NCERT Solutions Class 10 Maths Chapter 7- Coordinate Geometry

Exercise7.1

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)
Solution:
Distance formula to determine the distance between 2 points (x_1, y_1) & (x_2, y_2) is given as;
d\ =\ \sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}
Or
d\ =\ \sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}
d\ =\ \sqrt{{(4-2)}^2+{(1-3)}^2}=\sqrt8=2\sqrt2
d\ =\ \sqrt{{(-1+5)}^2+{(3-7)}^2}=\sqrt{32}=4\sqrt2
d\ =\ \sqrt{{(-a-a)}^2+{(-b-b)}^2}=\sqrt{{(-2a)}^2+{(-2b)}^2}=2\sqrt{a^2+b^2}

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, town A at point (0, 0). Thus, town B will be at point (36, 15).
Distance between point (0, 0) & point (36, 15) is
d\ =\ \sqrt{{(36-0)}^2+{(15-0)}^2}=\sqrt{1296+225}=\sqrt{1521}=39
In section 7.2, A is (4, 0) and B is (6, 0)
{AB}^2\ =\ {(6-4)}^2+{(0-0)}^2=\ 4
The distance between town A & B IS 39 km. The distance between two towns A & B discussed in Section 7.2 is 4 km.

3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution:
The sum of lengths of any two-line segments is equal to the length of 3rd line segment then all the three points are collinear.
Let, A = (1, 5) B = (2, 3) & C = (-2, -11)
Distance AB, BC and CA is given as;
AB=\ \sqrt{{(2-1)}^2+{(3-5)}^2}=\sqrt{1+4}=\sqrt5\\ BC=\ \sqrt{{(-2-2)}^2+{(-11-3)}^2}=\sqrt{16+196}=\sqrt{212}\\ CA=\ \sqrt{{(-2-1)}^2+{(-11-5)}^2}=\sqrt{9+256}=\sqrt{265}
As AB + BC ≠ CA
Thus, points (1, 5), (2, 3), & (– 2, – 11) are not collinear.

4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Solution:
As 2 sides of any Isosceles Triangle are equal. To determine whether the given points are vertices of an isosceles triangle, we need to find the distance between all the points.
Suppose that the points (5, – 2), (6, 4), & (7, – 2) are represents the vertices A, B, and C.
AB=\ \sqrt{{(6-5)}^2+{(4+2)}^2}=\sqrt{37}\\ BC=\ \sqrt{{(7-6)}^2+{(-2-4)}^2}=\sqrt{1+36}=\sqrt{37}\\ CA=\ \sqrt{{(7-5)}^2+{(-2-2)}^2}=2
Hence,
AB = BC
Therefore, the given points are the vertices of an isosceles triangle.

5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. 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:
From the figure, the coordinates of the points A, B, C & D are (3, 4), (6, 7), (9, 4) and (6,1).
Distance is given as;
AB=\ \sqrt{{(6-3)}^2+{(7-4)}^2}=\sqrt{9+9}=3\sqrt2\\ BC=\ \sqrt{{(9-6)}^2+{(4-7)}^2}=\sqrt{9+9}=3\sqrt2\\ CD=\ \sqrt{{(6-9)}^2+{(1-4)}^2}=\sqrt{9+9}=3\sqrt2\\ DA=\ \sqrt{{(6-3)}^2+{(1-4)}^2}=\sqrt{9+9}=3\sqrt2\\ Diagonal\ AC=\ \sqrt{{(3-9)}^2+{(4-4)}^2}=6\\ Diagonal\ BD=\ \sqrt{{(6-6)}^2+{(7-1)}^2}=6
The length of all the sides is equal. Hence, ABCD is a square & therefore, Champa was 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)
Solution:
Let us consider, the points (- 1, – 2), (1, 0), ( – 1, 2), & ( – 3, 0) be represents the vertices A, B, C, & D of the given quadrilateral.
AB=\ \sqrt{{(1+1)}^2+{(0+2)}^2}=\sqrt{4+4}=2\sqrt2\\ BC=\ \sqrt{{(-1-1)}^2+{(2-0)}^2}=\sqrt{4+4}=2\sqrt2\\ CD=\ \sqrt{{(-3+1)}^2+{(0-2)}^2}=\sqrt{4+4}=2\sqrt2\\ DA=\ \sqrt{{(-3+1)}^2+{(0-2)}^2}=\sqrt{4+4}=2\sqrt2\\ Diagonal\ AC=\ \sqrt{{(-1+1)}^2+{(2+2)}^2}=4\\ Diagonal\ BD=\ \sqrt{{(-3-1)}^2+{(0-0)}^2}=4
Length of Side = AB = BC = CD = DA = 2√2
Diagonal = AC = BD = 4
Thus, the given points are vertices of square.

Let the points (- 3, 5), (3, 1), (0, 3), and (– 1, – 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
AB=\ \sqrt{{(-3-3)}^2+{(1-5)}^2}=\sqrt{36+16}=2\sqrt{13}\\ BC=\ \sqrt{{(0-3)}^2+{(3-1)}^2}=\sqrt{9+4}=\sqrt{13}\\ CD=\ \sqrt{{(-1-0)}^2+{(-4-3)}^2}=\sqrt{1+49}=5\sqrt2\\ DA=\ \sqrt{{(-1+3)}^2+{(-4-5)}^2}=\sqrt{4+81}=\sqrt{85}
Points A, B, and C are also shown to be collinear.
As a result, the specified points can only form a triangle with three sides, rather than a quadrilateral with four sides.
As a result, the supplied locations cannot form a quadrilateral in general.

Let us consider, the points (4, 5), (7, 6), (4, 3), & (1, 2) be represents the vertices A, B, C, & D of the given quadrilateral.
AB=\ \sqrt{{(7-4)}^2+{(6-5)}^2}=\sqrt{9+1}=\sqrt{10}\\ BC=\ \sqrt{{(4-7)}^2+{(3-6)}^2}=\sqrt{9+9}=\sqrt{18}\\ CD=\ \sqrt{{(1-4)}^2+{(2-3)}^2}=\sqrt{9+1}=\sqrt{10}\\ DA=\ \sqrt{{(1-4)}^2+{(2-5)}^2}=\sqrt{9+9}=\sqrt{18}\\ Diagonal\ AC=\ \sqrt{{(4-4)}^2+{(3-5)}^2}=2\\ Diagonal\ BD=\ \sqrt{{(1-7)}^2+{(2-6)}^2}=13\sqrt2
This quadrilateral’s opposite sides are the same length. The diagonals, on the other hand, are of varying lengths. As a result, the specified points are the parallelogram’s vertices.

7. Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Solution:
To find a point on the x-axis, the y-coordinate should be 0. Let us consider, the point on x-axis be (x,0).
Let, A = (x, 0); B = (2, – 5) & C = (- 2, 9).
AB=\ \sqrt{{(2-x)}^2+{(-5-0)}^2}=\sqrt{{(2-x)}^2+25}\\ AC=\ \sqrt{{(-2-x)}^2+{(9-0)}^2}\ =\ \sqrt{{(-2-x)}^2+81}
As both the distance are equal, hence;
{(2-x)}^2+25\ =\ {(-2-x)}^2+81\\ x^2+\ 4\ – 4x + 25 = x2+ 4 + 4x + 81\\ 8x\ =\ 25\ – 81 = -56
x\ =\ -7
Hence, the point 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.
Solution:
Given: Distance between points (2, – 3) & (10, y) is 10.
Use distance formula, then
PQ=\ \sqrt{{(10-2)}^2+{(y+3)}^2}=\sqrt{{(8)}^2+{(y+3)}^2}\\ PQ\ =10\\ \sqrt{{(8)}^2+{(y+3)}^2}=10\\ {(8)}^2+{(y+3)}^2=100\\ {(y+3)}^2=36\\ y+3=\pm6
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 distance QR and PR.
Solution:
Given: point Q (0, 1) is equidistant from point P (5, – 3) & R (x, 6), thus PQ = QR
Now, determine the distance between PQ & QR by using distance formula,
PQ=\ \sqrt{{(5-0)}^2+{(-3-1)}^2}=\sqrt{25+16}=\sqrt{41}\\ QR=\ \sqrt{{(0-x)}^2+{(1-6)}^2}=\sqrt{x^2+25}
Now, use PQ = QR, then we get
\sqrt{x^2+25}=\sqrt{41}\\ x^2+25=41\\ x=\pm4
Coordinates of the Point R should be R (4, 6) or R (-4, 6),
If R (4, 6), then
QR=\ \sqrt{{(0-4)}^2+{(1-6)}^2}=\sqrt{16+25}=\sqrt{41}\\ PR=\ \sqrt{{(5-4)}^2+{(-3-6)}^2}=\sqrt{1+81}=\sqrt{82}
If R (-4, 6), then
QR=\ \sqrt{{(0+4)}^2+{(1-6)}^2}=\sqrt{16+25}=\sqrt{41}\\ PR=\ \sqrt{{(5+4)}^2+{(-3-6)}^2}=\sqrt{81+81}=9\sqrt2

10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Solution:
Point (x, y) is equidistant from the points (3, 6) and (– 3, 4).
\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}
Squaring both sides, then we get;
{(x-3)}^2+{(y-6)}^2={(x+3)}^2+{(y-4)}^2\\ x^2\ +\ 9\ – 6x + y^2+ 36 – 12y = x^2 + 9 + 6x + y^2+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.
Solution:
Let us consider, P (x, y) be the needed point. Now, use the section formula, then we get
x= (2\times 4 + 3\times(-1))(2 + 3) = (8 – 3)/5 = 1\\ y =(2\times-3 + 3\times 7)/(2 + 3) = (-6 +21)/5\ = 3
Thus, the point is (1, 3).

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

Let us consider, P (x_1, y_1) & Q (x_2, y_2) are the points of trisection of line segment joining the given points that is, AP = PQ = QB
Hence, point P divides AB internally in the ratio of 1:2.
x_2\ =\ (1\times(-2)\ +\ 2\times4)/3\ =\ (-2\ +\ 8)/3\ =\ 6/3\ =\ 2\\ y_1\ =\ (1\times(-3)\ +\ 2\times(-1))/(1\ +\ 2)\ =\ (-3\ – 2)/3 = -5/3
Thus;
P\ (x_1,\ y_1)\ =\ P(2,\ -5/3)
Point Q divides AB internally in the ratio 2:1.
x_2\ =\ (2\times(-2)\ +\ 1\times4)/(2\ +\ 1)\ =\ (-4\ +\ 4)/3\ =\ 0\\ y_2\ =\ (2\times(-3)\ +\ 1\times(-1))/(2\ +\ 1)\ =\ (-6\ – 1)/3 = -7/3
The coordinates of the point Q are (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 1/4 th 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 flag exactly halfway between the line segment joining the two flags, where should she post her flag?

From given instruction, we seen that Niharika posted green flag at 1/4th of the distance AD i.e., (1/4 \times 100) m = 25 m from the starting point of second line. Thus, the coordinates of this point are (2, 25).
Preet, too, placed a red flag at 1/5 of the distance AD, i.e., (1/5 \times 100) m = 20m from the 8th line’s starting point. As a result, this point’s coordinates are (8, 20).
Distance between the flags is determined by using the distance formula,
Distance=\ \sqrt{{(8-2)}^2+{(20-25)}^2}=\sqrt{36+25}=\sqrt{61}\ m
The point where Rashmi posts her blue flag is the mid-point of line joining these points. Let us consider this point as P (x, y).
x=(2 + 8)/2 = 10/2 = 5 & y =(20 + 25)/2 =45/2
Hence, P (x, y) = (5, 45/2)
Therefore, Rashmi will post her blue flag at 45/2 = 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:
Let us consider the ratio in which the line segment joining ( -3, 10) & (6, -8) is divided by the point ( -1, 6) be k :1.
Thus,
-1\ =\ (\ 6k-3)/(k+1)\\ –k – 1 = 6k -3\\ 7k\ =\ 2\\ k\ =\ 2/7
Therefore, the 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.
Solution:
Let us consider, the ratio in which line segment joining A (1, – 5) and B (– 4, 5) is divided by the x-axis is k : 1. Thus, the coordinates of point of division, let P(x, y) is ((-4k+1)/(k+1),\ (5k-5)/(k+1)) .
P(x,y)=\frac{-4k+1}{k+1},\ \frac{5k-5}{k+1}
We know, y – Coordinate of any point on the x-axis is 0.
Therefore,
(5k\ – 5)/(k + 1) = 0\\ 5k\ =\ 5\\ or\ k\ =\ 1
Hence, x – axis divides line segment in the ratio of 1:1.
Now, the coordinates of point of division is:
P\ (x,\ y)\ =\ ((-4(1)+1)/(1+1)\ ,\ (5(1)-5)/(1+1))\ =\ (-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.
Solution:
Let us consider, A, B, C & D are the points of a parallelogram such that A(1, 2), B(4, y), C(x, 6) and D(3, 5).

As the diagonals of parallelogram bisect one another, hence, the midpoint is same.
For the value of x & y, we need to solve the midpoint first.
Midpoint of AC\ =\ (\ (1+x)/2\ ,\ (2+6)/2\ )\ =\ ((1+x)/2\ ,\ 4)
Midpoint of \ BD\ =\ ((4+3)/2\ ,\ (5+y)/2\ )\ =\ (7/2\ ,\ (5+y)/2)
Midpoint of AC & BD are same, therefore,
(1+x)/2\ =\ 7/2\ and\ 4\ =\ (5+y)/2\\ x\ +\ 1\ =\ 7\ and\ 5\ +\ y\ =\ 8\\ x\ =\ 6\ and\ y\ =\ 3

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).
Solution:
Let us consider, the coordinates of the point A be (x, y).
Mid – Point of AB is (2, – 3), which is also the centre of circle.
Coordinates of B = (1, 4)
(2,\ -3)\ =((x+1)/2\ ,\ (y+4)/2)\\ (x+1)/2\ =\ 2\ and\ (y+4)/2\ =\ -3\\ x\ +\ 1\ =\ 4\ and\ y\ +\ 4\ =\ -6\\ x\ =\ 3\ and\ y\ =\ -10
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 = 3/7 AB and P lies on the line segment AB.
Solution:

Coordinates of the point A & B are (-2, -2) & (2, -4). As AP\ =\ 3/7\ AB
Therefore, AP:\ PB\ =\ 3:4
Point P divides line segment AB in the ratio of 3:4.
coordinate\ of\ P\ =\frac{3(2)+4(-2)}{3+4},\ \frac{3(-4)+4(-2)}{3+4}\ =\ -\frac{2}{7}\ ,\ -\frac{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.
Solution:
Draw a line dividing by 4 points.

From the figure, points X, Y, Z are dividing the line segment in the ratio of 1:3, 1:1, 3:1.
coordinate\ of\ X\ =\frac{1(2)+3(-2)}{1+3},\ \frac{1(8)+3(2)}{1+3}\ =\ (-1\ ,\ \frac{7}{2}) \\ coordinate\ of\ P\ =\frac{2(1)\ -\ 2(1)}{1+1},\ \frac{2(1)+8(1)}{1+1}\ =\ (0\ ,\ 5) \\ coordinate\ of\ P\ =\frac{3(2)+1(-2)}{1+3},\ \frac{3(8)+1(2)}{1+3}\ =\ (1\ ,\ \frac{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)
Solution:
Let us consider, A(3, 0), B (4, 5), C( – 1, 4) & D ( – 2, – 1) are the vertices of rhombus ABCD.

Length\ of\ Diagonal\ AC=\ \sqrt{{(3-(-1))}^2+{(0-4)}^2}=\sqrt{16+16}=4\sqrt2 \\ Length\ of\ Diagonal\ DB=\ \sqrt{{(4-(-2))}^2+{(5-(-1))}^2}=\sqrt{36+36}=6\sqrt2
Therefore, area of rhombus is
\frac{1}{2}\times4\sqrt2\times6\sqrt2=\ 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)
Solution:
Formula for Area of a triangle = 1/2 \times\ [x_1(y_2\ – y3) + x2(y3 – y1) + x3(y1 – y2)]
(i) Here,
x_1= 2, x_2 =-1, x_3 = 2, y_1 =3, y_2 = 0 & y_3 = -4
Put all the values in the formula, then we get;
Area\ of\ Triangle\ =\ 1/2\ [2\ {0-\ (-4)}\ +\ (-1)\ {(-4)\ – (3)} + 2 (3 – 0)]\\ =\ 1/2\ {8\ +\ 7\ +\ 6}\\ =\ 21/2
Hence, area of the triangle is 21/2 square units.

(ii) Here,
x_1=-5, x_2= 3, x_3 =5, y_1 =-1, y_2 =-5 & y_3 = 2
\\
Area of the triangle= 1/2 [-5 {\ (-5)-(2)} + 3(2-(-1))+ 5{-1 – (-5)}]
\\
=1/2{35 + 9 + 20}= 32

Hence, the area of triangle is 32 square 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)
Solution:
(i) For the Collinear Points, area of triangle is always zero.
Let the points (7, -2) (5, 1), & (3, k) are the vertices of a triangle.
Area\ of\ triangle\ =\ 1/2\ [7\ {\ 1-\ k}\ +\ 5(k-(-2))\ +\ 3{(-2)\ – 1}] = 0\\ 7\ – 7k + 5k +10 -9 = 0\\ -2k\ +\ 8\ =\ 0\\ k\ =\ 4

(ii) For the Collinear Points, area of triangle formed is zero.
Thus, for the points (8, 1), (k, – 4), & (2, – 5), area is 0
1/2\ [8\ {\ -4-\ (-5)}\ +\ k{(-5)-(1)}\ +\ 2{1\ -(-4)}] = 0\\ 8\ – 6k + 10 = 0\\ 6k\ =\ 18\\ 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.
Solution:
Let the vertices of triangle be A (0, -1), B (2, 1), & C (0, 3).
Let us consider, D, E, & F be the mid-points of the sides of the triangle.
Coordinates of D, E, & F are
D\ =\ (0+2/2,\ -1+1/2\ )\ =\ (1,\ 0)\\ E\ =\ (\ 0+0/2,\ -1+3/2\ )\ =\ (0,\ 1)
F\ =\ (\ 0+2/2,\ 3+1/2\ )\ =\ (1,\ 2)

Formula for Area of a triangle = 1/2\ \times\ [x_1(y_2\ – y3) + x2(y3 – y1) + x3(y1 – y2)] Area\ of\ \mathrm{\Delta DEF}\ =\ 1/2\ {1(2-1)\ +\ 1(1-0)\ +\ 0(0-2)}\ =\ 1/2\ (1+1)\ =\ 1
Area of ΔDEF is 1 Square Units
Area\ of\ \mathrm{\Delta ABC}\ =\ 1/2\ [0(1-3)\ +\ 2{3-(-1)}\ +\ 0(-1-1)] = 1/2 {8} = 4
Area of ΔABC is 4 Square Units
Thus, 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).
Solution:
Let the vertices of quadrilateral be A (- 4, – 2), B (– 3, – 5), C (3, – 2), & D (2, 3).
Join AC & divide the quadrilateral into 2 triangles.

We have 2 triangles ΔABC & ΔACD.
Formula for Area of a triangle = 1/2 \times [x_1(y_2 – y3) + x2(y3 – y1) + x3(y1 – y2)] \\
Area of \Delta ABC= 1/2[(-4) {(-5) – (-2)} + (-3) {(-2) – (-2)} + 3 {(-2) – (-5)}] \\ = 1/2 (12 + 0 +9)\\ = 21/2 Square Units
Area of \Delta ACD= 1/2 [(-4){(-2) – (3)} + 3{(3) – (-2)} + 2 {(-2) – (-2)}] \\ =1/2 (20 + 15 + 0)\\ =\ 35/2 square units
Area of Quadrilateral ABCD = Area of ABC+ Area of ACD
\\
= (21/2 + 35/2) square units = 28 square units

5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).
Solution:
Let us consider, the vertices of triangle be A (4, -6), B (3, -2), & C (5, 2).

Let D be the Mid – Point of side BC of the triangle ΔABC. Thus, the median in ΔABC is AD.
Coordinates of the Point D = Midpoint of BC = ((3+5)/2, (-2+2)/2) =(4,\ 0)
Formula for Area of a triangle = 1/2 \times\ [x_1(y_2 – y_3) + x_1(y_3 – y_1) + x_3(y_1 – y_2)]\\
Area of ABD = 1/2 [(4) {(-2) – (0)} + 3{(0) – (-6)} + (4) {(-6) – (-2)}]
=1/2 (-8 + 18 – 16)\\
= -3 Square Units

But the area cannot be negative. So, the area of ΔABD is 3 square units.

Area of ACD = 1/2 [(4) {0 – (2)} + 4{(2) – (-6)} + (5) {(-6) – (0)}]

= 1/2 (-8 + 32 – 30) = -3 Square Units

But the area cannot be negative. So, the area of ΔABD is 3 square units.
Area of both the sides is same. Hence, median AD divides ΔABC in two triangles of equal areas.Exercise 7.4

1. Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7).
Solution:
Consider the line 2x + y – 4 = 0 divides the line AB joined by two points A(2, -2) & B(3, 7) in ratio of k : 1.
Coordinates of the point of division is given as:
x = (2 + 3k)/(k + 1) & y = (-2 + 7k)/(k + 1)
put the values of x & y in the equation 2x\ +\ y\ – 4 = 0 , the we have
2{(2\ +\ 3k)/(k\ +\ 1)}\ +\ {(-2\ +\ 7k)/(k\ +\ 1)}\ – 4 = 0\\ (4\ +\ 6k)/(k\ +\ 1)\ +\ (-2\ +\ 7k)/(k\ +\ 1)\ =\ 4\\ 4\ +\ 6k\ – 2 + 7k = 4(k+1)\\ -2\ +\ 9k\ =\ 0\\ Or\ k\ =\ 2/9
Thus, the ratio is 2: 9.

2. Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
If given points are collinear then the area of the triangle formed by them should be zero.
Let us consider, (x, y), (1, 2) & (7, 0) are the vertices of a triangle,
Formula for Area of a triangle = 1/2 \times [x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)]
[x(2 – 0) + 1 (0 – y) + 7( y – 2)] = 0
2x – y + 7y – 14 = 0
2x+ 6y – 14 = 0
x+ 3y – 7 = 0.

3. Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).
Solution:
Suppose A = (6, -6), B = (3, -7), & C = (3, 3) are three points on a circle.
If O is the centre, then OA = OB = OC (radii are always equal)
If O = (x, y) then
OA=\ \sqrt{{(x-6)}^2+{(y+6)}^2}\\ OB=\ \sqrt{{(x-3)}^2+{(y+7)}^2}\\ OC=\ \sqrt{{(x-3)}^2+{(y-3)}^2}\\
Consider OA = OB, then we have after simplification;
-6x\ =\ 2y\ – 14 ….(1)
Similarly, OB = OC
\left(x-3\right)^2+\ \left(y+7\right)^2\ =\ \left(x-3\right)^2+\left(y-3\right)^2 \\ \left(y+7\right)^2\ =\ \left(y-3\right)^2\\ y^2+\ 14y\ +\ 49\ =\ y^2\ – 6y + 9\\ 20y\ =-40\\ or\ y\ =\ -2
Put the value of y in equation (1), then we get;
-6x\ =\ 2y\ – 14\\ -6x\ =\ -4\ – 14 = -18\\ x\ =\ 3
Therefore, the centre of circle located at the point (3, -2).

4. The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
Solution:
Let us consider, ABCD is a square, in which A(-1,2) & B(3,2). Point O is the point of intersection of AC & BD
To determine: Coordinates of the points B & D.

Step 1: Find the distance between A & C and the coordinates of the point O.
We know, the diagonals of a square are equal & bisect each other.
AC=\ \sqrt{{(3+1)}^2+{(2-2)}^2}=4
Coordinates of O
x\ =\ (3\ – 1)/2 = 1 and y = (2 + 2)/2 = 2
So, O (1,2)

Step 2: Find the side of square using the Pythagoras Theorem
Let us consider, a is the side of square & AC = 4
From Right Triangle, ACD,
a\ =\ 2\sqrt2
Therefore, each side of the square is 2\sqrt2

Step 3: Find the coordinates of the point D
Equate AD and CD
Let, the coordinates of D are (x_1,\ y_1)
AD=\ \sqrt{{(x_1+1)}^2+{(y_1-2)}^2}
Square both the sides,
{AD}^2={(x_1+1)}^2+{(y_1-2)}^2
Similarly,
{DC}^2={(x_1-3)}^2+{(y_1-2)}^2

As all the sides of a square are equal, therefore, AD = CD
{(x_1+1)}^2+{(y_1-2)}^2={(x_1-3)}^2+{(y_1-2)}^2\\ {x_1}^2+1+2x_1={x_1}^2+9-6x_1\\ 8x_1\ =\ 8\\ x_1=\ 1
From step 2, each side of the square = 2\sqrt2\\ {DC}^2={(x_1-3)}^2+{(y_1-2)}^2\\ 8={(1-3)}^2+{(y_1-2)}^2\\ 8\ =\ 4\ +\ {(y_1-2)}^2\\ y_1\ =\ 4
Hence, D = (1, 4)

Step 4: Find the coordinates of the point B
From the line segment, BOD
Coordinates of B;
O = (1, 2)
Let B (x_2,\ y_2)
For BD;
1 = (x_2 +1)/2\\ x_2= 1\\
&
2=( y_2\ + 4)/2\\ y_2 = 0
Therefore, the coordinates of the points are B = (1,0) & D = (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 lawn in the plot as shown in the fig. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of triangle PQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?

Solution:
(i) Taking A as origin, now, the coordinates of vertices P, Q & R are,
From the figure: P = (4, 6), Q = (3, 2), R (6, 5)
Here AD is x – axis & AB is y – axis.
Consider C as origin,
Coordinates of the vertices P, Q & R are ( 12, 2), (13, 6) & (10, 3).
Here CB is x – axis & CD is y – axis.
Area of triangle PQR when origin is A:
Formula for Area of a triangle = \frac{1}{2}\times [x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)]\\ = \frac{1}{2}\ [4(2\ – 5) + 3 (5 – 6) + 6 (6 – 2)]\\ =\ \ \frac{1}{2} (- 12 – 3 + 24 )\\ =9/2 sq \ unit

(ii) Area of triangle PQR in case of origin C:
Formula for Area of a triangle = \frac{1}{2}\ \times\ [x_1(y_2\ – y3) + x2(y3 – y1) + x3(y1 – y2)]
=\ \frac{1}{2}\ \ [\ 12(6\ – 3) + 13 ( 3 – 2) + 10( 2 – 6)]\\ =\ \frac{1}{2}\ (\ 36\ +\ 13\ – 40)\\ =\ 9/2\ sq\ unit
Therefore, Area of Triangle PQR at Origin A = Area of Triangle PQR at Origin C
Area is same in both the case as triangle remains the same no matter which point is taken as origin.

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 area of ∆ ABC. (Recall Theorem 6.2 and Theorem 6.6)
Solution:
Given: Vertices of a ∆ABC are A (4, 6), B (1, 5) & C (7, 2)

AD/AB\ =\ AE/AC\ =\ ¼ \\ AD/(AD\ +\ BD)\ =\ AE/(AE\ +\ EC)\ =\ 1/4
Point D & Point E divide AB & AC in ratio of 1 : 3.
Coordinates of D;
x\ =\ (m_1x_2\ +\ m_2x_1)/(m_1\ +\ m_2)\ and\ y\ =\ (m_1y_2\ +\ m_2y_1)/(m_1\ +\ m_2)
Here m_1 = 1 & m_2 =3
Consider the line segment AB that is divided by point D in the ratio of 1:3.
x\ =\ [3(4)\ +\ 1(1)]/4 = 13/4\\ y\ =\ [3(6)\ +\ 1(5)]/4 = 23/4
Similarly, Coordinates of E is given as:
x\ =\ [1(7)\ +\ 3(4)]/4 = 19/4\\ y\ =\ [1(2)\ +\ 3(6)]/4 = 20/4 = 5
Area of triangle:
Formula for Area of a triangle = \frac{1}{2}\ \times\ [x_1(y_2\ – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)]
Area of triangle ∆ ABC
=\ \ \frac{1}{2}\ [4(5\ – 2) + 1( 2 – 6) + 7( 6 – 5)]\\ =\ \frac{1}{2}\ \ (12\ – 4 + 7) = 15/2 square unit
Area of ∆ ADE
=\ \ \frac{1}{2}\ [4(23/4\ – 5) + 13/4 (5 – 6) + 19/4 (6 – 23/4)]\\ =\ \ \frac{1}{2}\ \ (3\ – 13/4 + 19/16)\\ =\ \ \frac{1}{2}\ \ (\ 15/16\ )\ =\ 15/32\ square\ unit
Therefore, ratio of area of triangle ADE to the area of triangle ABC 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 point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find the coordinates of point 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 called the centroid
and this point divides each median in the ratio 2 : 1.]
(v) If A (x_1,y_1),B(x_2,y_2) & C (x_3,y_3) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.
Solution:

(i) Coordinates of D:
Coordinates of D = (\ (6+1)/2,\ (5+4)/2\ )\ =\ (7/2,\ 9/2)
So, D is (7/2,\ 9/2)

(ii) Coordinates of P:
Coordinates of P = (\ [2(7/2)\ +\ 1(4)]/(2 + 1), [2(9/2) + 1(2)]/(2 + 1) ) = (11/3, 11/3)
Hence, P is (11/3,\ 11/3)

(iii) Coordinates of E:
Coordinates of E = (\ (4+1)/2,\ (2+4)/2\ )\ =\ (5/2,\ 6/2)\ =\ (5/2\ ,\ 3)
So, E is (5/2\ ,\ 3)
Point Q & point P would be coincident since medians of a triangle intersect each other at a common point which is called centroid. Coordinate of Q is given as:
Coordinates of Q = ([2(5/2)\ +\ 1(6)]/(2 + 1), [2(3) + 1(5)]/(2 + 1)) = (11/3, 11/3)
F is the mid - point of side AB
Coordinates of F = ((4+6)/2,\ (2+5)/2)\ =\ (5,\ 7/2)
Point R divides the side CF in the ratio of 2:1
Coordinates of R = (\ [2(5)\ +\ 1(1)]/(2 + 1), [2(7/2) + 1(4)]/(2 + 1) ) = (11/3, 11/3)

(iv) Coordinates of P, Q & R are same which shows that the medians intersect each other at the centroid of triangle.
(v) If A (x_1, y_1),B(x_2, y_2) & C (x_3, y_3) are vertices of the triangle ABC, the coordinates of centroid is given as:
x = (x_1 + x_2 + x_3)/3& y = ( y_1 + y_2+ y_3)/3

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

P is the mid - point of the side AB
Coordinate of P = ((-1\ – 1)/2, (-1 + 4)/2 ) = (-1, 3/2)
Similarly, Q, R & S are (As Q is mid - point of BC, R is mid - point of CD & S is midpoint of AD)
Coordinates of Q = (2, 4)
Coordinates of R = (5, 3/2)
Coordinates of S = (2, -1)
Now,
PQ=\ \sqrt{{(-1-2)}^2+{(3/2-4)}^2}=\sqrt{\frac{61}{4}}=\frac{\sqrt{61}}{2}\\ SP=\ \sqrt{{(2+1)}^2+{(-1-3/2)}^2}=\sqrt{\frac{61}{4}}=\frac{\sqrt{61}}{2}\\ QR=\ \sqrt{{(2-5)}^2+{(4-3/2)}^2}=\sqrt{\frac{61}{4}}=\frac{\sqrt{61}}{2}\\ RS=\ \sqrt{{(5-2)}^2+{(3/2+1)}^2}=\sqrt{\frac{61}{4}}=\frac{\sqrt{61}}{2}\\ Diagonal\ PR=\ \sqrt{{(-1-5)}^2+{(3/2-3/2)}^2}=6\\ Diagonal\ QS=\ \sqrt{{(2-2)}^2+{(4+1)}^2}=5
From above, PQ\ =\ SP\ =\ QR\ =\ RS\ =\ \sqrt61/2 , that is, all sides are equal.
But PR\ \neq\ QS that is, diagonals are not equal.
Thus, the given figure is a rhombus.