Question

if p (1,2) Q(1,0 R(0,1) are the mid points of the sides Ab bc and ca respectively of a triangle Abc find the cordinates of the vertices A,B,C and hence find the are of a triangle ABC

Answer 1

Answer:

Vertices of triangle are A(0,3), B(2,1) and C(0,-1) and area is 4 sq units

Step-by-step explanation:

Given P(1,2), Q(1,0) and R(0,1) are the mid points of the sides AB, BC and CA respectively of a triangle ABC we have to find the coordinates of the vertices A,B,C

Let the coordinates of A, B and C are $(a_{1},a_{2}), (b_{1},b_{2}) ,(c_{1},c_{2})$ respectively. Now,

Coordinates of Mid point of AB is $[\frac{(a_{1}+b_{1})}{2}, \frac{(a_{2}+b_{2})}{2}]=P(1,2)$

⇒  $\frac{(a_{1}+b_{1})}{2}=1$,   $\frac{(a_{2}+b_{2})}{2}=2$

⇒  $a_{1}+b_{1}=2$, $a_{2}+b_{2}=4$   →  (1)

Similarly,

Coordinates of Mid point of BC is $[\frac{(b_{1}+c_{1})}{2}, \frac{(b_{2}+c_{2})}{2}]=P(1,0)$

⇒  $\frac{(b_{1}+c_{1})}{2}=1$,   $\frac{(b_{2}+c_{2})}{2}=0$

⇒  $b_{1}+c_{1}=2$, $b_{2}+c_{2}=0$   →  (2)

Coordinates of Mid point of AC is $[\frac{(a_{1}+c_{1})}{2}, \frac{(a_{2}+c_{2})}{2}]=P(0,1)$

⇒  $\frac{(a_{1}+c_{1})}{2}=0$,   $\frac{(a_{2}+c_{2})}{2}=1$

⇒  $a_{1}+c_{1}=0$, $a_{2}+c_{2}=2$   →  (3)

Now, solving eq (1),(2) and (3) we get

$a_{1}=0, a_{2}=3, b_{1}=2, b_{2}=1, c_{1}=0, c_{2}=-1$

Hence, vertices of triangle are A(0,3), B(2,1) and C(0,-1)

Now, Area of triangle=

$\frac{1}{2}\left |a_{1}(b_{2}-c_{2})+b_{1}(c_{2}-a_{2})+c_{1}(a_{2}-b_{1})\right |$

                                   = $\frac{1}{2}\left |2(-1-3)\right |=4 squnits$

Answer 2

Answer:

Vertices of triangle are A(0,3), B(2,1) and C(0,-1) and area is 4 sq units

Step-by-step explanation:

Given P(1,2), Q(1,0) and R(0,1) are the mid points of the sides AB, BC and CA respectively of a triangle ABC we have to find the coordinates of the vertices A,B,C

Let the coordinates of A, B and C are (a_{1},a_{2}), (b_{1},b_{2}) ,(c_{1},c_{2})(a

1

,a

2

),(b

1

,b

2

),(c

1

,c

2

) respectively. Now,

Coordinates of Mid point of AB is [\frac{(a_{1}+b_{1})}{2}, \frac{(a_{2}+b_{2})}{2}]=P(1,2)[

2

(a

1

+b

1

)

,

2

(a

2

+b

2

)

]=P(1,2)

⇒ \frac{(a_{1}+b_{1})}{2}=1

2

(a

1

+b

1

)

=1 , \frac{(a_{2}+b_{2})}{2}=2

2

(a

2

+b

2

)

=2

⇒ a_{1}+b_{1}=2a

1

+b

1

=2 , a_{2}+b_{2}=4a

2

+b

2

=4 → (1)

Similarly,

Coordinates of Mid point of BC is [\frac{(b_{1}+c_{1})}{2}, \frac{(b_{2}+c_{2})}{2}]=P(1,0)[

2

(b

1

+c

1

)

,

2

(b

2

+c

2

)

]=P(1,0)

⇒ \frac{(b_{1}+c_{1})}{2}=1

2

(b

1

+c

1

)

=1 , \frac{(b_{2}+c_{2})}{2}=0

2

(b

2

+c

2

)

=0

⇒ b_{1}+c_{1}=2b

1

+c

1

=2 , b_{2}+c_{2}=0b

2

+c

2

=0 → (2)

Coordinates of Mid point of AC is [\frac{(a_{1}+c_{1})}{2}, \frac{(a_{2}+c_{2})}{2}]=P(0,1)[

2

(a

1

+c

1

)

,

2

(a

2

+c

2

)

]=P(0,1)

⇒ \frac{(a_{1}+c_{1})}{2}=0

2

(a

1

+c

1

)

=0 , \frac{(a_{2}+c_{2})}{2}=1

2

(a

2

+c

2

)

=1

⇒ a_{1}+c_{1}=0a

1

+c

1

=0 , a_{2}+c_{2}=2a

2

+c

2

=2 → (3)

Now, solving eq (1),(2) and (3) we get

a_{1}=0, a_{2}=3, b_{1}=2, b_{2}=1, c_{1}=0, c_{2}=-1a

1

=0,a

2

=3,b

1

=2,b

2

=1,c

1

=0,c

2

=−1

Hence, vertices of triangle are A(0,3), B(2,1) and C(0,-1)

Now, Area of triangle=

\frac{1}{2}\left |a_{1}(b_{2}-c_{2})+b_{1}(c_{2}-a_{2})+c_{1}(a_{2}-b_{1})\right |

2

1

∣a

1

(b

2

−c

2

)+b

1

(c

2

−a

2

)+c

1

(a

2

−b

1

)∣

= \frac{1}{2}\left |2(-1-3)\right |=4 squnits

2

1

∣2(−1−3)∣=4squnits