Question

1. Find the area of the triangle whose vertices are (5.2) (3,-5) and (-5, -1) ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that vertices of triangle are (5, 2), (3, - 5) and (- 5, - 1) respectively.

We know,

Area of a triangle having vertices is evaluated as

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle, then the area of triangle is given by

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

So, here

$\rm \: x_1 = 5$

$\rm \: x_2 = 3$

$\rm \: x_3 = - 5$

$\rm \: y_1 = 2$

$\rm \: y_2 = - 5$

$\rm \: y_3 = - 1$

So, on substituting the values, we get

$\begin{gathered}{\tt{ Area =\dfrac{1}{2}\bigg|5( - 5 + 1) + 3( - 1 - 2) - 5(2 + 5)\bigg|}} \\ \end{gathered}$

$\begin{gathered}{\tt{ Area =\dfrac{1}{2}\bigg|5( - 4) + 3( - 3) - 5(7)\bigg|}} \\ \end{gathered}$

$\begin{gathered}{\tt{ Area =\dfrac{1}{2}\bigg| - 20 - 9 - 35\bigg|}} \\ \end{gathered}$

$\begin{gathered}{\tt{ Area =\dfrac{1}{2}\bigg| - 64\bigg|}} \\ \end{gathered}$

$\rm\implies \:\begin{gathered}{\tt{ Area \: = \: 32 \: square \: units}} \\ \end{gathered}$

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ADDITIONAL INFORMATION

1. Distance Formula

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

$\begin{gathered}\boxed{\tt{ AB \: = \sqrt{ {(x_{1} - x_{2}) }^{2} + {(y_{2} - y_{1})}^{2} }}} \\ \end{gathered}$

2. Section formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane and C(x, y) be the point which divides AB internally in the ratio m₁ : m₂, then the coordinates of C is given by

$\begin{gathered} \boxed{\tt{ (x, y) = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)}} \\ \end{gathered}$

3. Mid-point formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane and C(x, y) be the mid-point of AB, then the coordinates of C is given by

$\begin{gathered}\boxed{\tt{ (x,y) = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}} \\ \end{gathered}$

4. Centroid of a triangle

Centroid of a triangle is defined as the point at which the medians of the triangle meet and is represented by the symbol G.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle and G(x, y) be the centroid of the triangle, then the coordinates of G is given by

$\begin{gathered}\boxed{\tt{ (x, y) = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)}} \\ \end{gathered}$

5. Condition for 3 points to be Collinear

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the coordinates in cartesian plane, then points A, B and C are collinear, then

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

Answer 2

${\pmb{\underline{ \sf{\red{Solution:-}}}}}$

• Answer in the above attachment.

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@Shivam

#BeBrainly