WBBSE Solutions For Class 9 Maths Coordinate Geometry Chapter 1 Distance Formulas

Co-Ordinate Geometry Chapter 1 Distance Formulas

Chapter 1 Distance Formulas Introduction

Coordinate Geometry:-

The branch of mathematics in which different problems of geometry are solved with the help of algebra is known as coordinate geometry.

Rene Descartes, a famous French philosopher and mathematician developed this branch of mathematics and by his name, it is also known as cartesian geometry. 

Coordinate geometry is of two types:

1. Two-dimensional geometry

2. Three-dimensional geometry or solid coordinate geometry. 

Read and Learn More WBBSE Solutions For Class 9 Maths

The geometry based on any plane region is called two-dimensional geometry, while the geometry based on any solid object or space is usually called solid geometry.

Here we shall confine our discussion to two-dimensional geometry only.

Let us consider a page of this book, which is purely a plane where say O is a fixed point and \(\overleftrightarrow{\mathrm{XOX}^{\prime}}\) and \(\overleftrightarrow{\mathrm{YOY}^{\prime}}\) are two straight lines perpendicular to each other, i.e., they have divided the page plane into four sections.

Each of these sections is called a quadrant. The sections XOY, YOX’, X’OY’, and Y’OX’ are called the 1st, 2nd, 3rd, and 4th quadrants respectively 

The fixed point O is called the origin and the straight lines together are called the coordinate axes.

Separately \(\overleftrightarrow{\mathrm{XOX}^{\prime}}\) is called the x-axis or abscissa and \(\overleftrightarrow{\mathrm{YOY}^{\prime}}\) is called the y-axis or ordinate.

To express any point in this plane, firstly the digit expressing the distance along the axis (or parallel to the x-axis) and then the digit expressing the distance along the y-axis (or parallel to the y-axis) are written in a bracket.

If the coordinates of any point A be (a, b), then it is quadrant assumed that point A is at a distance of a unit from the origin O along or parallel to the x-axis and at a distance of b units along or parallel to the y-axis.

Whether the point lies on the 1st, 2nd, 3rd, or 4th quadrant or not totally depends on the signs of a and b.

Here, a is called the abscissa or x-coordinate, and b is called the ordinate or y-coordinate of point A.

Clearly, in the 1st quadrant, both the coordinates of any point are positive. In the 2nd quadrant, the abscissa is negative, but the ordinate is positive.

In the 3rd quadrant, the abscissa and the ordinate of any point are both negative, and in the 4th quadrant, the abscissa is positive, but the ordinate of any point is negative.

 

WBBSE Solutions For Class 9 Maths Coordinate Geometry Chapter 1 Distance Formulas

 

Chapter 1 Distance Formulas To Find The Distance Between Two Points Located At Different Positions In A Plane

 

1. Distance between two points located on the positive sides of the x-axis:

Let A (a, 0) and B (b, 0) (b> a) be two points on the positive x-axis, and O (0, 0) is the origin.

Now, AB = OB – OA 

= b-a

 

WBBSE Solutions For Class 9 Maths Coordinate Geometry Chapter 1 Distance Formulas 1

 

2. Distance between two points located on the negative side of the x-axis :

Let A (-a, 0) and B (- b, 0)(a> b) be any two points on the negative x-axis and 0 (0, 0) is the origin.

Here, AB = OA – OB

= a-b

 

WBBSE Solutions For Class 9 Maths Coordinate Geometry Chapter 1 Distance Formulas 2

 

3. Distance between two points located on the positive side of the y-axis :

Let A (0, a) and B (0, b) (b> a) be any two points located on the positive y-axis and O (0, 0) is the origin. 

Here, AB =  OB – OA

= b-a

 

WBBSE Solutions For Class 9 Maths Coordinate Geometry Chapter 1 Distance Formulas 3

 

4. Distance between two points located on the negative side of the y-axis:

Let A (0,- a) and B (0, – b)(a > b) be any two points located on the negative y-axis, and O (0, 0) is the origin.

Here, AB = OA – OB

= a-b

 

WBBSE Solutions For Class 9 Maths Coordinate Geometry Chapter 1 Distance Formulas 4

 

5. Distance between two points located on both positive and negative sides of the x-axis :

Let A (a, 0) and B (b, 0) be any two points on the both positive and negative sides of the x-axis and O (0, 0) be the origin.

Here, AB = OA + OB

= a + b

 

WBBSE Solutions For Class 9 Maths Coordinate Geometry Chapter 1 Distance Formulas 5

 

6. Distance between two points located on both the positive and negative sides of y-axis :

Let A (0, a) and B (0, b) be any two points on both positive and negative sides of the y-axis and O (0, 0) is the origin

Here, AB = OA + OB

= a + b

 

 

Chapter 1 Distance Formulas Distance Between Any Two Points In The XY-Plane

 

Let P \(\left(x_1, y_1\right) \text { and } \mathrm{Q}\left(x_2, y_2\right)\) be any two points in the XY-plane.

O (0, 0) is the origin. PR and QS are two perpendiculars drawn from P and Q respectively to the x-axis 

Again, let us draw QT ⊥ PR.

Now, SR = OR – OS

= (Abscissa of P) (Abscissa of Q)

= \(x_1-x_2\)

PT = PR – TR

= PR – QS (∵ TR = QS)

= (Ordinate of P) (Ordinate of Q)

= \(y_1-y_2\)

Now the distance between P and Q = PQ

But, from the right-angled triangle PQT, we get,

PQ² = PT² + QT²

= \(\left(y_1-y_2\right)^2+\left(x_1-x_2\right)^2\)  [∵ QT = SR = \(x_1-x_2\) ]

∴ PQ = \(\sqrt{\left(y_1-y_2\right)^2+\left(x_1-x_2\right)^2}=\sqrt{\left(y_2-y_1\right)^2+\left(x_2-x_1\right)^2}\)

∴ In the XY-plane, the distance between any two points

= √( (square of the difference of the ordinates of the points)+(square of the difference of the abscissas of the points) )

 

WBBSE Solutions For Class 9 Maths Coordinate Geometry Chapter 1 Distance Formulas

 

 

Chapter 1 Distance Formulas Distance Of A Point In The XY-Plane From The Origin

 

Distance Of A Point In The XY-Plane From The Origin

Let P (x, y) be any point on the XY-plane, and O (0, 0) be the origin

Again, PM ⊥ OX.

∴ OM = x and PM = y

Now, from the right-angled triangle POM, we get,

OP² = OM²+ PM² = x²+ y². 

∴ OP = \(\sqrt{x^2+y^2}\)

Also, by the distance formula, OP² = (x – 0)² + ( y – 0)² = x² + y²

∴ OP = \(\sqrt{x^2+y^2}\)

 

WBBSE Solutions For Class 9 Maths Coordinate Geometry Chapter 1 Distance Formulas

 

Formulas at a glance :

1. Distance of a given point P (x, y) from the origin 0 (0, 0) is \(\overline{\mathrm{OP}}=\sqrt{x^2+y^2}\) units.

2. Distance between two given points P (\(x_1-x_1\)) and Q (\(y_2-y_2\)) is \(\overline{\mathrm{PQ}}=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\) units.

 

Chapter 1 Distance Formulas Select The Correct Answer (MCQ)

 

Question 1. The distance between the points (a + b, c – d) and (a – b, c + d) is

  1. \(2 \sqrt{a^2+c^2}\)
  2. \(2 \sqrt{b^2+d^2}\)
  3. \(\sqrt{a^2+c^2}\)
  4. \(\sqrt{b^2+d^2}\)

Solution: 

Required distance = \(\sqrt{[(a+b)-(a-b)]^2+[(c-d)-(c+d)]^2}\)

= \(\sqrt{(a+b-a+b)^2+(c-d-c-d)^2}\)

= \(\sqrt{4 b^2+4 d^2}=\sqrt{4\left(b^2+d^2\right)}=2 \sqrt{b^2+d^2}\)

Question 2. The distance between the points (x, -7) and (3, -3) is 5 units. Then the values of x are 

  1. 0 or 6
  2. 2 or 3
  3. 5 or 1
  4. – 6 or 0

Solution:

Given

The Distance Between The Points (x, -7) And (3, -3) is 5 Units

The required distance =\(\sqrt{(x-3)^2+(-7+3)^2}=\sqrt{(x-3)^2+16}\)

As per the question, \(\sqrt{(x-3)^2+16}=5 \Rightarrow(x-3)^2+16=25 \Rightarrow(x-3)^2=9\)

x – 3 = ± 3

Then,

x-3 =+3

x = 6 and x-3 = -3

x=0

x= 0 or 6.

 

Question 3. If the distance of the point (x, 4) from the origin is 5 units, then the value of x is

  1. ± 4
  2. ± 5
  3. ± 3
  4. None of these

Solution:

Given

The distance of (x, 4) from the origin is 5 units.

∴ \(\sqrt{x^2+4^2}=5 \Rightarrow x^2+16=25 \Rightarrow x^2=25-16=9 \Rightarrow x= \pm 3\)

 

Question 4. The coordinates of the center of a circle are (0, 0). If the coordinates of any point on its circumference be (3, 4), then the radius of the circle is

  1. 5 units
  2. 4 units
  3. 3 units
  4. None of these

Solution:

Given 

The coordinates of the center of a circle are (0, 0).

If the coordinates of any point on its circumference be (3, 4).

Since the point (3, 4) is on the circumference of the circle, the distance between (3, 4) and (0, 0) is the radius of the circle.

∴ the required radius = \(\sqrt{3^2+4^2} \text { units }=\sqrt{9+16} \text { units }\) 

= √25 units 

= 5 units.

Question 5. The distance of the point (a + b, a – b) from the origin is

  1. \(2 \sqrt{a^2-b^2}\)
  2. \(2 \sqrt{a^2+b^2}\)
  3. \(\sqrt{2\left(a^2+b^2\right)}\)
  4. None of these

Solution:

Given 

The distance of the point (a + b, a – b)

Required distance = \(\sqrt{(a+b)^2+(a-b)^2}=\sqrt{2\left(a^2+b^2\right)}\)

 

Chapter 1 Distance Formulas Short Answer Type Questions

 

Question 2. 

1. Determine the coordinates of a point on the y-axis from which the distance of the points (2, 3) and (1, 2) are equal.

Solution: 

Given:-

Coordinates of a point on the y-axis from which the distance of the points (2, 3) and (1, 2) are equal

Let the coordinates of the point on the y-axis be (0, k).

∴ the distance of (2, 3) from (0, k)=√(2-0)² + (3-k)² units.

Again, the distance of (-1, 2) from (0, k)= \(\sqrt{(2-0)^2+(3-k)^2}\) units.

As per the question, \(\sqrt{(-1-0)^2+(2-k)^2}\).

\(\sqrt{2^2+(3-k)^2}=\sqrt{1^2+(2-k)^2}\)

4+9+k² -6k = 1+4+k²-4k

⇒-6k+4k = 1-9

⇒ 2k=8 

⇒ k=4

the required point is (0, 4).

 

2. If the square of the distance between the points (-2, a) and (a,- 3) is 85, find the value of a.

Solution: 

Given:-

square of the distance between the points (-2, a) and (a,- 3) is 85

The distance between (-2, a) and (a, – 3) = \(\sqrt{(-2-a)^2+(a+3)^2}\)

As per the question, \((-2-a)^2+(a+3)^2=85\)

\((a+2)^2+(a+3)^2=85\)

a²+4+4a+a²+9+6a = 85

2a² + 10a +13 = 85 

2a²+10+13-85 = 0

2a²+10a-72 = 0

a²+5a-36 = 0

(a+9)(a-4)=0

a = -9, 4

the values of a are 9 and 4.

 

3. Show that the distance between (1, 1) and \(\left(\frac{2 m^2}{1+m^2}, \frac{(1-m)^2}{1+m^2}\right)\) is independent of m.

Solution:

Given:-

 The distance between (1, 1) and \(\left(\frac{2 m^2}{1+m^2}, \frac{(1-m)^2}{1+m^2}\right)\)

= \(\sqrt{\left(1-\frac{2 m^2}{1+m^2}\right)^2+\left[1-\frac{(1-m)^2}{1+m^2}\right]^2}\)

= \(\sqrt{\left(\frac{1+m^2-2 m^2}{1+m^2}\right)^2+\left[\frac{1+m^2-(1-m)^2}{1+m^2}\right]^2}\)

= \(\sqrt{\frac{\left(1-m^2\right)^2}{\left(1+m^2\right)^2}+\frac{\left(1+m^2-1-m^2+2 m\right)^2}{\left(1+m^2\right)^2}}\)

= \(\frac{1}{\left(1+m^2\right)} \sqrt{\left(1-m^2\right)^2+4 m^2}\)

= \(\frac{1}{\left(1+m^2\right)} \sqrt{\left(1+m^2\right)^2}=\frac{1}{\left(1+m^2\right)} \times\left(1+m^2\right)=1\)

the distance between (1, 1) and \(\left(\frac{2 m^2}{1+m^2}, \frac{(1-m)^2}{1+m^2}\right)\) is independent of m. (Proved).

 

4. The coordinates of one of the vertices of a triangle is (2, 0) and the coordinates of the mid- points of its opposite side is (5, 3). Find the length of the median.

Solution:

Given:-

The coordinates of one of the vertices of a triangle is (2, 0) and the coordinates of the mid- points of its opposite side is (5, 3)

The required length = \(\sqrt{(2-5)^2+(0-3)^2}\) units

= √9+9 units = √18 units = 3√2 units.

∴ length of the median = 3√2 units.

 

Chapter 1 Distance Formulas Long Answer Type Questions

 

Question 1. Prove that A (3, 3), B (8,-2), and C (-2,-2) are the vertices of a right-angled isosceles triangle. Also, find the length of the hypotenuse of ΔABC.

Solution:

Given That A (3, 3), B (8,-2), and C (-2,-2) are the vertices of a right-angled isosceles triangle

The three vertices of the given triangle are A (3, 3), B (8,-2), and C (-2,-2).

Now, AB =  \(\sqrt{(3-8)^2+(3+2)^2}\) units

= √25+25 units

= √50 units

BC = \(\sqrt{(8+2)^2+(-2+2)^2}\) units

=√100 units

= 10 units.

AC =  \(\sqrt{(3+2)^2+(3+2)^2}\) units

= √25+25 units

= √50 units.

AB²+ AC² = (√50)² + (√50)²

=50+50

= 100

= (10)²

= BC²       [∵ BC= 10]

Clearly, AB = AC         ∴ ΔABC is isosceles.

And AB²+ AC² = BC²         ∴ ΔABC is right-angled.

The given points are the vertices of a right-angled isosceles triangle, the length of whose hypotenuse is 10 units. (Proved)

 

Question 2. Show that the points (2, 1), (0, 0), (1, 2), and (1, 3) are the vertices of a square. 

Solution:

Given Points (2, 1), (0, 0), (1, 2), and (1, 3)

Let the given points be A (2, 1), B (0, 0), C (-1, 2), and D (1, 3).

\(\overline{\mathrm{AB}}=\sqrt{2^2+1^2}\)units

= √5 units

\(\overline{\mathrm{BC}}=\sqrt{1^2+(-2)^2}\) units

=√1+4 units

= √5 units

\(\overline{\mathrm{CD}}=\sqrt{(-1-1)^2+(2-3)^2}\) units

= √4+1 units

= √5 units;

\(\overline{\mathrm{AD}}=\sqrt{(2-1)^2+(1-3)^2}\) units

= √1+4 units

= √5 units;

∴ \(\overline{\mathrm{AB}}=\overline{\mathrm{BC}}=\overline{\mathrm{CD}}=\overline{\mathrm{AD}}\)

Again, \(\overline{\mathrm{AC}}=\sqrt{(2+1)^2+(1-2)^2}\) units

= √9+1 units

= √10 units

\(\overline{\mathrm{AC}}=\sqrt{(2+1)^2+(1-2)^2}\) units

= √9+1 units

= √10 units

\(\overline{\mathrm{BD}}=\sqrt{1^2+3^2}\)

= √1+9 units

= √10 units

\(\overline{\mathrm{AC}}=\overline{\mathrm{BD}}\)

Clearly, the four sides of the quadrilateral ABCD are equal and the two diagonals are also equal.

∴ ABCD is a square. (Proved)

 

Question 3. Calculate whether the three points 0 (0, 0), A (4, 3) and B (8, 6) are collinear or not.

Solution: 

The given three points are O (0, 0), A (4, 3), and B (8, 6).

\(\overline{\mathrm{OA}}=\sqrt{(+4)^2+(-3)^2}\) units = √16+9 units = √25 units = 5 units 

\(\overline{\mathrm{AB}}=\sqrt{(4-8)^2+(3-6)^2}\) units = √16+9 units = √25 units = 5 units

\(\overline{\mathrm{OB}}=\sqrt{(8)^2+(6)^2}\) units = √64+36 units = √100 units = 10 units

\(\overline{\mathrm{OA}}+\overline{\mathrm{AB}}=5+5=10=\overline{\mathrm{OB}}\)

∴ 0 (0, 0), A (4, 3), and B (8, 6) are collinear. (Proved)

 

Question 4. Show that the successive joining of the points (7, 2), (19, 8), (15, 6), and (-11, 12) produce a parallelogram.

Solution:

Given (7, 2), (19, 8), (15, 6), And (-11, 12)

Let the points are A (-7, 2), B (19, 8), C (15, 6), and D (11, 12)

\(\overline{\mathrm{AB}}=\sqrt{(-7-19)^2+(2-8)^2}\) units = √676+36 units

= √712 units 

= √4 x 178 units 

= 2√178 units

∴ \(\overline{\mathrm{BC}}=\sqrt{(19-15)^2+(8+6)^2}\) units= √16+ 196 units -√4×53 units -2√53 units

\(\overline{C D}=\sqrt{(15+11)^2+(-6+12)^2}\) units = √676+36 units = √712 units = 2√178 units

∴ \([\overline{\mathrm{AD}}=\sqrt{(-7+11)^2+(2+12)^2}/latex] units = √16+ 196 units = √212 units=2√53 units

[latex]\overline{\mathrm{AB}}=\overline{\mathrm{CD}} \text { and } \overline{\mathrm{AD}}=\overline{\mathrm{BC}}\)

Clearly, the opposite sides of the quadrilateral are equal.

∴ the quadrilateral is a parallelogram. (Proved)

 

Question 5. Show that the successive joining of the points (2, 5), (5, 9), (9, 12), and (6, 8) produce a rhombus.

Solution:

Given (2, 5), (5, 9), (9, 12), And (6, 8)

Let the given points be A (2, 5), B (5, 9), C (9, 12), and D (6, 8).

∴ \(\overline{\mathrm{AB}}=\sqrt{(2-5)^2+(5-9)^2}\)units = √9+16 units = √25 units = 5 units.

\(\overline{\mathrm{BC}}=\sqrt{(5-9)^2+(9-12)^2}\) units = √16+9 units = √25 units = 5 units. 

∴ \(\overline{C D}=\sqrt{(9-6)^2+(12-8)^2}\) units=√9+16 units = √25 units = 5 units.

∴ \(\overline{\mathrm{AD}}=\sqrt{(2-6)^2+(5-8)^2}\) units = √16+9 units = √25 units = 5 units.

\(\overline{\mathrm{AC}}=\sqrt{(2-9)^2+(5-12)^2}\) units = √49+49 units = √98 units.

\(\overline{\mathrm{BD}}=\sqrt{(5-6)^2+(9-8)^2}\) units = √1+1 units = √2 units.

∴ In the quadrilateral ABCD, \(\overline{\mathrm{AB}}=\overline{\mathrm{BC}}=\overline{\mathrm{CD}}=\overline{\mathrm{AD}} \text { and } \overline{\mathrm{AC}} \neq \overline{\mathrm{BD}}\).

i.e., the sides. of ABCD are equal, but the diagonals are not equal.

∴ ABCD is a rhombus. (Proved)

 

Question 6. The line segment joining the points (7, 1) and (9, 3) is the base of an isosceles triangle. If the abscissa of the triangle is 4, find the vertex.

Solution:

Given The line segment joining the points (7, 1) and (9, 3) is the base of an isosceles triangle

Let the vertex be (4, k).

Now, the distance between (4, k) and (7,-1) = \(\sqrt{(4-7)^2+(k+1)^2}\) units.

Again, the distance between (4, k) and (9, 3)=\(\sqrt{(4-9)^2+(k-3)^2}\) units. 

Since the given triangle is isosceles,

\(\sqrt{(4-7)^2+(k+1)^2}=\sqrt{(4-9)^2+(k-3)^2}\)

⇒ \(\sqrt{9+(k+1)^2}=\sqrt{25+(k-3)^2}\)

9+(k+1)² = 25+ (k − 3)²

9+k²+1+2k = 25+k²+9-6k

2k+6k = 34-10 

8k = 24

⇒ k = 3

∴ the required vertex is (4, 3)

 

Question 7. If the point (x, y) is equidistant from the points (a + b, b- a) and (a – b, a + b), then prove that bx = ay.

Solution: 

Given

The Point (x, y) is equidistant from the points (a + b, b- a) and (a – b, a + b)

Distance between (x, y) and (a + b, b- a) = \(\sqrt{[x-(a+b)]^2+[y-(b-a)]^2}\) units 

= \(\sqrt{[x-(a+b)]^2+[y+(a-b)]^2}\) units

Distance between (x, y) and (a – b, a + b) = \(\sqrt{[x-(a-b)]^2+[y-(a+b)]^2}\) units.

As per the question, \(\sqrt{[x-(a+b)]^2+[y+(a-b)]^2}=\sqrt{[x-(a-b)]^2+[y-(a+b)]^2}\)

[x-(a+b)]² +[y+(a−b)]² = [x-(a – b)]² +[y-(a+b)]²

x² +(a+b)² -2x(a+b) + y² + (a−b)² +2y(a-b)

x² +(a-b)² -2x(a – b) + y² +(a+b)² -2y(a+b) 

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

2x [a-b-a-b] = -2y [a+b+a-b]

⇒ 2bx=2ay

bx = ay (Proved)

 

Question 8. Prove that the circle of centre (4, 3) passes through the points (0, 0), (8, 0), (1, 7) and (1, 1). Find also the radius of the circle.

Solution:

Given The circle of centre (4, 3) passes through the points (0, 0), (8, 0), (1, 7) and (1, 1).

Let the centre of the circle be O (4, 3) and the points be A (0, 0), B (8, 0), C (1, 7) and D (1, 1).

Now, the distance of four points from the centre of the circle is

\(\sqrt{4^2+3^2}\) units=√25 units = 5 units

\(\sqrt{(4-8)^2+3^2}\) units = √25 units = 5 units

\(\sqrt{(4-1)^2+(3-7)^2}\) units = √9+16 units = √25 units = 5 units

\(\sqrt{(4-1)^2+(3+1)^2}\) units=√9+16 units = √25 units = 5 units

Clearly, the distances of all the given points from the centre of the circle are equal. 

∴ the circle passes through the given points. (Proved)

∴ The radius of the circle is 5 units.

 

Question 9. The centre of a circle is (5, 3) and its radius is 5 units. Determine the length of the chord of the circle, which is bisected at the point (3, 2).

Solution:

Given The centre of a circle is (5, 3) and its radius is 5 units

Let the centre of the circle be O (5, 3) and its chord AC is bisected at B (3,2) 

Now, \(\overline{\mathrm{OA}}\) = radius of the circle = 5 units;

\(\overline{\mathrm{OB}}=\sqrt{(5-3)^2+(3-2)^2}\) units=√4+1 units = √5 units.

From the right-angled triangle AOB, we get,

\(\mathrm{AB}^2=\mathrm{AO}^2-\mathrm{OB}^2\) units = (20 – 5)

= 20 units

AB = √20 units=2√5 units.

AC=2AB = 4√5 units

∴ the required length of the chord = 4√5 units.

 

WBBSE Solutions For Class 9 Maths Coordinate Geometry Chapter 1 Distance Formulas Question 9

 

Question 10. Prove that the points (2, 2), (2, 2) and (-2√3, 2√3) are the vertices of an equilateral triangle.

Solution: 

Given Points (2, 2), (2, 2) And (-2√3, 2√3)

Let the given points be A (2, 2), B (-2,-2) and C(-2√3, 2√3).

∴ \(\sqrt{(2+2)^2+(2+2)^2}\) units = √16+16 units = √32 units

\(\sqrt{(-2+2 \sqrt{3})^2+(-2-2 \sqrt{3})^2}\) units = \(\sqrt{2\left[(-2)^2+(2 \sqrt{3})^2\right]}\) units 

= \(\sqrt{2[4+12]}\) units 

= √32 units

\(\sqrt{(2+2 \sqrt{3})^2+(2-2 \sqrt{3})^2}\) units = v units

= \(\sqrt{2\left[(2)^2+(2 \sqrt{3})^2\right]}\) units 

= √2×16 units

= √32 units

Clearly, \(\overline{\mathrm{AB}}=\overline{\mathrm{BC}}=\overline{\mathrm{AC}}\)

∴ The given points are the vertices of an equilateral triangle. (Proved)

 

 

 

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