Subject: Optional Mathematics

The distance formula is used to determine the distance, d, between two points. If the coordinates of the two points are (x1, y1) and (x2, y2), the distance equals the square root of x2 − x1 squared + y2 − y1 squared. The distance formula is derived by creating a triangle and using the Pythagorean theorem to find the length of the hypotenuse. The hypotenuse of the triangle is the distance between the two points.

The calculation of a distance between any two points on a plane surface can be done.

Let P(x_{1},y_{1}) and Q(x_{2},x_{2}) be two given points in the coordinate plane. Draw perpendicular PL from P to the x-axis. Then , OL = x_{1} and PL = y_{1}. Draw perpendicular QM from Q to the x-axis.

Then, OM = x_{2}and QM =y_{2}

Again, draw perpendicular PR from P to the line segment QM.

Then, PR = LM = OM-OL =x_{2}-x_{1}

QR = QM - NM = QM - PL = y_{2}-y_{1}

From right-angled ΔPRQ, by Pythagoras theorem,

PQ^{2} = PR^{2}+ QR^{2} = (x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}

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

(Only square root is to be taken because PQ being the distance beyween two points is positive)

The distance of the point (x,y) from the origin (0,0) =\(\sqrt{(x-0)^2 + (y-0)^2}\) = \(\sqrt{x^2+y^2}\)

**Remarks**

- The formula remains the same if the point P (x
_{1},y_{1}) and Q(x_{2},y_{2}) are taken in different quadrants. - If a point lies on the x-axis, its ordinate is zero. Therefore, any point on x-axis can be taken as (x,0).
- If a point lies on Y-axis, its abscissa is zero. therefore, any point on y-axis can be taken as (0,y).
- To prove that a quadrilateral is a:

a) rhombus, show that all sides are equal.

b) square, show that all are equal and diagonals are also equal.

c) parallelogram, show that opposite sides are equal.

d) rectangle, show that opposite sides are equal and diagonals are also equal. - To prove that a triangle is a :

a) scalene, show that none of the sides are equal.

b) isosceles, show that two sides are equal.

c) equilateral, show that all sides are equal.

d) right-angled triangle, show that square on one side equals to the sum of squares of the other two sides.

To find the coordinates of the point which divides internally the line joining two points (x_{1},y_{1}) and (x_{2},y_{2}) in the given ratio m_{1}:m_{2}.

Let A(x, y) and B(x_{1}, y_{1}) be two given points. Let the point P (x_{2}, y_{2}) divide the line joining AB internally in the ratiom_{1}: m_{2}

Then AP : PB = m_{1}: m_{2}

Draw perpendiculars AL, PN and BM from A, P and B respectively to the x-axis. Then, OL = x_{1}, ON = x, OM = x_{2}, AL = y_{1}, PN = y and BM = y_{2}. Again draw respectively AQ and PR from A and P to the line segments PN and BM respectively.

The AQ =LN = ON -OL = x - x_{1} PR = NM = OM - ON = x_{2}-x

PQ = PN - QN = PN -AL = y - y_{1 }BR = BM - RM = BM - PN = y_{2} - y

In Δ^{s} PQA and BRA, ∠PQA = ∠ BRP, ∠QAP = ∠RPB and ∠APQ = ∠PBR

So, Δ^{s }PQA and BRP are similar.

Then,

\(\frac{AP}{PB}\)

= \(\frac{AQ}{PR}\)

= \(\frac{QP}{RB}\)

= \(\frac{m_1}{m_2}\)

= \(\frac{x-x_1}{x_2-x}\)

= \(\frac{y-y_1}{y_2y}\)

Taking first two relation, we get,

or, \(\frac{m_1}{m_2}\) = \(\frac{x-x_1}{x_2-x}\)

or, m_{2}x - m_{2}x_{1} = m_{1}x_{2}-m_{1}x

or, m_{2}x + m_{1}x = m_{1}x_{2} + m_{2}x_{1}

or, x(m_{1}+m_{2}) = m_{1}x_{2}+m_{2}x_{1}

or, x = \(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}\)

Similarly, from the relations

\(\frac{m_1}{m_2}\) = \(\frac{y-y_1}{y_2-y}\)

we get

y = \(\frac{m_1y_2 + m_2y_1}{m_1 + m_2}\)

∴ The coordinates of P are \(\frac{m_1x_2 + m_2x_1}{m_1 +m_2}\),\(\frac{m_1y_2 + m_2y_1}{m_1+m_2}\)

If the point P(x, y) divides AB externally in the ratio of m_{1}:m_{2} then the divided segment BP is measured in opposite direction and hence m_{2} is taken as negative.

\(\therefore\) The section formulae for external division is,

(x, y) = (\(\frac{m_1x_2 - m_2x_1}{m_1 - m_2}\)),(\(\frac{m_1y_2 - m_2y_1}{m_1 - m_2}\))

In special case, the midpoint formulae is also used'.

m_{1}:m_{2} = 1:1 i.e. m_{1} = m_{2}\(\therefore\) x = \(\frac{x_1 + x_2}{2}\) and y = \(\frac{y_1 + y_2}{2}\)

Thus, co-ordinates P(x, y) are P(\(\frac{x_1 + x_2}{2}\),\(\frac{y_1 + y_2}{2}\)) which is called mid-point formulae.

If the point P(x, y) divides the joining two points A(x_{1}, y_{1}) and B(x_{2}, y_{2}) in the ratio k : 1, then

x = \(\frac{k . x_2 + 1 . x_1}{k + 1}\) = \(\frac{kx_2 + x_1}{k + 1}\)

and, y = \(\frac{k . y_2 + 1 . y_1}{k + 1}\) = \(\frac{ky_2 + y_1}{k + 1}\)

∴ Coordinate of P are (\(\frac{kx_2 + x_1}{k + 1}, \frac{ky_2 + y_1}{k + 1}\))

For the problems in which it is required to find the ratio when a given point divides the join of two given points, it is convenient to take the ratio k : 1.

Let P(x_{1}, y_{1}), Q(x_{2}, y_{2}) and R(x_{3}, y_{3}) are the vertices of a triangle PQR. Let S, T and U are the mid - points of sides QR, RP and PQ respectively. Then, PS, QT and RU are called medians of the triangle PQR. If these medians intersect each other at a point N, then N is called the centroid of the triangle PQR.

Since S is the middle point of the side QR, then its coordinates are (\(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\))

From plane geometry, we know that the centroid of a triangle divides the median in the ratio 2 : 1. Now, if coordinates of N are (x, y), then by section formula.

x = \(\frac{2 . \frac{x_2 + x_3}{2} + 1 . x_1}{2 + 1}\) = \(\frac{x_1 + x_2 + x_3}{3}\) and y = \(\frac{2 . \frac{y_2 + y_3}{2} + 1 . y_1}{2 + 1}\) = \(\frac{y_1 + y_2 + y_3}{3}\)

Hence, coordinates of N are (\(\frac{x_1 + x_2 + x_3}{3},\frac{y_1 + y_2 + y_3}{3}\))

**→** The formula remains the same if the points P (x_{1},y_{1}) and Q (x_{2},y_{2}) are taken in different quadrants.

**→ **If a point lies on x-axis**, **its ordinate is zero. Therefore, any point on x-axis can be taken as (x,0).

**→ **If a point lies on y-axis**, **its abscissa is zero. Therefore, any point on y-axis can be taken as (0,y).

**→** To prove that a quadrilateral is a

** **1) rhombus, show that all sides are equal.

** **2) square, show that all sides are equal and diagonals are equal.

** **3) parallelogram, show that opposite sides are equal.

4) rectangle, show that opposite sides are equal and diagonals are equal.

**→ **To prove that a triangle is a

1) scalene, show that none of the sides are equal.

2) isosceles, show that two sides are equal.

3) equilateral, show that all sides are equal.

4) right-angled triangle, show that square on one side equals

** **

** **

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

A point which divides a line joining points (4 , 3) and (5 , 6) in the ratio 2 : 3.

Let , the given points be

(x_{1} , y_{1}) = (4 , 3) amd (x_{2} , y_{2}) = (5 ,6)

ratio m_{1}:m_{2} = 2 : 3 or m_{1} = 2 and m_{2} = 3

x = \(\frac{m1x2+m2x1}{m1+m2}\) and y = \(\frac{m1y2 + m2y1}{m1 + m2}\)

\(\therefore\) x = \(\frac{2.5+3.4}{2+3}\) and y = \(\frac{2.6 + 3.3}{2 + 3}\)

= \(\frac{10 + 12}{5}\) = \(\frac{22}{5}\) and y = \(\frac{12+9}{5}\) = \(\frac{21}{5}\)

\(\therefore\) (x , y) = \(\frac{22}{5}\) , \(\frac{21}{5}\) Ans.

Find the co-ordinates of :

A points which divides a line joining points (4,3) and (5,6) in the ratio 2:3.

Let the fiven points be

(x_{1} y_{1})=(4,3) and (x_{2} y_{2})=(5,6)

ratio m_{1}:m_{2}=2:3 or m_{1}=2 and m_{2}=3

Using section formula

x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\) and y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\)

x=\(\frac{2.5+3.4}{2+3}\) and y=\(\frac{2.6+3.3}{2+3}\)

=\(\frac{10+12}{5}\) =\(\frac{22}{5}\) and y=\(\frac{12+9}{5}\) =\(\frac{21}{5}\)

∴(x,y)=(\(\frac{22}{5}\),\(\frac{21}{5}\)).Ans.

Find the co-ordinates of :

A points which divides a line joining points (-3,4) and (-8,7) in the ratio 3:4.

Let the fiven points be

(x_{1} y_{1})=(-3,-4) and (x_{2} y_{2})=(-8,7)

ratio m_{1}:m_{2}=3:4 or m_{1}=3 and m_{2}=4

Using section formula

x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\) and y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\)

x=\(\frac{3×-8+4×-3}{3+4}\) ,y=\(\frac{3×7+4×-4}{3+4}\)

=\(\frac{-24-12}{7}\) , y=\(\frac{21-16}{7}\)

∴(x,y)=(\(\frac{-36}{7}\),\(\frac{5}{7}\)).Ans.

Find the co-ordinates of :

A points which divides a line joining points (2,-3) and (-6,5) in the ratio 4:5.

Let the fiven points be

(x_{1} y_{1})=(2,-3) and (x_{2} y_{2})=(-6,5)

ratio m_{1}:m_{2}=4:5

Using section formula

x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\) and y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\)

x=\(\frac{4 ×-6+5×2}{4+5}\) ,y=\(\frac{4 × 5 +5 ×-3}{4+5}\)

=\(\frac{-24+10}{9}\) , y=\(\frac{20-15}{9}\)

=\(\frac{-14}{9}\), \(\frac{5}{9}\)

∴(x,y)=(\(\frac{-14}{9}\),\(\frac{5}{9}\)).Ans.

Find the co-ordinates of :

A points which divides a line joining points (-3,-4) and (-8,7) in the ratio 3:(-4).

Let the given points be

(x_{1} y_{1})=(-3,-4) and (x_{2} y_{2})=(-8,7)

ratio m_{1}:m_{2}==3:(-4) or m_{1}=3 and m_{2}=-4

Using section formula

x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\) and y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\)

x=\(\frac{3×-8+(-4)×-3}{3+(-4)}\) ,y=\(\frac{3×7+(-4)×-4}{3+(-4)}\)

=\(\frac{-24+12}{3-4}\) , y=\(\frac{21+16}{3-4}\)

=\(\frac{-12}{-1}\), =\(\frac{37}{-1}\)

= 12 , =-37

∴(x,y)=(12,-37).Ans.

Find the co-ordinates of a point which divides the line joining the points (1,-2) and (4,7) in the following ratio:

- internally in the ratio1:2
- externally in the ratio 2:3

- Let the given points be

(x_{1} y_{1})=(1,-2) and (x_{2} y_{2})=(4,7)

Internal ratio m_{1}:m_{2}=1:2

Using formula,

x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\) and y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\)

or, x=\(\frac{1 × 4 + 2×1 }{1+2}\) ,y=\(\frac{1 ×7+2×-2}{1+2}\)

=\(\frac{4+2}{3}\) , =\(\frac{7-4}{3}\)

=\(\frac{6}{3}\), =\(\frac{3}{3}\)

= 2 , =1

∴(x,y)=The required point=(2,1).Ans.

- Let the given points be
(x

_{1}y_{1})=(1,-2) and (x_{2}y_{2})=(4,7)Let (x,y) divides externally in the ratio m

_{1}:m_{2}=2:3 or m_{1}=2 and m_{2}=3Using formula,

x=\(\frac{m_1x_2-m_2x_1}{m_1-m_2}\) and y=\(\frac{m_1y_2-m_2y_1}{m_1-m_2}\)

or, x=\(\frac{2 × 4 - 3 ×1 }{2-3}\) ,y=\(\frac{2 ×7-3×-2}{2-3}\)

=\(\frac{8-3}{-1}\) , =\(\frac{14+6}{-1}\)

=\(\frac{5}{-1}\), =\(\frac{20}{-1}\)

= -5 , =-20

∴The required point=(-5,-20).Ans.

What are the cco-ordinates of the points which divide the line joining the points (4,-5) and (6,3) internally and externally in the ratio 2:5?

Here, Let the given points be (x_{1},y_{1})= (4,-5) and (x_{2},y_{2}) = (6,3) and m_{1}:m_{2}=2:5

Let (x,y) divides the given points internally then

Using formula,

x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\) and y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\)

x=\(\frac{2×6+5×4}{2+5}\) ,y=\(\frac{2×3+5×-5}{2+5}\)

=\(\frac{12+20}{7}\) , y=\(\frac{6-25}{7}\)

=\(\frac{32}{7}\), =\(\frac{-19}{7}\)

∴The required point =(\(\frac{32}{7}\),\(\frac{-19}{7}\)).Ans.

If (x^{1},y^{1}) divides externally then using formula

x'=\(\frac{m_1x_2-m_2x_1}{m_1-m_2}\) and y'=\(\frac{m_1y_2-m_2y_1}{m_1-m_2}\)

x'=\(\frac{2×6-5×4}{2-5}\) ,y'=\(\frac{2×3-5 ×-5)}{2-5}\)

=\(\frac{12-20}{-3}\) , y=\(\frac{6+25}{-3}\)

=\(\frac{-8}{-3}\), =\(\frac{31}{-3}\)

∴The required point=\(\frac{8}{3}\), =\(\frac{-31}{3}\) Ans.

Find the co-ordinates of the midpoint of the line joining the points A(-3,-6) and B(1,-2).

Here,

Let the given points be (x_{1},y_{1}) =A(-3,-6) and (x_{2},y_{2}) = B(1,-2) If (x,y) be the mid point, then using

x=\(\frac{x_1+x_2}{2}\) and y=\(\frac{y_1+y_2}{2}\)

∴ x =\(\frac{-3+1}{2}\) and y=\(\frac{-6-2}{2}\)

=\(\frac{-2}{2}\), =\(\frac{-8}{2}\)

∴x=-1, =-4

∴(x,y)=(-1,-4).Ans.

Find the distance of the following points.

a. (2 , 3) and (4 , 3)

Here , given P(x1 , y1) = (2 , 3) and Q(x2 , y2) = (4 , 3)

Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)

Distance , PQ = \(\sqrt{(4-2)^{2} + (3-3)^{2}}\) = \(\sqrt{2^{2} + 0^{2}}\) = \(\sqrt{2^{2}}\) = 2.

PQ = 2 units. Ans.

If the co-ordinates of the midpoint of a line joining the points M(1,4) and N(x,y) is (-2,2),what are the value of x and y?

Here,

Let the mid point of M(1,4) and N(x,y) be (-2,2)

-2=\(\frac{1+x}{2}\) and 2=\(\frac{4+y}{2}\)

or, -4=1+x, 4=4+y

or, -4=-1=x, 4-4=y

or, -5=x, 0=y

∴(x,y)=(-5,0).Ans.

Find the distance between the given pairs of points.

(-1 , 3) and (5 , 1)

Here given , P(x1, ,y1) = (-1 , 3) and Q(x2 , y2) = (5 , 1)

Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)

Distance PQ = \(\sqrt{(5 -(-1)^{2} + (1 -3)^{2}}\) = \(\sqrt{6^{2} + 2^{2}}\) = \(\sqrt{36 + 4}\)

\(\therefore\) PQ = \(\sqrt{40}\) = \(\sqrt{4 \times 10}\) = 2\(\sqrt{10}\) Units. Ans.

If one end and the midpoint of a line are (4,4) and (-3,2) respectively, find the co-ordinates of the other end point of the line.

Here,

Let one end (x_{1}y_{1})=(4,4) they let other end =(x_{2}y_{2})

Midpoint (x,y)=(-2,2)

Using formula,

x=\(\frac{x_1+x_2}{2}\) and y=\(\frac{y_1+y_2}{2}\)

-2=\(\frac{4+x_2}{2}\), 2=\(\frac{y_1+y_2}{2}\)

or, -4=4+x_{2} 4=4+y_{2}

or, -4-4=x_{2} 0=y_{2}

or, -8=x_{2}, 0=y_{2}

∴(x_{2}y_{2})=(-8,0).Ans.

Find the distance between the given pairs of points.

(1 , -2) and (-2 , 2)

Here given , P(x1 , y1) = (1 , -2) and Q(x2, , y2) = (-2 , 2)

Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)

Distance PQ = \(\sqrt{(-1 -3)^{2} + [-1 - (-1)]^{2}}\)

= \(\sqrt{(-4)^{2} + (-1 = 1)^{2}}\)

= \(\sqrt{4^{2} + 0^{2}}\) = \(\therefore\) PQ = \(\sqrt{4^{2}}\) = 4 units. Ans.

Find the distance between the given pairs of points.

(-6 , 7) and (-1 , -5)

Here given , P(x1, y1) = (-6 , 7) and Q(x2 , y2) = (-1 , -5)

Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)

Distance PQ = \(\sqrt{[-1 -(-6)]^{2} + (-5 -7)^{2}}\)

= \(\sqrt{-1 ^{2} + 6^{2} + (-12)^{2}}\) = \(\sqrt{5 ^{2} + 144}\)

= \(\sqrt{25 + 144}\) = \(\sqrt{169}\)

\(\therefore\) PQ = 13 Units Ans.

In what ratio is the line joining the points (2,-4) and (-3,6) is divided by

x-axis?

Here given points,A(2,-4) and B(-3,6)

Let the points on x-axis be P(x,0) which divides AB in the ratio k:1 then,

Using formula y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\)

0=\(\frac{k×6+1×-4}{k+1}\)

or, 0×(k+1)=6k-4

or, 6k-4=0

or, 6k=4, or k=\(\frac{4}{6}\)=\(\frac{2}{3}\)

∴This shows that x-axis divides the line AB internally in the ratio 2:3.Ans.

Find the distance bewteen the given pairs of points.

(4 , 3) and (-2 , 2)

Here given , P(x1 , y1) = (4 , 3) and Q(x2 , y2) = (3 , -6)

Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1) ^{2}}\)

Distance PQ = \(\sqrt{(-2 , -4)^{2} + (2 - 3 )^{2}}\) = \(\sqrt{(-6)^{2} + (-1)^{2}}\)

= \(\sqrt{36 + 1}\) = \(\sqrt{37}\) Units. Ans.

Find the distance from the given pairs of points.

(-2 , 6) and (3 , -6)

Here , given P(x1 , y1) = (-2 , 6) and Q(x2 , y2) = (3 , -6)

Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)

Distance PQ = \(\sqrt{[ 3 - (-2)^{2} + (-6 -6)^{2}}\)

= \(\sqrt{(3 + 2)^{2} + (-12)^{2}}\) = \(\sqrt{5 ^{2} + 144}\)

= \(\sqrt{25 + 144}\) = \(\sqrt{169}\) = 13 Units. Ans.

Find the points whose x-coordinates is 3 and which is on the line joining the points P(7,-3) and Q(-2,-5).

Here, Let (3,y) be a point in the line PQ with co-ordinates P(7,-3) and Q(-2,-5) and as it divided PQ in the ratio k:1 then,

Using formula, x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\)

3=\(\frac{k×-2+1×7}{k+1}\) or, 3k +3=-2k+7

or, 3k +2k=7-3,or, 5k = 4 ∴k=\(\frac{4}{5}\)

∴Ratio m_{1}:m_{2}=4:5

Again using y =\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\)

or, y=\(\frac{4 ×-5+5 ×-3}{4+5}\) =\(\frac{-20-15}{9}\)

=\(\frac{-35}{9}\)

∴ Requierd point (x,y)=(3,\(\frac{-35}{9}\)).Ans.

If A(0 , 0) , B(3 , -4) , C(-3 , 4) , D(-2 , 2) and E (10 , - 3) are five points , find the distance between the following point.

1. A and B

Here, given A(0 , 0) and B(3 , -4)

\(\therefore\) AB = \(\sqrt{(3 - 0)^{2} + (4 - 0)^{2}}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5 units.

If A(0 , 0) , B(3 , -4) , C(-3 , 4) , D(-2 , 2) and E (10 , - 3) are five points , find the distance between the following point.

1. B and C

Here given , A(0 , 0) amd C(-3 , 4)

\(\therefore\) AC = \(\sqrt{(-3 -0)^{2} + (4 - 0)^{2}}\)

= \(\sqrt{(-3)^{2} + (4)^{2}}\)

= \(\sqrt{9 + 16}\)

= \(\sqrt{25}\) = 5 units. Ans.

If A(0 , 0) , B(3 , -4) , C(-3 , 4) , D(-2 , 2) and E (10 , - 3) are five points , find the distance between the following point.

1. A and C

Here given , A(0 , 0) and C (-3 , 4)

\(\therefore\) AC = \(\sqrt{(-3 - 0)^{2}+(4 - 0)^{2} }\)

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

In what ratio is the line joining the points (2,-4) and (-3,6) is divided by

y-axis

Let the point Q(0,y) on y-axis divides AB in the ratio k:1,then,

Using formula y=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\)

0=\(\frac{k×-3+1×2}{1+2}\)

or, 0=3k+2

or, 3k=2∴k=\(\frac{2}{3}\)

∴y-axis divides internally in the ratio 2:3.Ans.

Prove that a triangle with the vertices (2 , 4) (6 , 4) and (6 , 7) is a right angled triangle.

Let , ABC be a triangle whose vertices are A(2 , 4) B(6 , 4) , C(6 , 7)

Using distance formula ,

AB = \(\sqrt{(6 - 2)^{2} + (4 - 4)^{2}}\) = \(\sqrt{4^{2}}\) = 4

BC = \(\sqrt{(6 - 6)^{2} + (7 - 4)^{2}}\) = \(\sqrt{3^{2}}\) = 3

CA= \(\sqrt{(2 -6)^{2} + (4 - 7)^{2}}\) = \(\sqrt{16 + 9}\) = \(\sqrt{25}\) = 5

Again here ,

AC \(^2\) = AB\(^2\) + BC\(^2\)

or , 5\(^2\) = 4\(^2\) + 3\(^2\)

\(\therefore\) 25 = 25

Since , h\(^2\) = p + b\(^2\) is satisfied so by pythagoras theorem \(\triangle\) ABC is a right angled triangle.

Here,A(-2,1) and B(4,3) be any two points.
Using distance formula,we get
(AB)2 =(x2-x1)2+(y2-y1)2
=(-2-4)2 + (1-3)2
=(-6)2 +(-2)2
=36+4 =40

© 2019-20 Kullabs. All Rights Reserved.