Question

show that the points A(3,-5) B(4,3) and cc11,-4) are vertices of an isosceles triangle​

Answer 1

Question :-

Show that the points A (3,-5), B (4,3) and C 11,-4) are vertices of an isosceles triangle.

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

Given that,

Coordinates of A is (3,-5)

Coordinates of B is (4,3)

and

Coordinates of C is (11,-4)

Now, we have to show that these are the vertices of isosceles triangle, i. e. we have to show that length of any two sides of a triangle are equal.

We know,

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

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

So, using this, we have

Distance between A (3, - 5) and B (4, 3)

$\rm \: AB = \sqrt{ {(4 - 3)}^{2} + {(3 + 5)}^{2} }$

$\rm \: = \: \sqrt{ {(1)}^{2} + {(8)}^{2} }$

$\rm \: = \: \sqrt{ 1 + 64 }$

$\rm \: = \: \sqrt{ 65 }$

$\rm\implies \:AB \: = \: \sqrt{65}$

Now,

Consider, Distance between A (3, - 5) and C (11, - 4)

$\rm \: AC = \sqrt{ {(11 - 3)}^{2} + {( - 4 + 5)}^{2} }$

$\rm \: = \: \sqrt{ {8}^{2} + {1}^{2} }$

$\rm \: = \: \sqrt{1 + 64}$

$\rm \: = \: \sqrt{65}$

$\rm\implies \:AC = \sqrt{65}$

Hence, from this, we concluded that

$\rm\implies \:AB = AC = \sqrt{65}$

$\rm\implies \: \triangle \: ABC \: is \: isosceles$

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Additional Information :-

1. 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

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

2. 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

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

3. 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

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

4. Area of a triangle

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

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

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

Step-by-step explanation:

This is a correct answer.

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