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Section Formula – Vector Algebra
  • Last Updated : 12 May, 2021

Physical quantities are divided into two categories – scalar and vector quantities. The quantities which have only magnitude and not any fixed direction are called Scalar quantities. Eg. Mass, volume, density, etc. Quantities that have both magnitude and direction. Such quantities are called vector quantities or vectors. Eg. Displacement, velocity, acceleration, momentum, etc.

Vectors are represented by line segments that are directed such as, the direction of an arrow marked at one end, which denotes direction and the length of the line segment is the magnitude of the vector. Eg. \overrightarrow{AB}

It denotes two points A and B, such that the magnitude of the vector is the length of the straight line AB and its direction is from A to B. Here, point A is called the initial point of vector \overrightarrow{AB}    and point B is called the terminal point(or endpoint).

The magnitude or modulus of a vector \overrightarrow{AB} is a positive number(>0) which is the measure of its length and is denoted by |\overrightarrow{AB}|.

Types of Vectors

Zero Vector 

A vector whose initial and terminal(end) points coincide with each other, is called a zero vector (or null vector), and denoted as \overrightarrow{0}   . Zero vector can not be assigned a defined direction as it has zero magnitude. Or, in other words, it may be defined as having any direction. 

The vectors \overrightarrow{AA}   \overrightarrow{BB}    represent the zero vector.

Unit Vector

A vector whose magnitude is unity (=1) is called as unit vector. The unit vector \overrightarrow{a}    is denoted by \hat{a}

and |\hat{a}|    = 1

Coinitial Vectors 

Two or more vectors having the same initial point or starting point are called coinitial vectors.

Collinear Vectors 

If two or more vectors are parallel to each other, irrespective of their magnitudes and directions. Then they will be called collinear vectors.

Equal Vectors 

If two vectors \overrightarrow{a}    and \overrightarrow{b}    have the same magnitude and direction regardless of the positions of their initial points will be called as equal vectors, and it can be represented as \overrightarrow{a} = \overrightarrow{b}

Negative of a Vector 

A vector whose magnitude(length) is equal to of given vector (say \overrightarrow{AB}   ) but the direction is opposite to that of it(initial and terminal points are reversed), is called negative of the given vector. 

For example, vector \overrightarrow{BA}    is negative of the vector \overrightarrow{AB}    and written as, \overrightarrow{BA} = - \overrightarrow{AB}

Position Vector

Consider a point P in 3D space, having coordinates as (x, y, z) with respect to the origin O(0, 0, 0). And, the vector \overrightarrow{OP}    is having O as the initial point and P as its terminal point is called the position vector of the point P with respect to O.

Using the distance formula, the magnitude(or length) of \overrightarrow{OP}   is

|\overrightarrow{OP}|    = \sqrt{(x-0)^2+(y-0)^2+(z-0)^2}

|\overrightarrow{OP}| = \sqrt{x^2+y^2+z^2}

Note: The vector’s definition defined above are such type of vectors that can be subjected to its parallel displacement without changing its magnitude and direction. Such vectors are called free vectors.

Vector Addition

A vector \overrightarrow{AB}    simply denotes the displacement of anything from point A to point B. 

Triangle law of vector addition

Consider a situation that a person moves from A to B and then from B to C. The net displacement made by the person from point A to point C, is given by the vector and expressed as

\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}

This is known as the triangle law of vector addition.

As, \overrightarrow{AC} = -\overrightarrow{CA}    

- \overrightarrow{CA} = \overrightarrow{AB} + \overrightarrow{BC}

\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = 0

When the sides of a triangle are taken in order, then it leads to zero resultant(no displacement result) as the initial and terminal points get coincided with each other.

Parallelogram Law of Vector Addition

If two vectors \overrightarrow{a}    and \overrightarrow{b}    are represented in magnitude and direction by the two adjacent sides of the parallelogram, then their sum(\overrightarrow{a}+\overrightarrow{b}   \overrightarrow{c}    is represented by the diagonals of the given parallelogram which is coinitial with the given vectors \overrightarrow{a}    and \overrightarrow{b}   .

Now, let’s consider the parallelogram ABCD

where, \overrightarrow{AB} = \overrightarrow{a}    and \overrightarrow{BC} = \overrightarrow{b}

then by using the triangle law of vector addition, from triangle ABC, we have

\overrightarrow{AC} = \overrightarrow{a} +\overrightarrow{b}    ………………….(1)

Now, as the opposite sides of a parallelogram are equal and parallel

\overrightarrow{AB} = \overrightarrow{DC} = \overrightarrow{a}    and \overrightarrow{AD} = \overrightarrow{BC} = \overrightarrow{b}

Again using triangle law of vector addition, from triangle ADC, we have

\overrightarrow{AC} = \overrightarrow{b} +\overrightarrow{a}    …………………………….(2)

From (1) and (2), we get commutative property

\overrightarrow{a} +\overrightarrow{b} = \overrightarrow{b} +\overrightarrow{a}

Note: The magnitude \overrightarrow{a} +\overrightarrow{b}    is not equal to the sum of the magnitude(length) of \overrightarrow{a}    and \overrightarrow{b}   .\overrightarrow{a} +\overrightarrow{b} \neq |\overrightarrow{a} +\overrightarrow{b}|

(ii) Associative property

Using triangle law, from triangle PQR, we have

\overrightarrow{PR} = \overrightarrow{PQ} +\overrightarrow{QR}

\overrightarrow{PR} = \overrightarrow{a} +\overrightarrow{b}

Using triangle law, from triangle QRS, we have

\overrightarrow{QS} = \overrightarrow{QR} +\overrightarrow{RS}

\overrightarrow{QS} = \overrightarrow{b} +\overrightarrow{c}

Using triangle law, from triangle PRS, we have

\overrightarrow{PS} = \overrightarrow{PR} +\overrightarrow{RS}

\overrightarrow{PS} = (\overrightarrow{a} +\overrightarrow{b})+\overrightarrow{c}

Using triangle law, from triangle PQS, we have

\overrightarrow{PS} = \overrightarrow{PQ} +\overrightarrow{QS}

\overrightarrow{PS} = \overrightarrow{a} +(\overrightarrow{b}+\overrightarrow{c})


(\overrightarrow{a} +\overrightarrow{b})+\overrightarrow{c} = \overrightarrow{a} +(\overrightarrow{b}+\overrightarrow{c})

(iii) Additive identity

For any vector \overrightarrow{a}

\overrightarrow{a} + \overrightarrow{0} = \overrightarrow{0} + \overrightarrow{a} = \overrightarrow{a}            

(iv) Additive inverse

For any vector \overrightarrow{a}

\overrightarrow{a} + (\overrightarrow{-a}) =0 = (\overrightarrow{-a}) + \overrightarrow{a}

Problem 1: If \overrightarrow{PO} + \overrightarrow{OQ} = \overrightarrow{QO} + \overrightarrow{OR}   , show that the points P, Q and R are collinear.


We have,

\overrightarrow{PO} + \overrightarrow{OQ} = \overrightarrow{QO} + \overrightarrow{OR}

Using, converse of triangle law of addition of vectors, we get

\overrightarrow{PQ} = \overrightarrow{QR}

\overrightarrow{PQ}    and \overrightarrow{QR}    ae either parallel or collinear. But, Q is a point common to them. 

So, \overrightarrow{PQ}    and \overrightarrow{QR}    are collinear. Hence, points P, Q, R are collinear.

Problem 2: B, P, Q, R and A are five points in a plane. Show that the sum of vectors \overrightarrow{AP}, \overrightarrow{AQ}, \overrightarrow{AR}, \overrightarrow{PB}, \overrightarrow{QB}    and \overrightarrow{RB}    in 3 \overrightarrow{AB}   .


Using triangle law addition of vectors in △APB

\overrightarrow{AB}=\overrightarrow{AP} + \overrightarrow{PB}        ……………(1)

Using triangle law addition of vectors in △AQB

\overrightarrow{AB}=\overrightarrow{AQ} + \overrightarrow{QB}        …………………..(2)

Using triangle law addition of vectors in △ARB

\overrightarrow{AB}=\overrightarrow{AR} + \overrightarrow{RB}        …………….(3)

Adding (1), (2) and (3), we get

\overrightarrow{AP} + \overrightarrow{PB}+\overrightarrow{AQ} + \overrightarrow{QB}+\overrightarrow{AR} + \overrightarrow{RB}=3\hspace{0.1cm}\overrightarrow{AB}

Hence, the sum of the vectors is 3 \overrightarrow{AB}

Problem 3: Let O be the centre of a regular hexagon CDEFAB. Find the sum of the vectors \overrightarrow{OA}, \overrightarrow{OB}, \overrightarrow{OC}, \overrightarrow{OD}, \overrightarrow{OE}    and \overrightarrow{OF}   .


As hexagon property states that, the centre of a regular hexagon bisects all the diagonals passing through it.


\overrightarrow{OA}=-\overrightarrow{OD}, \overrightarrow{OB}=-\overrightarrow{OE}    and \overrightarrow{OC}=-\overrightarrow{OF}

\overrightarrow{OA}+\overrightarrow{OD}=0         ………………..(1)

\overrightarrow{OB}+\overrightarrow{OE}=0         …………………..(2)

\overrightarrow{OC}+\overrightarrow{OF}=0         …………………..(3)

By adding (1), (2) and (3), we get

\overrightarrow{OA}+ \overrightarrow{OB}+ \overrightarrow{OC}+ \overrightarrow{OD}+ \overrightarrow{OE}+ \overrightarrow{OF} = 0

Section formula

Here, the points P and Q are the two points represented by the position vectors \overrightarrow{OP}    and \overrightarrow{OQ}   , respectively, with respect to the origin O. Then the line segment joining the points P and Q can be divided by a third point, here we say R, in two ways as follows:

Here, we intend to find the position vector \overrightarrow{OR}    for the point R with respect to the origin O. We take the two cases one by one. 


When R divides PQ internally. If R divides \overrightarrow{PQ}    such that

\frac{\overrightarrow{PR}}{\overrightarrow{RQ}} = \frac{m}{n}

where m and n are positive values, we specify that the point R divides \overrightarrow{PQ}    internally in the ratio of m : n. 

Now from triangles ORQ and OPR, we have

\overrightarrow{PR} = \overrightarrow{OR} - \overrightarrow{OP} = \overrightarrow{r} - \overrightarrow{a}

\overrightarrow{RQ} = \overrightarrow{OQ} - \overrightarrow{OR} = \overrightarrow{b} - \overrightarrow{r}

Hence, we can conclude that,

(\overrightarrow{b} - \overrightarrow{r})    = n (\overrightarrow{r} - \overrightarrow{a})

On simplification, we get

\mathbf{\overrightarrow{r} = \frac{m\overrightarrow{b}+n\overrightarrow{a}}{m+n}}

When R is the midpoint PQ

then m = n

So, we get

\mathbf{\overrightarrow{r} = \frac{\overrightarrow{b}+\overrightarrow{a}}{2}}


When R divides PQ externally. If R divides \overrightarrow{PQ}    such that

\frac{\overrightarrow{PR}}{\overrightarrow{QR}} = \frac{m}{n}

where m and n are positive values, we say that the point R divides \overrightarrow{PQ}    externally in the ratio of the m : n.

Now from triangles ORQ and OPR, we have

\overrightarrow{PR} = \overrightarrow{OR} - \overrightarrow{OP} = \overrightarrow{r} - \overrightarrow{a}

\overrightarrow{RQ} = \overrightarrow{OR} - \overrightarrow{OQ} = \overrightarrow{r} - \overrightarrow{b}

Hence, we can conclude that,

m (\overrightarrow{r} - \overrightarrow{b}) = n (\overrightarrow{r} - \overrightarrow{a})

m\overrightarrow{r} - n\overrightarrow{r} = m\overrightarrow{b} - n\overrightarrow{a}

(m-n) \overrightarrow{r} = m\overrightarrow{b} - n\overrightarrow{a}

On simplification, we get

\mathbf{\overrightarrow{r} = \frac{m\overrightarrow{b}-n\overrightarrow{a}}{m-n}}

Problem 1: Find the position vectors of the points which divide the join of the points 2\overrightarrow{a}-3\overrightarrow{b}     and 3\overrightarrow{a}-2\overrightarrow{b}    internally and externally in the ratio 2 : 3.


Let A and B be the given points with the position vectors 2\overrightarrow{a}-3\overrightarrow{b}    and  3\overrightarrow{a}-2\overrightarrow{b}    respectively.

Let P divide the \overrightarrow{AB}   in the ratio 2 : 3 internally

m = 2 and n = 3

Using internally section formula,

Position vector of P =  \frac{m(2\overrightarrow{a}-3\overrightarrow{b})+n(3\overrightarrow{a}-2\overrightarrow{b})}{m+n}

Position vector of P =  \frac{3(2\overrightarrow{a}-3\overrightarrow{b})+2(3\overrightarrow{a}-2\overrightarrow{b})}{2+3}         

Position vector of P =  \frac{6\overrightarrow{a}-9\overrightarrow{b}+6\overrightarrow{a}-4\overrightarrow{b}}{5}

Position vector of P =  \frac{12\overrightarrow{a}}{5}-\frac{13\overrightarrow{b}}{5}

Now, Let P divide the \overrightarrow{AB}    in the ratio 2 : 3 externally

m = 2 and n = 3

Using externally section formula,

Position vector of P =  \frac{m(2\overrightarrow{a}-3\overrightarrow{b})-n(3\overrightarrow{a}-2\overrightarrow{b})}{m-n}

Position vector of P =  \frac{3(2\overrightarrow{a}-3\overrightarrow{b})-2(3\overrightarrow{a}-2\overrightarrow{b})}{3-2}

Position vector of P =  \frac{6\overrightarrow{a}-9\overrightarrow{b}-6\overrightarrow{a}+4\overrightarrow{b}}{1}    

Position vector of P = -5\overrightarrow{b}

Problem 2: If \overrightarrow{a}    and \overrightarrow{b}    are position vectors of points A and B respectively, then find the position vector of points of trisection of AB.


Let P and Q be points of trisection. Then, AP = PQ = QB = k (constant variable)

PB = PQ + QB = k + k = 2k

\frac{AP}{PB} = \frac{k}{2k} = \frac{1}{2}

P divides AB in the ratio 1 : 2

Using internally section formula, where m=1 and n=2

Position vector of P =  \frac{m(\overrightarrow{b})+n(\overrightarrow{a})}{m+n}

Position vector of P =  \frac{1(\overrightarrow{b})+2(\overrightarrow{a})}{1+2}

Position vector of P =  \frac{\overrightarrow{b}+2\overrightarrow{a}}{3}

Now, we can clearly see that Q is the mid-point of PB.

Apply mid-point section formula we have,

Position vector of Q =  \frac{\frac{\overrightarrow{b}+2\overrightarrow{a}}{3}+\overrightarrow{b}}{2}

Position vector of Q =  \frac{4\overrightarrow{b}+2\overrightarrow{a}}{6}    

Position vector of Q =  \frac{\overrightarrow{a}+2\overrightarrow{b}}{3}

Problem 3: If D is the mid-point of the side BC of a triangle ABC, prove that \overrightarrow{AB} + \overrightarrow{AC} = 2 \overrightarrow{AD}


Let A be the origin here and position vectors of B and C be \overrightarrow{b}    and \overrightarrow{c}    respectively.

As D is the mid-point of BC.

2 \overrightarrow{BD} = \overrightarrow{BC}

Applying mid-point section formula, we get

Position vector of D =  \frac{\overrightarrow{b}+\overrightarrow{c}}{2}

As, \overrightarrow{AB} + \overrightarrow{AC} = \overrightarrow{b} + \overrightarrow{c}

\overrightarrow{AB} + \overrightarrow{AC} = 2 (\frac{\overrightarrow{b} + \overrightarrow{c}}{2})

\overrightarrow{AB} + \overrightarrow{AC} = 2 (\overrightarrow{AD})

Hence, Proved!!

Components of a vector

Let’s take the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on the x-axis, y-axis and z-axis, respectively. Then,

|\overrightarrow{OA}|    = 1, |\overrightarrow{OB}|    = 1 and |\overrightarrow{OC}|    = 1

The vectors, |\overrightarrow{OA}|   |\overrightarrow{OB}|    and |\overrightarrow{OC}|    are having magnitude unity or 1, which are called unit vectors along the axes OX, OY and OZ, respectively, and denoted by \hat{i}   \hat{j}    and \hat{k}    respectively.

Lets, consider the position vector \overrightarrow{OP}   , of a point P(x, y, z). Let P1 be the foot of the perpendicular from P on the XY plane.

Hence we see that P1P is parallel to z-axis. As \hat{i}   \hat{j}    and \hat{k}    are the unit vectors along the axes x, y and z, respectively, and by the definition of the coordinates of P, we have

\overrightarrow{P_1P} = \overrightarrow{OR} = z\hat{k}

\overrightarrow{QP_1} = \overrightarrow{OS} = y\hat{j}            

\overrightarrow{OQ} = x\hat{i}

Using triangle law, from triangle OQP1, we have

\overrightarrow{OP_1} = \overrightarrow{OQ} +\overrightarrow{QP_1} = x\hat{i}+y\hat{j}    

Using triangle law, from triangle OP1P, we have

\overrightarrow{OP} = \overrightarrow{OP_1} +\overrightarrow{P_1P}=x\hat{i}+y\hat{j}+z\hat{k}

Hence, the position vector of P with reference to O is as follows:

\overrightarrow{OP} (or\overrightarrow{r})=x\hat{i}+y\hat{j}+z\hat{k}


The length of vector \overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}    , 

OP12 = OQ2 + QP12    (Using Pythagoras theorem)

OP12 = x2 + y2

and in the right-angled triangle OP1 P, we have

 OP2 = OP12 + PP12 

OP2 = x2 + y2 + z 

OP = \sqrt{x^2+y^2+z^2}

Hence, the length of vector \overrightarrow{r}=|\overrightarrow{r}|=|x\hat{i}+y\hat{j}+z\hat{k}|    = \sqrt{x^2+y^2+z^2}

If \overrightarrow{a}    and \overrightarrow{b}    are two vectors as \overrightarrow{a} = a_1\hat{i}+b_1\hat{j}+c_1\hat{k}    and \overrightarrow{b} = a_2\hat{i}+b_2\hat{j}+c_2\hat{k}    then,


 \overrightarrow{a}+\overrightarrow{b} = (a_1+a_2)\hat{i}+(b_1+b_2)\hat{j}+(c_1+c_2)\hat{k}


\overrightarrow{a}-\overrightarrow{b} = (a_1-a_2)\hat{i}+(b_1-b_2)\hat{j}+(c_1-c_2)\hat{k}

The vectors \overrightarrow{a} = \overrightarrow{b}   , iff

a1 = a2, b1 = b2 and c1 = c2

The multiplication of vector  by any scalar k is given by

k\overrightarrow{a} = (ka_1)\hat{i}+(kb_1)\hat{j}+(kc_1)\hat{k}

Vector joining two points

If P(x1, y1, z1) and Q(x2, y2, z2) are any two points, then the vector joining P and Q is the vector \overrightarrow{PQ}

Joining the points P and Q with the origin O, and applying triangle law, from the triangle OPQ, we have


Using the properties of vector addition, the above equation become




Hence, the magnitude of \overrightarrow{PQ}    is as follows:

|\overrightarrow{PQ}| = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}

Problem 1: Find the value of x, y and z so that the vectors \overrightarrow{a} = x\hat{i}+2\hat{j}+z\hat{k}    and \overrightarrow{b} = 2\hat{i}+y\hat{j}+\hat{k}    are equal.


Two vectors \overrightarrow{a} = a_1\hat{i}+b_1\hat{j}+c_1\hat{k}    and \overrightarrow{b} = a_2\hat{i}+b_2\hat{j}+c_2\hat{k}    are equal iff

a1 = a2 , b1 = b2 and c1 = c2

Hence, the values of x = 2, y = 2 and z =1.

Problem 2: Find the magnitude of the vector \overrightarrow{a} = 3\hat{i}-2\hat{j}+6\hat{k}


As, we have \overrightarrow{a} = 3\hat{i}-2\hat{j}+6\hat{k}

|\overrightarrow{a}| = \sqrt{3^2+(-2)^2+6^2}

|\overrightarrow{a}| = \sqrt{49}    = 7

Problem 3: Find the unit vector of given vector as  3\hat{i}-6\hat{j}+2\hat{k}


Let, \overrightarrow{a} = 3\hat{i}-6\hat{j}+2\hat{k}

|\overrightarrow{a}| = \sqrt{3^2+(-6)^2+2^2}

|\overrightarrow{a}| = \sqrt{49}    = 7

So, the unit vector in the direction of \overrightarrow{a}    is given by,

\hat{a} = \frac{\overrightarrow{a}}{|\overrightarrow{a}|}

\hat{a} = \frac{3\hat{i}-6\hat{j}+2\hat{k}}{7}

\hat{a} = \frac{3}{7}\hat{i}-\frac{6}{7}\hat{j}+\frac{2}{7}\hat{k}

Problem 4: Find the unit vector of given vector as \overrightarrow{PQ}   , where the points P(1,2,3) and Q(4,5,6).


Position vector of P(1,2,3) = \hat{i}+2\hat{j}+3\hat{k}

Position vector of Q(4,5,6) = 4\hat{i}+5\hat{j}+6\hat{k}

\overrightarrow{PQ}        = \overrightarrow{Q}        – \overrightarrow{P}

\overrightarrow{PQ} = (4\hat{i}+5\hat{j}+6\hat{k}) - (\hat{i}+2\hat{j}+3\hat{k})

\overrightarrow{PQ} = 3\hat{i}+3\hat{j}+3\hat{k}

Now, magnitude of \overrightarrow{PQ}

|\overrightarrow{PQ}| = \sqrt{3^2+3^2+3^2}

|\overrightarrow{PQ}| = \sqrt{27}        = 3√3

So, the unit vector in the direction of \overrightarrow{PQ}        is given by,

\hat{PQ} = \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}

\hat{PQ} = \frac{3\hat{i}+3\hat{j}+3\hat{k}}{3\sqrt{3}}

\hat{PQ} = \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}

Problem 5: Show that the vector 2\hat{i}-3\hat{j}+4\hat{k}    and -4\hat{i}+6\hat{j}-8\hat{k}    are collinear.


Let, \overrightarrow{a} = 2\hat{i}-3\hat{j}+4\hat{k}    and \overrightarrow{b} = -4\hat{i}+6\hat{j}-8\hat{k}

As, we can see

\overrightarrow{b} = -4\hat{i}+6\hat{j}-8\hat{k} = -2 (2\hat{i}-3\hat{j}+4\hat{k})

This implies,

\overrightarrow{b} = -2 \overrightarrow{a}

Hence, \overrightarrow{a}    and \overrightarrow{b}    are collinear.

Problem 6: If A(2,0,0), B(0,1,0), C (0,0,2) have position vectors, show that △ABC is isosceles triangle.


Position vector of A(2,0,0) = 2\hat{i}+0\hat{j}+0\hat{k}

Position vector of B(0,1,0) = 0\hat{i}+\hat{j}+0\hat{k}

Position vector of C(0,0,2) = 0\hat{i}+0\hat{j}+2\hat{k}

\overrightarrow{AB}   \overrightarrow{B}-\overrightarrow{A}

\overrightarrow{AB} = (0\hat{i}+\hat{j}+0\hat{k}) - (2\hat{i}+0\hat{j}+0\hat{k})

\overrightarrow{AB} = -2\hat{i}+\hat{j}+0\hat{k}

Now, magnitude of \overrightarrow{AB}

|\overrightarrow{AB}| = \sqrt{(-2)^2+1^2+0^2}

|\overrightarrow{AB}| = \sqrt{5}

\overrightarrow{BC}    = \overrightarrow{C}-\overrightarrow{B}

\overrightarrow{BC} = (0\hat{i}+0\hat{j}+2\hat{k}) - (0\hat{i}+\hat{j}+0\hat{k})

\overrightarrow{BC} = 0\hat{i}-\hat{j}+2\hat{k}

Now, magnitude of \overrightarrow{BC}

|\overrightarrow{BC}| = \sqrt{0^2+(-1)^2+2^2}

|\overrightarrow{BC}| = \sqrt{5}

Clearly, AB = BC.

Hence, △ABC is isosceles triangle.

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