Question

The area of the triangle formed by point A(0,1),B(0,5) and C(3,4) is ( in sq.units

Answer 1

ar=1/2×base×alt
=1/2×4×3
=6square units

Answer 2

Answer:

$4.105 unit^2$

Step-by-step explanation:

We are given a triangle which is formed by the points A(0,1),B(0,5) and C(3,4)

So, the sides of the triangle will be AB,BC,AC

First to find the length of AB

$(x_1,y_1)=(0,1)$

$(x_2,y_2)=(0,5)$

Now we will use distance formula :

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

Substitute the values

$d=\sqrt{(0-0)^2 +(5-1)^2}$

$d=\sqrt{(0)^2 +(4)^2}$

$d=\sqrt{16}$

$d=4$

So, length of AB is 4 units

Now to find the length of BC

$(x_1,y_1)=(0,5)$

$(x_2,y_2)=(3,4)$

Now we will use distance formula :

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

Substitute the values

$d=\sqrt{(3-0)^2 +(4-5)^2}$

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

$d=\sqrt{9+1}$

$d=\sqrt{9+1}$

$d=3.16$

So, length of BC is 3.16 units

Now to find the Length of AC

$(x_1,y_1)=(0,1)$

$(x_2,y_2)=(3,4)$

Now we will use distance formula :

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

Substitute the values

$d=\sqrt{(3-0)^2 +(4-1)^2}$

$d=\sqrt{(3)^2 +(3)^2}$

$d=\sqrt{9+9}$

$d=\sqrt{18}$

$d=4.24$

So, length of AC is 4.24 units .

So, AB = 4 units

BC= 3.16 units

AC= 4.24 units

Now to find area of triangle we will use heron's formula :

$Area= \sqrt{s(s-a)(s-b)/(s-c)}$

Where $s=\frac{a+b+c}{2}$

AB = a = 4 units

BC= b = 3.16 units

AC= c = 4.24 units

Substitute the values

$s=\frac{4+3.16+4.24}{2}$

$s=5.7$

$Area= \sqrt{5.7(5.7-4)(5.7-3.16)/(5.7-4.24)}$

$Area= 4.105 unit^2$

Hence The area of the triangle formed by point A(0,1),B(0,5) and C(3,4) is  $4.105 unit^2$