonals.

AC =NCERT solutions for class 10 maths/image025.png

BD =NCERT solutions for class 10 maths/image026.png

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 =NCERT solutions for class 10 maths/image027.png

BC =NCERT solutions for class 10 maths/image028.png

CD =NCERT solutions for class 10 maths/image029.png

DA =NCERT solutions for class 10 maths/image030.png

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 =NCERT solutions for class 10 maths/image031.png

BC =NCERT solutions for class 10 maths/image032.png

CD =NCERT solutions for class 10 maths/image033.png

DA =NCERT solutions for class 10 maths/image034.png

Here opposite sides of quadrilateral ABCD are equal.

We can now find out the lengths of diagonals.

AC =NCERT solutions for class 10 maths/image035.png

BD =NCERT solutions for class 10 maths/image036.png

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:

NCERT solutions for class 10 maths/image037.png

NCERT solutions for class 10 maths/image038.png

Squaring both sides, we get

NCERT solutions for class 10 maths/image039.png

(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

NCERT solutions for class 10 maths/image041.png

NCERT solutions for class 10 maths/image042.png

⇒ 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

NCERT solutions for class 10 maths/image047.png

NCERT solutions for class 10 maths/image047.png

⇒√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 =NCERT solutions for class 10 maths/image054.png

 QR =NCERT solutions for class 10 maths/image055.png

When point R is (- 4,6)

PR = NCERT solutions for class 10 maths/image056.png

 QR =NCERT solutions for class 10 maths/image057.png

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

NCERT solutions for class 10 maths/image060.png

NCERT solutions for class 10 maths/image061.png

⇒ (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

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

chapter 7-Coordinate Geometry Exercise 7.2/18.PNG

chapter 7-Coordinate Geometry Exercise 7.2/18.PNG

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:

chapter 7-Coordinate Geometry Exercise 7.2/18.PNG

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.

chapter 7-Coordinate Geometry Exercise 7.2/18.PNG

chapter 7-Coordinate Geometry Exercise 7.2/18.PNG

Therefore P(x1,y1) = (2, -5/3)

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

chapter 7-Coordinate Geometry Exercise 7.2/18.PNG

chapter 7-Coordinate Geometry Exercise 7.2/18.PNG

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?

math 10

Answer:

chapter 7-Coordinate Geometry Exercise 7.2/18.PNG

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.

chapter 7-Coordinate Geometry Exercise 7.2/fig-2.jpg

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  be the coordinate of .

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

now, as centre is the mid-point of .

-coordinate of centre 

-coordinate of centre 

But given that centre of circle is .

Therefore,

Thus the coordinate of  is .

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 

Then, ratio  of  and 
Let the coordinates of  be .
And 
 Coordinates of 

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:
chapter 7-Coordinate Geometry Exercise 7.2/fig-4.PNG
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.

EXERCISE 7.3

1. Find the area of the triangle whose vertices are:

(i) (2, 3), (–1, 0), (2, –4)

(ii) (–5, –1), (3, –5), (5, 2)

Answer:

(i) (2, 3), (–1, 0), (2, –4)

Area of Triangle is given by

Area of Triangle =chapter 7-Coordinate Geometry Exercise 7.3/image001.png

Area of  the given triangle = ½[2 {0 − (−4)} – 1 (−4 − 3) + 2 (3 − 0)]

=½ (8 + 7 + 6) = 21/2 sq. units

(ii) (–5, –1), (3, –5), (5, 2)

Area of the given triangle = ½ [−5 (−5 − 2) + 3 {2 − (−1)} + 5 {−1 − (−5)}]

= ½(35 + 9 + 20)

= 32 sq. units

2. In each of the following find the value of ‘k’, for which the points are collinear.

(i) (7, –2), (5, 1), (3, k)

(ii) (8, 1), (k, –4), (2, –5)

Answer:

(i) (7, –2), (5, 1), (3, k)

Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.

Therefore, for points (7, -2) (5, 1), and (3,k), area = 0

⇒ ½[7 (1 − k) + 5 {k − (−2)} + 3 (−2 − 1)]

= ½(7 − 7k + 5k + 10 − 9) = 0

⇒ ½(7 − 7k + 5k + 1) = 0

⇒ ½(8 − 2k) = 0

⇒ 8 − 2k = 0

⇒ 2k = 8

⇒ k = 4

(ii) (8, 1), (k, –4), (2, –5)

Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.

Therefore, for points (8, 1) (k, -4), and (2,-5), area = 0

⇒ ½[8 {−4 − (−5)} + k (−5 − 1) + 2 {1 − (−4)}]

= ½(8 − 6k + 10) = 0

⇒ ½(18 − 6k) = 0

⇒18 − 6k = 0

⇒ 18 = 6k

⇒ k = 3

3. Find the area of the triangle formed by joining the mid–points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Answer:

chapter 7-Coordinate Geometry Exercise 7.3/image006.png

Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).

Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by

D = (0+2/2 , -1+1/2) = (1,0)

E = (0+0/2 , -3-1/2) = (0,1)

Area of a triangle = 1/2 {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)}

Area of ΔDEF = 1/2 {1(2-1) + 1(1-0) + 0(0-2)}

                        = 1/2 (1+1) = 1 square units

Area of ΔABC = 1/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)]

                         = 1/2 {8} = 4 square units

Therefore, the required ratio is 1:4.

4. Find the area of the quadrilateral whose vertices taken in order are (–4, –2), (–3, –5), (3, –2) and (2, 3).

Answer:

chapter 7-Coordinate Geometry Exercise 7.3/image016.jpg

Let the vertices of the quadrilateral be A ( – 4, – 2), B ( – 3, – 5), C (3, – 2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

Area of a triangle = 1/2 {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)}

Area of ΔABC =1/2[−4 (−5 − 2) – 3 {-2 − (−2)} + 3 {−2 − (−5)}]

=1/2 [12 + 0 + 9]

= 21/2 sq. units

Again using formula to find area of triangle:

Area of △ACD = [−4 (−2 − 3) + 3 {3 − (−2)} + 2 {−2 − (-2)}]

= 1/2 [20 + 15 + 0]

= 35/2 sq. units

Area of ☐ABCD = Area of ΔABC + Area of ΔACD

                           = 21/2 + 35/2 = 28 sq. units

5. We know that median of a triangle divides it into two triangles of equal areas. Verify this result for △ABC whose vertices are A (4, –6), B (3, –2) and C (5, 2).

Answer:

chapter 7-Coordinate Geometry Exercise 7.3/image018.jpg

Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2).

Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.

Coordinates of point D = (3+5/2, -2+2/2) = (4,0)

Area of a triangle = 1/2 {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)}

Area of △ABD = 1/2[4 (−2 − 0) + 3 {0 − (−6)} + 4 {−6 − (−2)}]

=1/2(−8 + 18 −16)

=1/2 (−6) = −3 sq units

Area cannot be in negative.

Therefore, we just consider its numerical value.

Therefore, area of △ABD = 3 sq units

Again using formula to find area of triangle:

Area of △ABD = 1/2 [4 (0 − 2) + 4{2 − (−6)} + 5 {−6 −0 )}]

 = 1/2 (- 8 + 32 −30) = ½ (-6) = -3 sq units.

However, area cannot be negative. Therefore, area of ΔABD is 3 square units.

The area of both sides is same. Thus, median AD has divided ΔABC in two triangles of equal areas

Hence Proved.

Exercise 7.4

1. Determine t]“1he ratio in which the line 2x + y -4 =0 divides the line segment joining the points A(2,-2) and B(3,7).

Answer:

Let the given line divide the line segment joining the points A(2, −2) and B(3, 7) in a ratio k : 1

Coordinates of the point of division =chapter 7-Coordinate Geometry Exercise 7.4/image006.jpg

This point also lies on 2x + y – 4 = 0

chapter 7-Coordinate Geometry Exercise 7.4/image006.jpg

chapter 7-Coordinate Geometry Exercise 7.4/image007.png

⇒  9k – 2 = 0

⇒  k = 2/9

Therefore, the ratio in which the line 2x + y – 4 = 0  divides the line segment joining the points A(2, −2) and B(3, 7) is 2:9.

2. Find a relation between x and y if the points (x,y),(1, 2)and (7, 0)are collinear.

Answer:

If the given points are collinear, then the area of triangle formed by these points will be 0.

Area of triangle = chapter 7-Coordinate Geometry Exercise 7.4/image008.png

 Area = chapter 7-Coordinate Geometry Exercise 7.4/image008.png

0 = 1/2 [2x – y + 7y – 14]

0 =1/2 [2x + 6y – 14]

[2x + 6y – 14] = 0

x + 3y – 7 = 0

This is the required relation between x & y.

3. Find the centre of a circle passing through the points (6,-6), B(3,-7)and (3, 3).

Answer:

 Let O (x,y) be the centre of the circle. And let the points (6, −6), (3, −7), and (3, 3) be representing the points A, B, and C on the circumference of the circle.

chapter 7-Coordinate Geometry Exercise 7.4/image025.png

chapter 7-Coordinate Geometry Exercise 7.4/image008.png

chapter 7-Coordinate Geometry Exercise 7.4/image008.png

However OA = OB (Radii of same circle)

⇒ chapter 7-Coordinate Geometry Exercise 7.4/image008.png

=>x² + 36 – 12x + y² + 36 + 12y = x² + 9 – 6x + y² + 49 -14y

⇒ -6x + 2y + 14 = 0

⇒ 3x + y = 7 ….1

Similary OA = OC (Radii of same circle)

chapter 7-Coordinate Geometry Exercise 7.4/image008.png

=>x² + 36 – 12x + y² + 36 + 12y = x² + 9 – 6x + y² + 9 – 6y

⇒ -6x + 18y + 54 = 0

⇒ -3x + 9y = -27  …..(2)

On adding equation (1) and (2), we obtain

10y = – 20

y = – 2

From equation (1), we obtain

3x − 2 = 7

3x = 9

x = 3

Therefore, the centre of the circle is (3, −2).

4. The two opposite vertices of a square are (-1,2) and (3,2). Find the coordinates of the other two vertices.

Answer

chapter 7-Coordinate Geometry Exercise 7.4/image043.png

Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively. Let (x,y), (x1,y1) be the coordinate of vertex B and D respectively.

We know that the sides of a square are equal to each other.

∴ AB = BC

chapter 7-Coordinate Geometry Exercise 7.4/image045.png

=>x² + 2x + 1 + y²  -4y + 4 = x² + 9  -6x + y² + 4 – 4y

⇒ 8x = 8

⇒ x = 1

We know that in a square, all interior angles are of 90°.

In ΔABC,

AB² + BC² = AC²

chapter 7-Coordinate Geometry Exercise 7.4/image008.png

⇒4 + y² + 4 − 4y + 4 + y² + 4 − 4y = 16

⇒2y² + 16 − 8 = 16

⇒2y² – 8 = 0

⇒y(y – 4) = 0

⇒y = 0 or 4

We know that in a square, the diagonals are of equal length and bisect each other at 90°. Let O be the mid-point of AC. Therefore, it will also be the mid-point of BD

Coordinate of point O = ((-1+3)/2, (2+2)/2)

chapter 7-Coordinate Geometry Exercise 7.4/image008.png

⇒1 + x1/2 = 1

⇒1 + x1 = 2

⇒x1 = 1

and y + y1 /2 = 2

⇒ y + y1= 4

⇒If y = 0

⇒y1= 4

⇒If y = 4

⇒y1= 0

Therefore, the required coordinates are (1, 0) and (1, 4).

5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

chapter 7-Coordinate Geometry Exercise 7.4/image059.png

(i) Taking A as origin, find the coordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of △ PQR if C is the origin? Also calculate the area of the triangle in these cases. What do you observe?

Answer:

(i) Taking A as the origin, AD and AB as the coordinate axes. Clearly, the points P, Q and

R are (4, 6), (3, 2) and (6, 5) respectively.

(ii)Taking C as the origin, CB and CD as the coordinate axes. Clearly, the points P, Q and R are given by (12, 2), (13, 6) and (10, 3) respectively.

We know that the area of the triangle (POR) =chapter 7-Coordinate Geometry Exercise 7.4/image060.png

= 1/2 [4(2-5) + 3(5-6) + 6(6-2)]

= 1/2 [-12 -3 +24]

= 9/2 sq. units

(ii) Taking C as origin, CB as x- axis, and CD as y- axis, the coordinates of vertices P, Q, and R are (12, 2), (13, 6), and (10, 3) respectively.

area of the triangle (PQR) =chapter 7-Coordinate Geometry Exercise 7.4/image060.png

= 1/2[12(6-3) + 13(3-2) + 10(2-6)

= 1/2 [36+13-40]

= 9/2 sq. units

Hence, the areas are same in both the cases.

6. The vertices of a △ ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively such that AD/AB = AE/AC = 1/4 .Calculate the area of the △ ADE and compare it with the area of △ ABC.

Answer:

chapter 7-Coordinate Geometry Exercise 7.4/image070.png

Given that (AD)/(AB) = (AE)/(AC) = 1/4

(AD)/(AD+DB) = (AE)/(AE+EC) = 1/4

(AD)/(DB)=(AE)/(EC) = 1/3

Therefore, D and E are two points on side AB and AC respectively such that they divide side AB and AC in a ratio of 1:3

Coordinates of Point D =chapter 7-Coordinate Geometry Exercise 7.4/image070.png= (13/4, 23/4)

Coordinates of Point E =chapter 7-Coordinate Geometry Exercise 7.4/image070.png = (19/4, 20/4)

Area of triangle =chapter 7-Coordinate Geometry Exercise 7.4/image060.png

Area of ΔADE =chapter 7-Coordinate Geometry Exercise 7.4/image060.png

 =chapter 7-Coordinate Geometry Exercise 7.4/image060.png

Area of ΔABC = 1/2[4(5-2) + 1(2-6) + 7(6-5)]

= 1/2[12 – 4 + 7]

= 15/2

Clearly, the ratio between the areas of ΔADE and ΔABC is 1:16.

7. Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of △ ABC.

(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP: PD = 2: 1.

(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ: QE = 2: 1 and CR : RF = 2 : 1.

(iv) What do you observe?

(Note: The point which is common to all the three medians is calledcentroid and this point divides each median in the ratio 2: 1)

(v) If A(x1,y1), B(x2,y2),and C(x3,y3),are the vertices of △ ABC, find the coordinates of the centroid of the triangle.

Answer:

chapter 7-Coordinate Geometry Exercise 7.4/image050.png

(i) Median AD of the triangle will divide the side BC in two equal parts.

Therefore, D is the mid-point of side BC

Coordinate of D = (6 +1/2, 5 + 4/2) =(7/2,9/2)

(ii) Point P divides the side AD in a ratio 2:1.

Coordinate of P = chapter 7-Coordinate Geometry Exercise 7.4/image050.png= (11/3 , 11/3)

(iii) Median BE of the triangle will divide the side AC in two equal parts.

Therefore, E is the mid-point of side AC.

Coordinate of E = (4+1/2, 2+4/2) = (5/2,3)

Point Q divides the side BE in a ratio 2:1.

Coordinate of Q =chapter 7-Coordinate Geometry Exercise 7.4/image050.png= (11/3 , 11/3)

Median CF of the triangle will divide the side AB in two equal parts. Therefore, F is the mid-point of side AB

Coordinate of F = (4+6/2, 2+5/2) = (5,7/2)

Point R divides the side CF in a ratio 2:1.

Coordinate of R =chapter 7-Coordinate Geometry Exercise 7.4/image050.png= (11/3 , 11/3)

(iv) It can be observed that the coordinates of point P, Q, R are the same.

Therefore, all these are representing the same point on the plane i.e., the centroid of the triangle.

(v) Consider a triangle, ΔABC, having its vertices as A(x1,y1), B(x2,y2) and C(x3,y3)

Median AD of the triangle will divide the side BC in two equal parts. Therefore, D is the mid-point of side BC.

Coordinate of D =chapter 7-Coordinate Geometry Exercise 7.4/image050.png

Let the centroid of this triangle be O.

Point O divides the side AD in a ratio 2:1.

Coordinate of O =chapter 7-Coordinate Geometry Exercise 7.4/image050.png

chapter 7-Coordinate Geometry Exercise 7.4/image050.png

8. ABCD is a rectangle formed by joining points A(-1,-1), B (-1,4), C (5,4) and D (5,-1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? Or a rhombus? Justify your answer.

Answer:

chapter 7-Coordinate Geometry Exercise 7.4/image104.png

P is the mid point of side AB

Therefore the coordinates of P are ((-1-1)/2,(-1+4)/2) = (-1, 3/2)

Similary the coordinates of Q , R and S are  (2,4),(5, 3/2), and (2, -1) respectilvely

Length of PQ =chapter 7-Coordinate Geometry Exercise 7.4/image105.png

Length of QR =chapter 7-Coordinate Geometry Exercise 7.4/image105.png

Length of RS =chapter 7-Coordinate Geometry Exercise 7.4/image105.png

Length of SP =chapter 7-Coordinate Geometry Exercise 7.4/image105.png

Length of PR =chapter 7-Coordinate Geometry Exercise 7.4/image105.png= 6

Length of QS =chapter 7-Coordinate Geometry Exercise 7.4/image105.png= 5

It can be observed that all sides of the given quadrilateral are of the same measure. However, the diagonals are of different lengths. Therefore, PQRS is a rhombus.