Coordinates Geometry

Given coordinate (10,-3) lies in Quadrant IV because as shown in the figure that those coordinate who have (+, -) sign lies in IV quadrants, or can say whose x-axis is “+” and y-axis is “–“ lies in IV Quadrant.

Given coordinate (-4,-6) lies in Quadrant III because as shown in the figure that those points which have a sign like this (-, -) or can say whose both x-axis and y-axis is “–“ lies in III quadrants. So (-4,-6) lies in III Quadrant.

Given coordinate (-8,6) lies in Quadrant II because as shown in the figure that those points which have a sign like this (-, +) lies in II quadrants. So (-8,6) lies in III Quadrant. Here also x-axis point is “-” and y-axis point is “+” so (-8,6) lies in Quadrant II.

Given coordinate lies, in Quadrant I because as shown in the figure that those coordinate who have (+, +) sign lies in IV quadrants or can say whose x-axis is “+” and y-axis is also “+“ lies in I Quadrant.

Given coordinate is (3,0). This point lies on the x-axis because its y-axis is on origin. Therefore, it lies on the x-axis between the Quadrant I and Quadrant IV.

Given coordinate is (0, -5). This point lies on the y-axis because its x-axis is on origin. Therefore, it lies on the y-axis between the Quadrant III and Quadrant IV.

Here is the graph for coordinate (4,5)

The point having -5 lies on the on the y-axis on the negative side because here x-axis is 0 and when x-axis is 0 then points lies on the y-axis and when y-axis is 0 then point lies on the x-axis.

We can show it on graph with points (0, -5).

Here we have three vertices of rectangle say A (-2, 0) B (2, 0) and C (2, 1) so when we start graphing it on the graph as shown in the graph below then after plotting all three vertices you will get something like this.

Therefore, after joining all the vertices with a line segment, we will get our fourth vertex because in rectangle opposites sides are parallel and all the angles are right angle so, by joining all the lines according to properties of the rectangle you will get the fourth vertex. So fourth vertex of a rectangle is (-2, 1).

It is easy to draw a triangle when it’s all vertices are given. We have to just locate all the given points on the graph and join them with a line as shown in the graph below.

Step 1. Locate all the vertices on the graph.

Step 2. Joins all the vertices with a line and it will form a triangle.

Given that the base of the equilateral triangle is on the y-axis and mid-point of the base is at the origin, so its figure will be like this as shown.

So here, O(0,0) is the midpoint of the base.

An equilateral triangle has all sides equal so if O is the mid-point of the base BC, so B and C are the two vertices of the triangle. Now we have two vertices of the triangle, which is the base of equilateral triangle lying on the y-axis. Now if base in on y-axis then x-axis are as bisector of the base and so our third vertices will be on the x-axis either left or right.

So now in right ∆BOA

Pythagoras Theorem: In a right-angled triangle the square of the biggest side(hypotenuse) equals the sum of the squares of the other two sides(Perpendicular and base).

BO² + OA² = AB² { By Pythagoras theorem}

a² + OA² = (2a) ²

OA² = 4a²- a²

OA² = 3a²

OA = ±a√3

So vertices of triangle are A(±a√3,0) B(0,a)and C(0,-a).

Since AB and AD both have as an endpoint, we can find the coordinate of A by finding the intersection of the two sides.

So the coordinates of A will be where the x-axis and y-axis intersect.

As we know AB lies on the x-axis so the coordinates of B can be found by using the coordinates of A and changing the x-coordinate by the measure of AB.

As we know the opposite side of a rectangle, AD and BC are congruent. Now if we have a measure of BC, we can simply find the y-coordinate of D.

Since AB and AD are the x and y-axes, A is at(0,0), B is at (10,0), C is at (10,8), and D is at (0,8).

Square ABCD. Center = O(0,0) Origin.

AB = BC = 20 units.

Y-coordinates of AB = = -10

Y-coordinates of AD = = 10

∴the coordinates are :-

A(-10,-10)

B(10,-10)

C(10,10)

D(-10,10)

Given points are (0, 0) and (- 5, 12)

We need to find the distance between these two points.

We know that distance(S) between the points (x₁, y₁) and (x₂, y₂) is

⇒ S = √169

⇒ S = 13

∴ The distance between the points (0, 0) and (- 5, 12) is 13 units.

Given points are (4, 5) and (- 3, 2)

We need to find the distance between these two points.

We know that distance(S) between the points (x₁, y₁) and (x₂, y₂) is

⇒ S = √58

∴ The distance between the points (4, 5) and (- 3, 2) is √58 units.

Given points are (5, - 12) and (9, - 9)

We need to find the distance between these two points.

We know that distance(S) between the points (x₁, y₁) and (x₂, y₂) is

⇒ S = √25

⇒ S = 5

∴ The distance between the points (5, - 12) and (9, - 9) is 5 units.

Given points are (- 3, 4) and (3, 0)

We need to find the distance between these two points.

We know that distance(S) between the points (x₁, y₁) and (x₂, y₂) is

⇒ S = √52

⇒

⇒ S = 2√13

∴ The distance between the points (- 3, 4) and (3, 0) is 2√13 units.

Given points are (2, 3) and (4, 1)

We need to find the distance between these two points.

We know that distance(S) between the points (x₁, y₁) and (x₂, y₂) is

⇒ S = √8

⇒

⇒ S = 2√2

∴ The distance between the points (2, 3) and (4, 1) is 2√2 units.

Given points are (a, b) and (- a, - b)

We need to find the distance between these two points.

We know that distance(S) between the points (x₁, y₁) and (x₂, y₂) is

∴ The distance between the points (a, b) and (- a, - b) is .

Given that we need to show that the points (1, - 1), (- 5, 7) and (2, 6) are equidistant from the point (- 2, 3).

We know that distance between two points (x₁, y₁) and (x₂, y₂) is

Let S₁ be the distance between the points (1, - 1) and (- 2, 3)

⇒ S_{1 =} 5 ..... (1)

Let S₂ be the distance between the points (- 5, 7) and (- 2, 3)

⇒ S_{2 =} 5 ..... (2)

Let S₃ be the distance between the points (2, 6) and (- 2, 3)

⇒ S_{3 =} 5 ..... (3)

From (1), (2), and (3) we got S_{1 =} S_{2 =} S₃ which tells us that (1, - 1), (5, 7) and (2, 5) are equidistant from (- 2, 3).

Given that the distance between the points (a, 2) and (3, 4) is 8.

We need to find the value of a.

We know that distance(S) between the points (x₁, y₁) and (x₂, y₂) is

⇒ 8^{2 =} (a - 3)² + (- 2)²

⇒ 64 = (a - 3)^{2 +} 4

⇒ (a - 3)^{2 =} 60

⇒ a – 3 = ± √60

⇒ a = 3 ± √60

∴ The values of a are 3±√60.

Given that the line has length of 10 units and one of its ends is (- 2, 3).

It is also given that the ordinate of the other end is 9. Let us assume the other end is (x, 9).

We know that distance(S) between the points (x₁, y₁) and (x₂, y₂) is

⇒

⇒ 10 = (x + 2)² + (- 6)²

⇒ 100 = (x + 2)² + 36

⇒ (x + 2)² = 64

⇒ x + 2 = ±8

⇒ x = - 2 + 8 (or) x = - 2 - 8

⇒ x = 6 or 10

∴ Thus proved.

Given that the distance between the points P(2, - 3) and Q(10, y) is 10.

We need to find the value of y.

We know that distance(S) between the points (x₁, y₁) and (x₂, y₂) is

⇒

⇒ 10² = (- 8)² + (3 + y)²

⇒ 100 = 64 + (3 + y)²

⇒ (3 + y)² = 36

⇒ 3 + y = ±6

⇒ y = 3 + 6 (or) y = 3 - 6

⇒ y = 9 (or) y = - 3

∴ The values of y are 9, - 3.

Given points are (at₁², 2at₁) and (at₂², 2at₂).

We need to find the distance between these two points.

We know that distance(S) between the points (x₁, y₁) and (x₂, y₂) is

∴ The distance between the points (at₁², 2at₁) and (at₂², 2at₂) is .

Given points are (a - b, b - a) and (a + b, a + b).

We need to find the distance between these two points.

We know that distance(S) between the points (x₁, y₁) and (x₂, y₂) is

∴ The distance between the points (a - b, b - a) and (a + b, a + b) is .

Given points are (cosθ, sinθ) and (sinθ, cosθ).

We need to find the distance between these two points.

We know that distance(S) between the points (x₁, y₁) and (x₂, y₂) is

∴ The distance between the points (cosθ, sinθ) and (sinθ, cosθ) is .

Given points are A(7, 6) and B(- 3, 4).

We need to find a point on x - axis which is equidistant from these points.

Let us assume the point on x - axis be S(x, o).

We know that distance between the points (x₁, y₁) and (x₂, y₂) is .

From the problem,

⇒ SA = SB

⇒ SA² = SB²

⇒ (x - 7)² + (0 - 6)² = (x - (- 3))² + (0 - 4)²

⇒ (x - 7)² + (- 6)² = (x + 3)² + (- 4)²

⇒ x² - 14x + 49 + 36 = x² + 6x + 9 + 16

⇒ 20x = 60

⇒ x = 3

∴ The point on x - axis is (3, 0).

Given points are A(3, 2) and B(- 5, - 2).

We need to find a point on x - axis which is equidistant from these points.

Let us assume the point on x - axis be S(x, o).

We know that distance between the points (x₁, y₁) and (x₂, y₂) is .

From the problem,

⇒ SA = SB

⇒ SA² = SB²

⇒ (x - 3)² + (0 - 2)² = (x + 5)² + (0 - (- 2))²

⇒ (x - 3)² + (- 2)² = (x + 5)² + (2)²

⇒ x² - 6x + 9 + 4 = x² + 10x + 25 + 4

⇒ 16x = - 16

⇒ x = - 1

∴ The point on x - axis is (- 1, 0).

Given points are A(2, - 5) and B(- 2, 9).

We need to find a point on x - axis which is equidistant from these points.

Let us assume the point on x - axis be S(x, o).

We know that distance between the points (x₁, y₁) and (x₂, y₂) is .

From the problem,

⇒ SA = SB

⇒ SA² = SB²

⇒ (x - 2)² + (0 - (- 5))² = (x - (- 2))² + (0 - 9)²

⇒ (x - 2)² + (5)² = (x + 2)² + (- 9)²

⇒ x² - 4x + 4 + 25 = x² + 4x + 4 + 81

⇒ 8x = - 56

⇒ x = - 7

∴ The point on x - axis is (- 7, 0).

Given points are A(- 5, - 2) and B(3, 2).

We need to find a point on y - axis which is equidistant from these points.

Let us assume the point on y - axis be S(0, y).

We know that distance between the points (x₁, y₁) and (x₂, y₂) is .

From the problem,

⇒ SA = SB

⇒ SA² = SB²

⇒ (0 - (- 5))² + (y - (- 2))² = (0 - 3)² + (y - 2)²

⇒ (5)² + (y + 2)² = (- 3)² + (y - 2)²

⇒ 25 + y² + 4y + 4 = 9 + y² - 4y + 4

⇒ 8y = - 16

⇒ y = - 2

∴ The point on y - axis is (0, - 2).

Given points are A(6, 5) and B(- 4, 3).

We need to find a point on y - axis which is equidistant from these points.

Let us assume the point on y - axis be S(0, y).

We know that distance between the points (x₁, y₁) and (x₂, y₂) is .

From the problem,

⇒ SA = SB

⇒ SA² = SB²

⇒ (0 - 6)² + (y - 5)² = (0 - (- 4))² + (y - 3)²

⇒ (- 6)² + (y - 5)² = (4)² + (y - 3)²

⇒ 36 + y² - 10y + 25 = 16 + y² - 6y + 9

⇒ 4y = 36

⇒ y = 9

∴ The point on y - axis is (0, 9).

Given points are A(3, 5), B(1, 1) and C(- 2, - 5).

We need to check whether these points are collinear.

We know that for three points A, B and C to be collinear, the criteria to be satisfied is AC = AB + BC.

Let us find the distances first,

We know that distance between two points (x₁, y₁) and (x₂, y₂) is

⇒ AC = 5√5 ..... (1)

⇒ AB = 2√5 ..... (2)

⇒ BC = 3√5 ..... (3)

From (1), (2), (3) we can see that AB + BC = AC.

∴ The three points are collinear.

Given points are A(5, 1), B(1, - 1) and C(11, 4).

We need to check whether these points are collinear.

We know that for three points A, B and C to be collinear, the criteria to be satisfied is a linear relationship between AB, BC and AC.

Let us find the distances first,

We know that distance between two points (x₁, y₁) and (x₂, y₂) is

⇒ AC = 3√5 ..... (1)

⇒ AB = 2√5 ..... (2)

⇒ BC = 5√5 ..... (3)

From (1), (2), (3) we can see that AB + AC = BC.

∴ The three points are collinear.

Given points are A(0, 0), B(9, 6) and C(3, 2).

We need to check whether these points are collinear.

We know that for three points A, B and C to be collinear, the criteria to be satisfied is a linear relationship between AB, BC and AC.

Let us find the distances first,

We know that distance between two points (x₁, y₁) and (x₂, y₂) is

⇒ AC = √13 ..... (1)

⇒ AB = 3√13 ..... (2)

⇒ BC = 2√13 ..... (3)

From (1), (2), (3) we can see that AB = BC + AC.

∴ The three points are collinear.

Given points are A(- 1, 2), B(5, 0) and C(2, 1).

We need to check whether these points are collinear.

We know that for three points A, B and C to be collinear, the criteria to be satisfied is a linear relationship between AB, BC and AC.

Let us find the distances first,

We know that distance between two points (x₁, y₁) and (x₂, y₂) is

⇒ AC = √10 ..... (1)

⇒ AB = 2√10 ..... (2)

⇒ BC = √10 ..... (3)

From (1), (2), (3) we can see that AC + BC = AB.

∴ The three points are collinear.

Given points are A(1, 5), B(2, 3) and C(- 2, - 11).

We need to check whether these points are collinear.

We know that for three points A, B and C to be collinear, the criteria to be satisfied is AC = AB + BC.

Let us find the distances first,

We know that distance between two points (x₁, y₁) and (x₂, y₂) is

⇒ AC = √265 ..... (1)

⇒ AB = √5 ..... (2)

⇒ BC = 2√53 ..... (3)

From (1), (2), (3) we can see that we cannot get any linear relationship.

∴ The three points are not collinear.

Given points are A(6, 1), B(1, 3) and C(x, 8). We need to find the value of x such that AB = BC.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is

⇒ AB = BC

⇒ AB² = BC²

⇒ (6 - 1)² + (1 - 3)² = (1 - x)² + (3 - 8)²

⇒ (5)² + (- 2)² = (1 - x)² + (- 5)²

⇒ 25 + 4 = 1 - 2x + x² + 25

⇒ x² - 2x - 3 = 0

⇒ x² - 3x + x - 3 = 0

⇒ x(x - 3) + 1(x - 3) = 0

⇒ (x + 1)(x - 3) = 0

⇒ x + 1 = 0 (or) x - 3 = 0

⇒ x = - 1 (or) x = 3

∴ The values of x are - 1 or 3.

Given points are A(a + rcosθ, b + rsinθ) and B(a, b).

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is

⇒ AB = r

We can see that AB is independent of θ.

∴ Thus proved.

Given points are A(cosec²θ, 0), B(0, sec²θ ) and C(1, 1).

We need to check whether these points are collinear.

We know that for three points A, B and C to be collinear, the criteria to be satisfied is AB = AC + BC.

Let us find the distances first,

We know that distance between two points (x₁, y₁) and (x₂, y₂) is

..... (1)

.... - (2)

⇒

⇒

⇒ ..... (3)

Now,

⇒ AC + BC = AB

∴ The three points are collinear.

Given that circle passes through the points A(6, 2), B(0, 4), C(4, 6).

Let us assume O(x, y) be the centre of the circle.

We know that distance from the centre to any point on h circle is equal.

So, OA = OB = OC

We know that distance between two points (x₁, y₁) and (x₂, y₂) is

Now,

⇒ OA = OB

⇒ OA² = OB²

⇒ (x - 6)² + (y - 2)² = (x - 0)² + (y - 4)²

⇒ x² - 12x + 36 + y² - 4y + 4 = x² + y² - 8y + 16

⇒ 12x - 4y = 24

⇒ 3x - y = 6 ..... (1)

Now,

⇒ OB = OC

⇒ OB² = OC²

⇒ (x - 0)² + (y - 4)² = (x - 4)² + (y - 6)²

⇒ x² + y² - 8y + 16 = x² - 8x + 16 + y² - 12y + 36

⇒ 8x + 4y = 36

⇒ 2x + y = 9 .... - (2)

On solving (1) and (2), we get

⇒ x = 3 and y = 3

∴ (3, 3) is the centre of the circle.

We know radius is the distance between the centre and any point on the circle.

Let ‘r’ be the radius of the circle.

⇒ r = √10

∴ The radius of the circle is √10.

Given points are A(2, 3) and B(6, - 1). It is told that S(x, y) is equidistant from A and B.

So, we get SA = SB,

We know that distance between two points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒ SA = SB

⇒ SA² = SB²

⇒ (x - 2)² + (y - 3)² = (x - 6)² + (y - (- 1))²

⇒ (x - 2)² + (y - 3)² = (x - 6)² + (y + 1)²

⇒ x² - 4x + 4 + y² - 6y + 9 = x² - 12x + 36 + y² + 2y + 1

⇒ 8x - 8y = 24

⇒ x - y = 3

∴ The relation between x and y is x - y = 3.

Given points are A(7, 1) and B(3, 5). It is told that S(x, y) is equidistant from A and B.

So, we get SA = SB,

We know that distance between two points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒ SA = SB

⇒ SA² = SB²

⇒ (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²

⇒ x² - 14x + 49 + y² - 2y + 1 = x² - 6x + 9 + y² - 10y + 25

⇒ 8x - 8y = 16

⇒ x - y = 2

∴ The relation between x and y is x - y = 2.

Given points are A(3, 6) and B(- 3, 4). It is told that S(x, y) is equidistant from A and B.

So, we get SA = SB,

We know that distance between two points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒ SA = SB

⇒ SA² = SB²

⇒ (x - 3)² + (y - 6)² = (x - (- 3))² + (y - 4)²

⇒ (x - 3)² + (y - 6)² = (x + 3)² + (y - 4)²

⇒ x² - 6x + 9 + y² - 12y + 36 = x² + 6x + 9 + y² - 8y + 16

⇒ 12x + 4y = 20

⇒ 3x + y = 5

∴ Thus proved.

Given points are A(a + b, b - a) and B(a - b, a + b). It is told that S(x, y) is equidistant from A and B.

So, we get SA = SB,

We know that distance between two points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒ SA = SB

⇒ SA² = SB²

⇒ (x - (a + b))² + (y - (b - a))² = (x - (a - b))² + (y - (a + b))²

⇒ x² - 2(a + b)x + (a + b)² + y² - 2(b - a)y + (b - a)² = x² - 2(a - b)x + (a - b)² + y² - 2(a + b)y + (a + b)²

⇒ x(- 2a - 2b + 2a - 2b) = y(2b - 2a - 2a - 2b)

⇒ x(- 4b) = y(- 4a)

⇒ x(b) = y(a)

⇒

Applying componendo and dividendo,

⇒

∴ Thus proved.

Given points are A(3, 4), B(8, - 6) and C(13, 9).

Let us find the distance between sides AB, BC and CA.

We know that distance between the two points (x₁, y₁) and (x₂, y₂) is .

⇒ AB = √125

⇒ BC = √250

⇒ CA = √125

Now,

⇒ AB² + CA² = 125 + 125

⇒ AB² + CA² = 250

⇒ AB² + CA² = BC²

∴ The given points form a right angled isosceles triangle.

Given points are A(1, 1), B(- √3, √3) and C(- 1, - 1).

Let us find the distance between sides AB, BC and CA.

We know that the distance between the two points (x₁, y₁) and (x₂, y₂) is

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AB = BC = CA

∴ The given points form an equilateral triangle.

Given points are A(0, 2), B(7, 0) and C(2, 5).

Let us find the distance between sides AB, BC and CA.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AB≠BC≠CA

∴ The given points form a scalene triangle.

Given points are A(- 2, 5), B(7, 10) and C(3, - 4).

Let us find the distance between sides AB, BC and CA.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AB = CA

Now,

⇒

⇒ AB² + CA² = 106 + 106

⇒ AB² + CA² = 212

⇒

⇒ AB² + CA² = BC²

∴ The given points form a right angles isosceles triangle.

Given points are A(4, 4), B(3, 5) and C(- 1, - 1).

Let us find the distance between sides AB, BC and CA.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Now,

⇒

⇒ AB² + CA² = 2 + 50

⇒ AB² + CA² = 52

⇒

⇒ AB² + CA² = BC²

∴ The given points form a right - angled triangle.

Given points are A(1, 2), B(3, 0) and C(- 1, 0).

Let us find the distance between sides AB, BC and CA.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒

⇒

⇒ AB = 4

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒ CA = 4

We got AB = BC = CA

∴ The given points form an equilateral triangle.

Given points are A(0, 6), B(- 5, 3) and C(3, 1).

Let us find the distance between sides AB, BC and CA.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AB = CA

Now,

⇒

⇒

⇒ AB² + CA² = 68

⇒

⇒ AB² + CA² = BC²

∴ The given points form a right - angled isosceles triangle.

Given points are A(5, - 2), B(6, 4) and C(7, - 2).

Let us find the distance between sides AB, BC and CA.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AB = BC≠CA

∴ The given points form an isosceles triangle.

Given points are A(at², 2at), B and C(a, 0).

Let us find the distance between sides AB, BC and CA.

We know that distance between the two points (x₁, y₁) and (x₂, y₂) is .

⇒ AC = at² + a

Now,

is independent of t.

Given that A(0, 0) and B(3, ) are two vertices of an equilateral triangle.

Let us assume C(x, y) be the third vertex of the triangle.

We have AB = BC = CA

We know that the distance between the two points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒ BC = CA

⇒ BC² = CA²

⇒ (3 - x)² + ( - y)² = (x - 0)² + (y - 0)²

⇒ x² - 6x + 9 + 3 + y² - 2y = x² + y²

⇒ 6x = 12 - 2y

..... - (1)

⇒ AB = BC

⇒ AB² = BC²

⇒ (0 - 3)² + (0 - )² = (3 - x)² + ( - y)²

⇒ 9 + 3 = 9 - 6x + x² + 3 - 2y + y²

From (1)

⇒

⇒ 48y² - 48√3y - 288 = 0

⇒ y²–√3y –6 = 0

⇒ y² - 2√3y + √3y - 6 = 0

⇒ y(y - 2√3) + √3(y - 2√3) = 0

⇒ (y + √3)(y - 2√3) = 0

⇒ y + √3 = 0 (or) y - 2√3 = 0

⇒ y = - √3 (or) y = 2√3

From (1), for y =

⇒ x = 3

From (1), for y = 2

⇒ x = 0

∴ The third vertex of equilateral triangle is (0, 2√3) and (3, √3).

Given that we need to find the circum - centre and circum - radius of the triangle whose vertices are A(- 2, 3), B(2, - 1), C(4, 0).

Let us assume O(x, y) be the Circum - centre of the circle.

We know that distance from circum - centre to any vertex is equal.

So, OA = OB = OC

We know that distance between two points (x₁, y₁) and (x₂, y₂) is

Now,

⇒ OA = OB

⇒ OA² = OB²

⇒ (x - (- 2))² + (y - 3)² = (x - 2)² + (y - (- 1))²

⇒ (x + 2)² + (y - 3)² = (x - 2)² + (y + 1)²

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

⇒ 8x - 8y = - 8

⇒ x - y = - 1 ..... (1)

Now,

⇒ OB = OC

⇒ OB² = OC²

⇒ (x - 2)² + (y - (- 1))² = (x - 4)² + (y - 0)²

⇒ (x - 2)² + (y + 1)² = (x - 4)² + (y)²

⇒ x² - 4x + 4 + y² + 2y + 1 = x² - 8x + 16 + y²

⇒ 4x + 2y = 11 .... - (2)

On solving (1) and (2), we get

⇒ and

∴ is the centre of the circle.

We know radius is the distance between the centre and any point on the circle.

Let ‘r’ be the circum - radius of the circle.

∴ The radius of the circle is .

Given that circle passes through the points A(6, - 6), B(3, - 7), C(3, 3).

Let us assume O(x, y) be the centre of the circle.

We know that distance from the centre to any point on h circle is equal.

So, OA = OB = OC

We know that distance between two points (x₁, y₁) and (x₂, y₂) is

Now,

⇒ OA = OB

⇒ OA² = OB²

⇒ (x - 6)² + (y - (- 6))² = (x - 3)² + (y - (- 7))²

⇒ (x - 6)² + (y + 6)² = (x - 3)² + (y + 7)²

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

⇒ 6x + 2y = 14

⇒ 3x + y = 7 ..... (1)

Now,

⇒ OB = OC

⇒ OB² = OC²

⇒ (x - 3)² + (y - (- 7))² = (x - 3)² + (y - 3)²

⇒ (x - 3)² + (y + 7)² = (x - 3)² + (y - 3)²

⇒ x² - 6x + 9 + y² + 14y + 49 = x² - 6x + 9 + y² - 6y + 9

⇒ 20y = - 40

⇒ y = - 2 .... - (2)

Substituting (2) in (1), we get

⇒ x = 3

∴ (3, - 2) is the centre of the circle.

We know radius is the distance between the centre and any point on the circle.

Let ‘r’ be the radius of the circle.

⇒ r = √(9 + 16)

⇒ r = √25

⇒ r = 5

∴ The radius of the circle is 5.

Given that the line segment joining the points A(a, b) and B(c, d) subtends a right angle at the origin O(0, 0)

So, AOB is a right angled triangle with right angle at O.

We got OA² + OB² = AB² [By Pythagoras Theorem]

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

⇒ OA² + OB² = AB²

⇒ (0 - a)² + (0 - b)² + (0 - c)² + (0 - d)² = (a - c)² + (b - d)²

⇒ a² + b² + c² + d² = a² + c² - 2ac + b² + d² - 2bd

⇒ 2ac + 2bd = 0

⇒ ac + bd = 0

Given that the circle has centre O(2x - 1, 3x + 1) and passes through the point A(- 3, - 1) and has a radius(r) of 10 units.

We know that the radius of the circle is the distance between the centre and any point on the circle.

So, we have r = OA

⇒ OA = 10

⇒ OA² = 100

⇒ (2x - 1 - (- 3))² + (3x + 1 - (- 1))² = 100

⇒ (2x + 2)² + (3x + 2)² = 100

⇒ 4x² + 8x + 4 + 9x² + 12x + 4 = 100

⇒ 13x² + 20x - 92 = 0

⇒ 13x² - 26x + 46x - 92 = 0

⇒ 13x(x - 2) + 46(x - 2) = 0

⇒ (13x + 46)(x - 2) = 0

⇒ 13x + 46 = 0 (or) x - 2 = 0

⇒ 13x = - 46 (or) x = 2

∴ The values of the x are or 2.

Given points are A(4, 3), B(6, 4), C(5, 6) and D(3, 5).

We need to prove that these are the vertices of a square.

We know that in the lengths of all sides are equal and the lengths of the diagonals are equal.

Let us find the lengths of the sides.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒ AB = √5

⇒ BC = √5

⇒ CD = √5

⇒ DA = √5

We got AB = BC = CD = DA, this may be square (or) rhombus.

Now we find the lengths of the diagonals.

⇒ AC = √10

⇒ BD = √10

We got AC = BD.

∴ The points form a square.

Given points are A(4, 3), B(6, 4), C(5, 6) and D(- 4, 4).

We need to prove that these are the vertices of a square.

We know that in the lengths of all sides are equal and the lengths of the diagonals are equal.

Let us find the lengths of the sides.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AB = BC≠CD≠DA,

∴ The points doesn’t form a square.

Given points are A(3, 2), B(6, 3), C(7, 6) and D(4, 5).

We need to prove that these are the vertices of a parallelogram.

We know that in the lengths of opposite sides are equal in a parallelogram.

Let us find the lengths of the sides.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒ AB = √10

⇒ BC = √10

⇒ CD = √10

⇒ DA = √10

We got AB = CD and BC = DA, these are the vertices of a parallelogram.

Now we find the lengths of the diagonals.

⇒ AC = √32

⇒ BD = √8

We got AC≠BD.

∴ The points doesn’t form a rectangle.

Given points are A(6, 8), B(3, 7), C(- 2, - 2) and D(1, - 1).

We need to prove that these are the vertices of a parallelogram.

We know that in the lengths of opposite sides are equal in a parallelogram and the lengths of diagonals are not equal.

Let us find the lengths of the sides.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AB = CD and BC = DA, these are the vertices of a parallelogram or rectangle.

Now we find the lengths of the diagonals.

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AC≠BD.

∴ The points form a parallelogram.

Given points are A(4, 8), B(0, 2), C(3, 0) and D(7, 6).

We need to prove that these are the vertices of a rectangle.

We know that in the lengths of opposite sides and lengths of diagonals are equal in a rectangle.

Let us find the lengths of the sides.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒ AB = √52

⇒ BC = √13

⇒ CD = √52

⇒ DA = √13

We got AB = CD and BC = DA, these are the vertices of a parallelogram or a rectangle.

Now we find the lengths of the diagonals.

We got AC = BD.

∴ The points form a rectangle.

Given points are A(1, 0), B(5, 3), C(2, 7) and D(- 2, 4).

We need to prove that these are the vertices of a rhombus.

We know that in the lengths of sides are equal in a rhombus and the length of diagonals are not equal.

Let us find the lengths of the sides.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒

⇒

⇒

⇒

⇒ AB = 5

⇒

⇒

⇒

⇒

⇒ BC = 5

⇒

⇒

⇒

⇒

⇒ CD = 5

⇒

⇒

⇒

⇒

⇒ DA = 5

We got AB = BC = CD = DA, these are the vertices of a square or a rhombus.

Now we find the lengths of the diagonals.

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AC = BD.

∴ The points form a square not rhombus.

Given points are A(4, 5), B(7, 6), C(4, 3) and D(1, 2).

Let us find the lengths of the sides.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AB = CD and BC = DA, this may be a parallelogram or a rectangle.

Now we find the lengths of the diagonals.

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AC≠BD.

∴ The points form a parallelogram.

Given points are A(- 1, - 2), B(1, 0), C(- 1, 2) and D(- 3, 0).

Let us find the lengths of the sides.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AB = BC = CD = DA, this may be square (or) rhombus.

Now we find the lengths of the diagonals.

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AC = BD.

∴ The points form a square.

Given points are A(- 3, 5), B(3, 1), C(0, 3) and D(- 1, - 4).

Let us find the lengths of the sides.

We know that the distance between the points (x₁, y₁) and (x₂, y₂) is .

Now,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We got AB≠BC≠CD≠DA, this may be a quadrilateral which is not of standard shape.

Now we find the lengths of the diagonals.

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒ AC + BC = AB

We got points ABC are collinear.

∴ The points doesn’t form a quadrilateral.

Given that A(- 1, 2) and C(3, 2) are the opposite vertices of a square.

Let us assume the other two vertices be B(x₁, y₁) and D(x₂, y₂) and the midpoint be M

We know that midpoint of AC = Midpoint of BD = M

⇒ M = (1, 2)

⇒ x₁ + x₂ = 2 ..... (1)

⇒ y₁ + y₂ = 4 ..... (2)

We know that lengths of the sides of the square are equal.

AB = BC = CD = DA.

We know that distance between two points (x₁, y₁) and (x₂, y₂) is

⇒ AB = BC

⇒ AB² = BC²

⇒ (x₁ - (- 1))² + (y₁ - 2)² = (x₁ - 3)² + (y₁ - 2)²

⇒ x₁² + 1 + 2x₁ + y₁² + 4 - 4y₁ = x₁² - 6x₁ + 9 + y₁² + 4 - 4y₁

⇒ 8x₁ = 8

⇒ x₁ = 1 ..... (3)

From (1)

⇒ x₂ = 2 - 1 = 1 ..... (4)

We know that points ABC form right angled isosceles triangle.

We have AB² + BC² = AC²

⇒ 2AB² = (- 1 - 3)² + (2 - 2)²

⇒ 2((1 - (- 1))² + (y₁ - 2)²) = (- 4)² + (0)²

⇒ 2(2² + (y₁ - 2)²) = 8

⇒ 4 + (y₁ - 2)² = 8

⇒ (y₁ - 2)² = 4

⇒ y₁ - 2 = ±2

⇒ y₁ = 2 - 2 (or) y₁ = 2 + 2

⇒ y₁ = 0 (or) y₁ = 4

From (2)

⇒ y₂ = 4 - 0

⇒ y₂ = 4

⇒ y₂ = 4 - 4

⇒ y₂ = 0

It is clear that the other two points are (1, 0) and (1, 4).

∴ The other two points are (1, 0) and (1, 4).

[Hint: Take A as the origin and AB and AD as x and y - axis respectively. Let AB = a, AD = b]

Let us assume A as the origin (0, 0) and AB and AD as x and y axis with length a and b units.

Then we get points B to be (a, 0), D to be (0, b) and C to be (a, b).

Let us assume P(x, y) be any point in a plane of the rectangle.

We need to prove PA² + PC² = PB² + PD².

We know that distance between two points (x₁, y₁) and (x₂, y₂) is .

Let us assume L.H.S,

⇒ PA² + PC² = ((x - 0)² + (y - 0)²) + ((x - a)² + (y - b)²)

⇒ PA² + PC² = x² + y² + x² - 2ax + a² + y² - 2by + b²

⇒ PA² + PC² = (x² - 2ax + a² + y²) + (x² + y² - 2by + b²)

⇒ PA² + PC² = ((x - a)² + (y - 0)²) + ((x - 0)² + (y - b)²)

⇒ PA² + PC² = PB² + PD²

⇒ L.H.S = R.H.S

∴ Thus proved.

Let us assume ABCD be a rectangle with A as the origin and AB and AD as x and y - axes having lengths a and b units.

We get the vertices of the rectangle as follows.

⇒ A = (0, 0)

⇒ B = (a, 0)

⇒ C = (a, b)

⇒ D = (0, b)

We need to prove the lengths of the diagonals are equal.

i.e., AC = BD

We know that distance between two points (x₁, y₁) and (x₂, y₂) is .

Let us find the individual lengths of diagonals,

⇒

⇒ ..... (1)

⇒

⇒ ..... (2)

From (1) and (2), we can clearly say that AC = BD.

∴ The diagonals of a rectangle are equal.

Let us assume ABCD be a rectangle with A as the origin and AB and AD as x and y - axes having lengths a and b units.

We get the vertices of the rectangle as follows.

⇒ A = (0, 0)

⇒ B = (a, 0)

⇒ C = (a, b)

⇒ D = (0, b)

We need to prove that the sum of squares of the diagonals of a rectangle is equal to the sum of squares of its sides.

i.e., AC² + BD² = AB² + BC² + CD² + DA²

We know that distance between two points (x₁, y₁) and (x₂, y₂) is .

Assume L.H.S

⇒ AC² + BD² = ((0 - a)² + (0 - b)²) + ((a - 0)² + (0 - b)²)

⇒ AC² + BD² = a² + b² + a² + b²

⇒ AC² + BD² = 2(a² + b²) ..... - (1)

Assume R.H.S

⇒ AB² + BC² + CD² + DA² = ((0 - a)² + (0 - 0)²) + ((a - a)² + (0 - b)²) + ((a - 0)² + (b - b)²) + ((0 - 0)² + (b - 0)²)

⇒ AB² + BC² + CD² + DA² = a² + 0 + 0 + b² + a² + 0 + 0 + b²

⇒ AB² + BC² + CD² + DA² = 2(a² + b²) .... (2)

From (1) and (2), we can clearly say that,

⇒ AC² + BD² = AB² + BC² + CD² + DA²

∴ Thus proved.

Let P(x,y) be the point which divides the line segment internally.

Using the section formula for the internal division, i.e.

…(i)

Here, m₁ = 1, m₂ = 3

(x₁, y₁) = (2, 4) and (x₂, y₂) = (6, 8)

Putting the above values in the above formula, we get

⇒ x = 3, y = 5

Hence, (3,5) is the point which divides the line segment internally.

Now, Let Q(x,y) be the point which divides the line segment externally.

Using the section formula for the external division, i.e.

…(i)

Here, m₁ = 1, m₂ = 3

(x₁, y₁) = (2, 4) and (x₂, y₂) = (6, 8)

Putting the above values in the above formula, we get

⇒ x = 0, y = 2

Hence, (0,2) is the point which divides the line segment externally.

Let P(x,y) be the point which divides the line segment internally.

Using the section formula for the internal division, i.e.

…(i)

Here, m₁ = 2, m₂ = 3

(x₁, y₁) = (-1, 7) and (x₂, y₂) = (4, -3)

Putting the above values in the above formula, we get

⇒ x = 1, y = 3

Hence, (1,3) is the point which divides the line segment internally.

Let P(x,y) be the point which divides the line segment internally.

Using the section formula for the internal division, i.e.

…(i)

Here, m₁ = 3, m₂ = 1

(x₁, y₁) = (4, -3) and (x₂, y₂) = (8, 5)

Putting the above values in the above formula, we get

⇒ x = 7, y = 3

Hence, (7,3) is the point which divides the line segment internally.

Let P and Q be the points of trisection of AB, i.e. AP = PQ = QB

∴ P divides AB internally in the ratio 1: 2.

∴ the coordinates of P, by applying the section formula, are

…(i)

Here, m₁ = 1, m₂ = 2

(x₁, y₁) = (2, 3) and (x₂, y₂) = (6, 5)

Putting the above values in the above formula, we get

Now, Q also divides AB internally in the ratio 2: 1. So, the coordinates of Q are

…(i)

Here, m₁ = 2, m₂ = 1

(x₁, y₁) = (2, 3) and (x₂, y₂) = (6, 5)

Putting the above values in the above formula, we get

Therefore, the coordinates of the points of trisection of the line segment joining A and B are

Let P and Q be the points of trisection of AB, i.e. AP = PQ = QB

∴ P divides AB internally in the ratio 1: 2.

∴ the coordinates of P, by applying the section formula, are

…(i)

Here, m₁ = 1, m₂ = 2

(x₁, y₁) = (1, -2) and (x₂, y₂) = (-3, 4)

Putting the above values in the above formula, we get

Now, Q also divides AB internally in the ratio 2: 1. So, the coordinates of Q are

…(i)

Here, m₁ = 2, m₂ = 1

(x₁, y₁) = (1, -2) and (x₂, y₂) = (-3, 4)

Putting the above values in the above formula, we get

Therefore, the coordinates of the points of trisection of the line segment joining A and B are

Given:

⇒ m₁ = 4 and m₂ = 3

and (x₁, y₁) = (1, 2) ; (x₂, y₂) = (2, 3)

Using the section formula for the internal division, i.e.

…(i)

Hence, the coordinates of P are

Given: D is the midpoint of BC. So, BD = DC

Then the coordinates of D are

⇒ x = 4 and y = 1

So, coordinates of D are (4, 1)

Now, we have to find the coordinates of P.

Given:

⇒ m₁ = 2 and m₂ = 1

and (x₁, y₁) = (4, -8) ; (x₂, y₂) = (4, 1)

Using the section formula for the internal division, i.e.

…(i)

⇒ x = 4, y = -2

Hence, the coordinates of P are P(x,y) = P(4, -2)

Given:

⇒ 7AP = 3AP + 3PB

⇒ 7AP – 3AP = 3PB

⇒ 4AP = 3PB

Hence, the point P divides AB in the ratio of 3:4

⇒ m₁ = 3 and m₂ = 4

and (x₁, y₁) = (-2, -2) ; (x₂, y₂) = (2, -4)

Using the section formula for the internal division, i.e.

…(i)

Hence, the coordinates of P are

Given: AP = AB + BP and AP = 10

Firstly, we find the distance between A and B

d(A,B) = √(x₂ – x₁)² + (y₂ – y₁)²

= √(4 – 1)² + (8 – 4)²

= √(3)² + (4)²

= √9 + 16

= √25

= 5

So, AB = 5

It is given that AP = AB + BP

⇒ 10 = 5 + BP

⇒ 10 – 5 = BP

⇒ BP = 5

⇒ A, B and P are collinear

and since AB = BP

⇒ B is the midpoint of AP

Let the coordinates of P = (x,y)

⇒ x + 1 = 8 and y + 4 = 16

⇒ x = 7 and y = 12

Hence, the coordinates of P are (7, 12)

Let P and Q be the required new ends

Coordinates of P

Let AP = k

∴ AB = AP = k

and PB = AP + AB = k + k = 2k

∴ P divides AB externally in the ratio 1:2

Using the section formula for the external division, i.e.

…(i)

Here, m₁ = 1, m₂ = 2

(x₁, y₁) = (2, 3) and (x₂, y₂) = (-3, 5)

Putting the above values in the above formula, we get

⇒ x = 7, y = 1

∴Coordinates of P are (7, 1)

Coordinates of Q.

Q divides AB externally in the ratio 2:1

Again, Using the section formula for the external division, i.e.

…(i)

Here, m₁ = 2, m₂ = 1

(x₁, y₁) = (2, 3) and (x₂, y₂) = (-3, 5)

Putting the above values in the above formula, we get

∴Coordinates of Q are (-8, 7)

Let P and Q be the required new ends

Coordinates of P

Let AP = k

∴ AB = 2AP = 2k

and PB = AP + AB = k + 2k = 3k

∴ P divides AB externally in the ratio 1:3

Using the section formula for the external division, i.e.

…(i)

Here, m₁ = 1, m₂ = 3

(x₁, y₁) = (6, 3) and (x₂, y₂) = (-1, -4)

Putting the above values in the above formula, we get

∴Coordinates of P are

Coordinates of Q.

Q divides AB externally in the ratio 3:1

Again, Using the section formula for the external division, i.e.

…(i)

Here, m₁ = 3, m₂ = 1

(x₁, y₁) = (6, 3) and (x₂, y₂) = (-1, -4)

Putting the above values in the above formula, we get

∴Coordinates of Q are

Now, Let P(x,y) be the point which lies on extended line AB

Using the section formula for the external division, i.e.

…(i)

Here, m₁ = 1, m₂ = 3

(x₁, y₁) = (-1, 4) and (x₂, y₂) = (5, 1)

Putting the above values in the above formula, we get

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

Let A = (5, -4) and B = (3, -2)

Here, the point divides the line segment in the ratio 4:3

So, m₁ = 4 and m₂ = 3

Using section formula,

…(i)

Hence, the coordinates of P are

Now, the distance from the origin (0,0) is

Consider a ΔABC with A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). If P(1, 1), Q(2, 3) and R(4, 1) are the midpoints of AB, BC, and CA. Then,

…(i)

…(ii)

…(iii)

…(iv)

…(v)

…(vi)

Adding (i), (iii) and (v), we get

x₁ + x₂ + x₂ + x₃ + x₁ + x₃ = 2 + 4 + 8

⇒ 2(x₁ + x₂ + x₃) = 14

⇒ x₁ + x₂ + x₃ = 7 …(vii)

From (i) and (vii), we get

x₃ = 7 – 2 = 5

From (iii) and (vii), we get

x₁ = 7 – 4 = 3

From (v) and (vii), we get

x₂ = 7 – 8 = -1

Now adding (ii), (iv) and (vi), we get

y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = 2 + 6 + 2

⇒ 2(y₁ + y₂ + y₃) = 10

⇒ y₁ + y₂ + y₃ = 5 …(viii)

From (ii) and (viii), we get

y₃ = 5 – 2 = 3

From (iv) and (vii), we get

y₁ = 5 – 6 = -1

From (vi) and (vii), we get

y₂ = 5 – 2 = 3

Hence, the vertices of ΔABC are A(3, -1), B(-1, 3) and C(5, 3)

Consider a ΔABC with A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). If P(10, 5), Q(8, 4) and R(6, 6) are the midpoints of AB, BC, and CA. Then,

…(i)

…(ii)

…(iii)

…(iv)

…(v)

…(vi)

Adding (i), (iii) and (v), we get

x₁ + x₂ + x₂ + x₃ + x₁ + x₃ = 20 + 16 + 12

⇒ 2(x₁ + x₂ + x₃) = 48

⇒ x₁ + x₂ + x₃ = 24 …(vii)

From (i) and (vii), we get

x₃ = 24 – 20 = 4

From (iii) and (vii), we get

x₁ = 24 – 16 = 8

From (v) and (vii), we get

x₂ = 24 – 12 = 12

Now adding (ii), (iv) and (vi), we get

y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = 10 + 8 + 12

⇒ 2(y₁ + y₂ + y₃) = 30

⇒ y₁ + y₂ + y₃ = 15 …(viii)

From (ii) and (viii), we get

y₃ = 15 – 10 = 5

From (iv) and (vii), we get

y₁ = 15 – 8 = 7

From (vi) and (vii), we get

y₂ = 15 – 12 = 3

Hence, the vertices of ΔABC are A(8, 7), B(12, 3) and C(4, 5)

Consider a ΔABC with A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). If P(3, 4), Q(4, 6) and R(5, 7) are the midpoints of AB, BC, and CA. Then,

…(i)

…(ii)

…(iii)

…(iv)

…(v)

…(vi)

Adding (i), (iii) and (v), we get

x₁ + x₂ + x₂ + x₃ + x₁ + x₃ = 6 + 8 + 10

⇒ 2(x₁ + x₂ + x₃) = 24

⇒ x₁ + x₂ + x₃ =12 …(vii)

From (i) and (vii), we get

x₃ = 12 – 6 = 6

From (iii) and (vii), we get

x₁ = 12 – 8 = 4

From (v) and (vii), we get

x₂ = 12 – 10 = 2

Now adding (ii), (iv) and (vi), we get

y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = 8 + 12 + 14

⇒ 2(y₁ + y₂ + y₃) = 34

⇒ y₁ + y₂ + y₃ = 17 …(viii)

From (ii) and (viii), we get

y₃ = 17 – 8 = 9

From (iv) and (vii), we get

y₁ = 17 – 12 = 5

From (vi) and (vii), we get

y₂ = 17 – 14 = 3

Hence, the vertices of ΔABC are A(4, 5), B(2, 3) and C(6, 9)

Given:

A(1, -2) and B(2, 5) are two points.

OC = 2OA …(i)

and OD = 2OB …(ii)

Adding (i) and (ii), we get

OC + OD = 2OA + 2OB

⇒ CD = 2[OA + OB]

⇒ CD = 2[AB] …(iii)

Now, we find the distance between A and B

d(A,B) = √(x₂ – x₁)² + (y₂ – y₁)²

= √(2 – 1)² + {5 – (-2)}²

= √(1)² + (5 + 2)²

= √1 + 49

= √50

= 5√2

Putting the value in eq. (iii), we get

CD = 2 × 5√2

= 10√2

Let the given points of a triangle be A(-1, 3), B(1, -1) and C(5,1)

Let D, E and F are the midpoints of the sides BC, CA and AB respectively.

The coordinates of D are:

D = (3, 0)

The coordinates of E are:

E = (2, 2)

The coordinates of F are:

F = (0, 1)

Now, we have to find the lengths of the medians.

d(A,D) = √(x₂ – x₁)² + (y₂ – y₁)²

= √{3 – (-1)²} + {0 – 3}²

= √(3 + 1)² + (-3)²

= √16 + 9

= √25

= 5 units

d(B,E) = √(x₂ – x₁)² + (y₂ – y₁)²

= √(2 – 1)² + {2 – (-1)}²

= √(1)² + (2 + 1)²

= √1 + 9

= √10 units

d(C,F) = √(x₂ – x₁)² + (y₂ – y₁)²

= √(5 – 0)² + {1 – 1}²

= √(5)² + (0)²

= √25

= 5 units

Hence, the length of the medians AD, BE and CF are 5, √10, 5 units respectively.

Given: A(1, 5), B(-2, 1) and C(4,1) are the vertices of ΔABC

Using angle bisector theorem, which states that:

The ratio of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC:{\displaystyle {\frac {|BD|}{|DC|}}={\frac {|AB|}{|AC|}},}

⇒ BD = DC

⇒ D is the midpoint of BC

So, the coordinates of D are:

D = (1, 1)

Now, AD = √(x₂ – x₁)² + (y₂ – y₁)²

= √(1 – 1)² + {5 – 1}²

= √(0)² + (4)²

= √16

= 4 units

Hence, AD = 4 units

Let P be the midpoint of the line segment joining (3, 4) and (k, 7)

So, the coordinates of P are:

Again,

2x + 2y + 1 = 0

⇒ 3 + k + 11 + 1 = 0

⇒ k + 15 = 0

⇒ k = -15

Let the coordinates of the other end be (x,y).

Since (-2, 5) is the midpoint of the line joining (2,3) and (x,y)

⇒ x + 2 = -4 and y + 3 = 10

⇒ x = -4 – 2 and y = 10 – 3

⇒ x = -6 and y = 7

Hence, the coordinates of the other end are (-6, 7)

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

Since 2, -3) is the midpoint of the line joining (1, 4) and (x,y)

⇒ x + 1 = 4 and y + 4 = -6

⇒ x = 4 – 1 and y = -6 – 4

⇒ x = 3 and y = -10

Hence, the coordinates of A are (3, -10)

Let the coordinates of B are (x, y)

It is given that the line segment divide in the ratio 3:4

So, m₁ = 3 and m₂ = 4

and (x’, y’) =(-1, 2); (x₁, y₁) = (2,5); (x₂, y₂) = (x, y)

Using section formula for the internal division, we get

…(i)

⇒ 3x + 8 = -7 and 3y + 20 = 14

⇒ 3x = -7 – 8 and 3y = 14 – 20

⇒ 3x = -15 and 3y = -6

⇒ x = -5 and y = -2

Hence, the coordinates of B are (-5, -2)

Let C(-8, 3) divides the line segment AB in the ratio m:n

Here, (x, y) = (-8, 3); (x₁, y₁) = (2, -2) and (x₂, y₂) = (-4,1)

So,

⇒ -8m -8n = -4m + 2n and 3m + 3n = m - 2n

⇒ -8m + 4m - 8n - 2n = 0 and 3m – m + 3n + 2n = 0

⇒ -4m – 10n = 0 and 2m + 5n = 0

⇒ -2m – 5n = 0 and 2m + 5n = 0

⇒ 2m + 5n = 0 and 2m + 5n = 0

⇒ 2m = -5n

Hence, the ratio is 5:2 and this negative sign shows that the division is external.

Let C(-4, 6) divides the line segment AB in the ratio m:n

Here, (x, y) = (-4, 6); (x₁, y₁) = (-6, 10) and (x₂, y₂) = (3, -8)

So,

⇒ -4m -4n = 3m - 6n and 6m + 6n = -8m + 10n

⇒ -4m – 3m – 4n + 6n = 0 and 6m + 8m + 6n – 10n = 0

⇒ -7m + 2n = 0 and 14m - 4n = 0

⇒ -7m = -2n and 14m = 4n

⇒ 7m = 2n and 7m = 2n

Hence, the ratio is 2:7 and the division is internal.

Let C(-1, 6) divides the line segment AB in the ratio m:n

Here, (x, y) = (-1, 6); (x₁, y₁) = (-3, 10) and (x₂, y₂) = (6, -8)

So,

⇒ -m – n = 6m - 3n and 6m + 6n = -8m + 10n

⇒ -m – 6m – n + 3n = 0 and 6m + 8m + 6n – 10n = 0

⇒ -7m + 2n = 0 and 14m - 4n = 0

⇒ 2n = 7m and 4n = 14m

⇒ 7m = 2n

Hence, the ratio is 2:7 and the division is internal.

Let C(x, 2) divides the line segment AB in the ratio m:n

Here, (x, y) = (x, 2); (x₁, y₁) = (-3, -4) and (x₂, y₂) = (3, 5)

So,

⇒ 2m + 2n = 5m – 4n

⇒ 2m – 5m = -4n – 2n

⇒ -3m = -6n

⇒ m = 2n

Now, the ratio is 2:1

Now,

⇒ 3x = 6 – 3

⇒ 3x = 3

⇒ x = 1

Hence, the ratio is 2:1 and the division is internal and the value of x = 1

Let the line segment A(2, -3) and B(5, 6) is divided at point P(x,0) by x-axis in ratio m:n

Here, (x, y) = (x, 0); (x₁, y₁) = (2, -3) and (x₂, y₂) = (5, 6)

So,

⇒ 0 = 6m – 3n

⇒ -6m = -3n

Hence, the ratio is 1:2 and the division is internal.

Let the line segment A(1, -5) and B(-4, 5) is divided at point P(x,0) by x-axis in ratio m:n

Here, (x, y) = (x, 0); (x₁, y₁) = (1, -5) and (x₂, y₂) = (-4, 5)

So,

⇒ 0 = 5m – 5n

⇒ 5m = 5n

Hence, the ratio is 1:1 and the division is internal.

Now,

Hence, the coordinates of the point of division is

Let the line segment A(5, -6) and B(-1, -4) is divided at point P(0, y) by y-axis in ratio m:n

Here, (x, y) = (0, y); (x₁, y₁) = (5, -6) and (x₂, y₂) = (-1, -4)

So,

⇒ 0 = -m + 5n

⇒ m = 5n

Hence, the ratio is 5:1 and the division is internal.

Now,

Hence, the coordinates of the point of division is

Here, x₁ = 2, x₂ = 6, x₃ = 2

and y₁ = 4, y₂ = 4, y₃ = 0

Let the coordinates of the centroid be(x,y)

So,

Hence, the centroid of a triangle is

Let (2, 2), (0, 6) and (8, 10) be the vertices A, B and C of the triangle respectively. Let AD, BE, CF be the medians

The coordinates of D are:

D = (4, 8)

The coordinates of E are:

E = (5, 6)

The coordinates of F are:

F = (1, 4)

Let P be the trisection point of the median AD which is nearer to the opposite side BC

∴ P divides DA in the ratio 1:2 internally

Let Q be the trisection point of the median BE which is nearer to the opposite side CA

∴ Q divides EB in the ratio 1:2 internally

Let R be the trisection point of the median CF which is nearer to the opposite side AB

∴ R divides FC in the ratio 1:2 internally

Therefore, Coordinates of required trisection points are

Let the third vertex of a triangle be(x,y)

Here, x₁ = 1, x₂ = 5, x₃ = x

and y₁ = 4, y₂ = 2, y₃ = y

and the coordinates of the centroid is (0, -3)

We know that

⇒ 6 + x = 0 and 6 + y = -9

⇒ x = -6 and y = -15

Hence, the third vertex of a triangle is (-6, -15)

Let the third vertex of a triangle be (x, y)

Here, x₁ = 2√3, x₂ = 2√3, x₃ = x

and y₁ = -1, y₂ = 5, y₃ = y

and the coordinates of the centroid is (√3, 2)

We know that

⇒ 4√3 + x = 3√3 and 4 + y = 6

⇒ x = -√3 and y = 2

Hence, the third vertex of a triangle is (-√3, 2)

The vertices of a triangle are A (9,2), B(1,10) and C(-7,-6)

Here, x₁ = 9, x₂ = 1, x₃ = -7

and y₁ = 2, y₂ = 10, y₃ = -6

Let the coordinates of the centroid be(x,y)

So,

= (1,2)

Hence, the centroid of a triangle is (1, 2)

Now,

Let D, E and F are the midpoints of the sides BC, CA and AB respectively.

The coordinates of D are:

D = (-3, 2)

The coordinates of E are:

E = (1, -2)

The coordinates of F are:

F = (5, 6)

Now, we find the centroid of a triangle formed by joining these middle points D, E, and F as shown in figure

Let P be the trisection point of the median AD which is nearer to the opposite side BC

∴ P divides DA in the ratio 1:2 internally

= (1, 2)

Let Q be the trisection point of the median BE which is nearer to the opposite side CA

∴ Q divides EB in the ratio 1:2 internally

= (1, 2)

Let R be the trisection point of the median CF which is nearer to the opposite side AB

∴ R divides FC in the ratio 1:2 internally

= (1, 2)

Yes, the triangle has the same centroid, i.e. (1,2)

Consider a ΔABC with A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). If P(1, 2), Q(0, -1) and R(2, -1) are the midpoints of AB, BC and CA. Then,

…(i)

…(ii)

…(iii)

…(iv)

…(v)

…(vi)

Adding (i), (iii) and (v), we get

x₁ + x₂ + x₂ + x₃ + x₁ + x₃ = 2 + 0 + 4

⇒ 2(x₁ + x₂ + x₃) = 6

⇒ x₁ + x₂ + x₃ = 3 …(vii)

From (i) and (vii), we get

x₃ = 3 – 2 = 1

From (iii) and (vii), we get

x₁ = 3 – 0 = 3

From (v) and (vii), we get

x₂ = 3 – 4 = -1

Now adding (ii), (iv) and (vi), we get

y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = 4 + (-2) + (-2)

⇒ 2(y₁ + y₂ + y₃) = 0

⇒ y₁ + y₂ + y₃ = 0 …(viii)

From (ii) and (viii), we get

y₃ = 0 – 4 = -4

From (iv) and (vii), we get

y₁ = 0 – (-2) = 2

From (vi) and (vii), we get

y₂ = 0 – (-2) = 2

Hence, the vertices of ΔABC are A(3, 2), B(-1, 2) and C(1, -4)

Now, we have to find the centroid of a triangle

The vertices of a triangle are A(3, 2), B(-1, 2) and C(1, -4)

Here, x₁ = 3, x₂ = -1, x₃ = 1

and y₁ = 2, y₂ = 2, y₃ = -4

Let the coordinates of the centroid be(x,y)

So,

= (1,0)

Hence, the centroid of a triangle is (1, 0)

Note that to show that a quadrilateral is a rhombus, it is sufficient to show that

(a) ABCD is a parallelogram, i.e., AC and BD have the same midpoint.

(b) a pair of adjacent edges are equal

(c) the diagonal AC and BD are not equal.

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

Coordinates of the midpoint of AC are

Coordinates of the midpoint of BD are

Thus, AC and BD have the same midpoint.

Hence, ABCD is a parallelogram

Now, using Distance Formula

d(A,B)= AB = √(-5 + 3)² + (-5 – 2)²

⇒ AB = √(-2)² + (-7)²

⇒ AB = √4 +49

⇒ AB = √53 units

d(B,C)= BC = √(-5 – 2)² + (-5 + 3)²

⇒ BC = √(-7)² + (-2)²

⇒ BC = √49 +4

⇒ BC = √53 units

d(C,D) = CD = √(4 – 2)² + (4 + 3)²

⇒ CD = √(2)² + (7)²

⇒ CD = √4 +49

⇒ CD = √53 units

d(A,D) = AD =√(4 + 3)² +(4 – 2)²

⇒ AD = √(7)² + (2)²

⇒ AD = √49 +4

⇒ AD = √53 units

Therefore, AB = BC = CD = AD = √53 units

Now, check for the diagonals

AC = √(2 + 3)² + (-3 – 2)²

= √(5)² + (-5)²

= √25 + 25

= √50

and

BD = √(4 + 5)² + (4 + 5)²

⇒ BD = √(9)² + (9)²

⇒ BD = √81 + 81

⇒ BD = √162

⇒ Diagonal AC ≠ Diagonal BD

Hence, ABCD is a rhombus.

Note that to show that a quadrilateral is a square, it is sufficient to show that

(a) ABCD is a parallelogram, i.e., AC and BD bisect each other

(b) a pair of adjacent edges are equal

(c) the diagonal AC and BD are equal.

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

Coordinates of the midpoint of AC are

Coordinates of the midpoint of BD are

Thus, AC and BD have the same midpoint.

Hence, ABCD is a parallelogram

Now, Using Distance Formula, we get

AB = √(x₂ – x₁)² + (y₂ – y₁)²

= √[(0 – 3)² + (5 - 2)²]

= √(-3)² + (3)²

= √(9 + 9)

= √18 units

BC = √[(-3 – 0)² + (2 - 5)²]

= √(-3)² + (-3)²

= √(9 + 9)

= √18 units

Therefore, AB = BC = √18 units

Now, check for the diagonals

AC = √(-3 – 3)² + (2 – 2)²

= √(-6)² + (0)²

= √36

= 6 units

and

BD = √(0 - 0)² + (-1 – 5)²

⇒ BD = √(0)² + (-6)²

⇒ BD = √36

⇒ BD = 6 units

∴ AC = BD

Hence, ABCD is a square.

Note that to show that a quadrilateral is a parallelogram, it is sufficient to show that the diagonals of the quadrilateral bisect each other.

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

Let M be the midpoint of AC, then the coordinates of M are given by

Let N be the midpoint of BD, then the coordinates of N are given by

Thus, AC and BD have the same midpoint.

In other words, AC and BD bisect each other.

Hence, ABCD is a parallelogram.

Note that to show that a quadrilateral is a rhombus, it is sufficient to show that

(a) ABCD is a parallelogram, i.e., AC and BD have the same midpoint.

(b) a pair of adjacent edges are equal

Let A(1, 0), B(5, 3), C(2, 7) and D(-2, 4) are the vertices of a rhombus.

Coordinates of the midpoint of AC are

Coordinates of the midpoint of BD are

Thus, AC and BD have the same midpoint.

Hence, ABCD is a parallelogram

Now, using Distance Formula

d(A,B)= AB = √(5 – 1)² + (3 – 0)²

⇒ AB = √(4)² + (3)²

⇒ AB = √16 + 9

⇒ AB = √25 = 5 units

d(B,C)= BC = √(2 – 5)² + (7 – 3)²

⇒ BC = √(-3)² + (4)²

⇒ BC = √9 + 16

⇒ BC = √25 = 5 units

Therefore, adjacent sides are equal.

Hence, ABCD is a rhombus.

Note that to show that a quadrilateral is a rectangle, it is sufficient to show that

(a) ABCD is a parallelogram, i.e., AC and BD bisect each other and,

(b) the diagonal AC and BD are equal

Let A(4, 8), B(0, 2), C(3, 0) and D(7, 6) are the vertices of a rectangle.

Coordinates of the midpoint of AC are

Coordinates of the midpoint of BD are

Thus, AC and BD have the same midpoint.

Hence, ABCD is a parallelogram

Now, check for the diagonals by using the distance formula

AC = √(3 – 4)² + (0 – 8)²

= √(-1)² + (-8)²

= √1 + 64

= √65 units

and

BD = √(7 - 0)² + (6 – 2)²

⇒ BD = √(7)² + (4)²

⇒ BD = √49 + 16

⇒ BD = √65 units

∴ AC = BD

Hence, ABCD is a rectangle.

Note that to show that a quadrilateral is a square, it is sufficient to show that

(a) ABCD is a parallelogram, i.e., AC and BD bisect each other

(b) a pair of adjacent edges are equal

(c) the diagonal AC and BD are equal.

Let the vertices of a quadrilateral are A(4, 3), B(6, 4), C(5, 6) and D(3, 5).

Coordinates of the midpoint of AC are

Coordinates of the midpoint of BD are

Thus, AC and BD have the same midpoint.

Hence, ABCD is a parallelogram

Now, Using Distance Formula, we get

AB = √(x₂ – x₁)² + (y₂ – y₁)²

= √[(6 – 4)² + (4 - 3)²]

= √(2)² + (1)²

= √(4 + 1)

= √5 units

BC = √[(5 – 6)² + (6 - 4)²]

= √(-1)² + (2)²

= √(1 + 4)

= √5 units

Therefore, AB = BC = √5 units

Now, check for the diagonals

AC = √(5 – 4)² + (6 – 3)²

= √(1)² + (3)²

= √1 + 9

= √10 units

and

BD = √(3 - 6)² + (5 – 4)²

⇒ BD = √(-3)² + (1)²

⇒ BD = √9 + 1

⇒ BD = √10 units

∴ AC = BD

Hence, ABCD is a square.

Let the coordinates of the fourth vertex D be (x, y).

We know that diagonals of a parallelogram bisect each other.

∴ Midpoint of AC = Midpoint of BD …(i)

Coordinates of the midpoint of AC are

Coordinates of the midpoint of BD are

So, according to eq. (i), we have

⇒ 3 + x = 4 and 7 + y = 6

⇒ x = 1 and y = -1

Thus, the coordinates of the vertex D are (1, -1)

Let the coordinates of the fourth vertex D be (x, y).

We know that diagonals of a rhombus bisect each other.

∴ Midpoint of AC = Midpoint of BD …(i)

Coordinates of the midpoint of AC are

Coordinates of the midpoint of BD are

So, according to eq. (i), we have

⇒ 2 + x = 3 and 7 + y = 7

⇒ x = 1 and y = 0

Thus, the coordinates of the vertex D are (1, 0)

Let the vertices of quadrilateral be P(-4,2), Q(2,6), R(8,5) and S(9,-7)

Let A, B, C and D are the midpoints of PQ, QR, RS and SP respectively.

Now, since A is the midpoint of P(-4, 2) and Q(2, 6)

∴ Coordinates of A are

Coordinates of B are

Coordinates of C are

and

Coordinates of D are

Now,

we find the distance between A and B

Now, since length of opposite sides of the quadrilateral formed by the midpoints of the given quadrilateral are equal .i.e.

AB = CD and AD = BC