CBSE - Coordinate Geometry

(i) Distance between the points is given by

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

(ii) Distance between (−5, 7) and (−1, 3) is given by

(iii) Distance between (a, b) and (− a, − b) is given by

Distance between points (0, 0) and (36, 15)

Yes, 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 39 km.

Let the points (1, 5), (2, 3), and (- 2,-11) be representing the vertices A, B, and C of the given triangle respectively.Let A = (1, 5), B = (2, 3) and C = (- 2,-11)

Since AB + BC ≠ CA
Therefore, the points (1, 5), (2, 3), and ( - 2, - 11) are not collinear.

Let the points (5, - 2), (6, 4), and (7, - 2) are representing the vertices A, B, and C of the given triangle respectively.

Therefore, AB = BC
As two sides are equal in length, therefore, ABC is an isosceles triangle.

Clearly from the figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1). 

By using distance formula, we get

It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length.
Therefore, ABCD is a square and hence, Champa was correct

Let the points ( - 1, - 2), (1, 0), ( - 1, 2), and ( - 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.

(ii) 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.

It can be observed that all sides of this quadrilateral are of different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc.

(iii) Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

It can be observed that opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.

We have to find a point on x-axis. Therefore, its y-coordinate will be 0.
Let the point on x-axis be (x,0)

(x - 2)² + 25 = (x - 2)² + 81
x² + 4 - 4x + 25 = x² + 4 + 4x + 81
8x = 25 -81
8x = -56
x = -7
Therefore, the point is ( - 7, 0)

It is given that the distance between (2, −3) and (10, y) is 10.

PQ = QR

41 = x² + 25
16 = x²
x = ±4
Therefore, point R is (4, 6) or ( - 4, 6).
When point R is (4, 6),

Point (x, y) is equidistant from (3, 6) and ( - 3, 4).

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

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

Therefore, the point is (1, 3).

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.

It can be observed that Niharika posted the green flag at 1/4th of the distance AD i.e., (1×100/4)m = 25m from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25).
Similarly, Preet posted red flag at 1/5 of the distance AD i.e., (1×100/5) m = 20m from the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20).
Distance between these flags by using distance formula = GR

The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (x, y).

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

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.

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.

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

Mid point of diagonal AC is 

and Mid point of Diagonal BD is 

Since the diagonals of a parallelogram bisect each other, the mid point of AC and BD are same.

∴ x+1/2 = 7/2 and 4 = 5+y/2 ⇒ x + 1 = 7 and 5 + y = 8

⇒ x = 6 and y = 3

Let the coordinates of point A be (x, y).
Mid-point of AB is (2, - 3), which is the center of the circle.

⇒ x + 1 = 4 and y + 4 = -6

⇒ x = 3 and y = -10
Therefore, the coordinates of A are (3,-10).

The coordinates of point A and B are (-2,-2) and (2,-4) respectively.

Since AP = 3/7 AB

Therefore, AP:PB = 3:4

Point P divides the line segment AB in the ratio 3:4.

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.

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

(i) Area of a triangle is given by

Area of triangle = 1/2 {x₁ (y₂ - y₃)+ x₂ (y₃ - y₁)+ x₃ (y₁ - y₂)}
Area of the given triangle = 1/2 [2 { 0- (-4)} + (-1) {(-4) - (3)} + 2 (3 - 0)]
                                          = 1/2 {8 + 7 + 6}
                                          = 21/2 square units.

(ii) Area of the given triangle = 1/2 [-5 { (-5)- (4)} + 3(2-(-1)) + 5{-1 - (-5)}]
                                                = 1/2{35 + 9 + 20}
                                                = 32 square units

(i) For collinear points, area of triangle formed by them is zero.
Therefore, for points (7, -2) (5, 1), and (3, k), area = 0
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 collinear points, area of triangle formed by them is zero.
Therefore, for points (8, 1), (k, - 4), and (2, - 5), area = 0
1/2 [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0
8 - 6k + 10 = 0
6k = 18
k = 3

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)

F = (2+0/2 , 1+3/2) = (1,2)

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.

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 square units
Area of ΔACD = 1/2 [(-4) {(-2) - (3)} + 3{(3) - (-2)} + 2 {(-2) - (-2)}]
                         = 1/2 (20+15+0)
                         = 35/2 square units
Area of ���ABCD  = Area of ΔABC + Area of ΔACD
                             = (21/2 + 35/2) square units = 28 square units

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 ΔABD = 1/2 [(4) {(-2) - (0)} + 3{(0) - (-6)} + (4) {(-6) - (-2)}]

                       = 1/2 (-8+18-16)
                       = -3 square units
However, area cannot be negative. Therefore, area of ΔABD is 3 square units.
Area of ΔABD = 1/2 [(4) {0 - (2)} + 4{(2) - (-6)} + (5) {(-6) - (0)}]
                         = 1/2 (-8+32-30)
                         = -3 square 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.