Question

Find the area of triangle whose vertices are (4,7) (1,13) (5,1)
10sq unit
11 sq unit
14 sq unit
16 sq unit​

Answer 1

Coordinate Geometry

The area of a ∆ABC with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is given by,

$\boxed{\tt{\;Area_{(triangle)} = \dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|\;}}$

We are given a triangle whose vertices are (4, 7), (1, 13) and (5, 1).

With this information, we are asked to find out the area of the triangle.

Let $A(4, 7)$, $B(1, 13)$ and $C(5, 1)$ be the vertices of the given $\triangle ABC$.

The coordinate of the vertices of the given triangle are $A(x_1 = 4, y_1 = 7)$, $B(x_2 = 1, y_2 = 13)$ and $C(x_3 = 5, y_3 = 1)$.

We know that,

$\;\;Area_{(triangle)} = \dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|$

By plugging the known values in the formula, the following results are achieved:

$\implies Area_{(triangle)} = \dfrac{1}{2}\bigg|4(13-1) + 1(1-7) + 5(7-13)\bigg| \\ \\ \implies Area_{(triangle)} = \dfrac{1}{2}\bigg|4(12) + 1(-6) + 5(-6)\bigg| \\ \\ \implies Area_{(triangle)} = \dfrac{1}{2}\bigg|48 + (-6) + (-30)\bigg| \\ \\ \implies Area_{(triangle)} = \dfrac{1}{2}\bigg|48-6-30\bigg| \\ \\ \implies Area_{(triangle)} = \dfrac{1}{2} \times 12 \\ \\ \implies\boxed{Area_{(triangle)} =6}$

Hence, the area of the given triangle is 6 sq units.

$\rule{300}{2}$

Explore More

1. Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane, then the distance between A and B is given by,

$\boxed{\tt{\;AB = \sqrt{ {(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2}\;}}}$

2. The distance of the point P(x, y) from the origin O(0, 0) is given by,

$\boxed{\tt{\;OP = \sqrt{x^2 + y^2\;}}}$

3. The area of a ∆ABC with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is given by,

$\boxed{\tt{\;Area = \dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|\;}}$

4. Three points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are collinear only when,

$\boxed{\tt{\;x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) = 0\;}}$

5. Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane and C(x, y) be the point which divides AB internally in the ratio m₁ : m₂. Then, the coordinates of C will be:

$\boxed{\tt{\:C = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)\;}}$

6. Let us consider a line segment joining the points A and B and Let C (x, y) be the midpoint of AB, then coordinates of C(x, y) is given by,

$\boxed{\tt{\;(x, y) = \bigg(\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2}\bigg)\;}}$

7. Any point on the x-axis is of the form (x, 0).

8. Any point on the y-axis is of the form (0, y).

Answer 2

Information provided with us:

• ➡ The vertices of any Triangle are as given below

$\rm \implies \:x_{1}= 4$

$\rm \implies \:x_{2} = 1$

$\rm \implies \:x_{3} = 5$

and

$\rm \implies \:y_{1} = 7$

$\rm \implies \:y_{2}= 13$

$\rm \implies \:y_{3}= 1$

━━━━━━━━━━━━━━━━━━━━━━━━━━━

What we have to find:

• The required area of that triangle

━━━━━━━━━━━━━━━━━━━━━━━━━━━

We know that :

$\longrightarrow\sf\qquad \: \dfrac{1}{2} \times \Bigg\{{x_1(y_2 - y_3) + x_2(y_3 -y_1) + x_3(y_1 - y_2)}\Bigg\}$

$\longrightarrow\qquad\underline{\boxed{\pmb{\frak{ \therefore \: Area \: of \: Triangle}}}}$

$\longrightarrow\sf\qquad \: \dfrac{1}{2} \times \Bigg\{{4(13 - 1) + 1(1 -7) + 5(7- 13)}\Bigg\}$

$\longrightarrow\sf\qquad \: \dfrac{1}{2} \times \Bigg\{{4(12) + 1( - 6) + 5( - 6)}\Bigg\}$

$\longrightarrow\sf\qquad \: \dfrac{1}{2} \times \Bigg\{{(48) +( - 6) + ( - 30)}\Bigg\}$

$\longrightarrow\sf\qquad \: \dfrac{1}{2} \times \Bigg\{{48 - 6- 30}\Bigg\}$

$\longrightarrow\sf\qquad \: \dfrac{1}{2} \times \Bigg\{{42- 30}\Bigg\}$

$\longrightarrow\sf\qquad \: \dfrac{1}{2} \times 12$

$\longrightarrow\sf\qquad \: \dfrac{1 \times 12}{2}$

$\longrightarrow\sf\qquad \: \dfrac{ 12}{2}$

$\longrightarrow\bf\qquad \: 6 \: sq \: units$

Therefore,

$\therefore$ Hence, the required answer is 6 sq units .