Question

If the point [x, y] is equidistant from the points [a+b,b-a] and [a-b, a+b], then​

Answer 1

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

Given that,

The point (x, y) is equidistant from the points (a+b,b-a) and (a-b, a+b)

Let assume that

The point P (x, y) is equidistant from the points A (a+b,b-a) and B (a-b, a+b).

So, we have

$\rm \: PA=PB$

$\rm \: PA^{2} =PB^{2}$

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

$\rm \: (a + b - x)^{2} + {(b - a - y)}^{2} = {(a - b - x)}^{2} + {(a + b - y)}^{2}$

Now, open the squares, we get

$\rm \: \cancel{(a + b)}^{2} + \cancel {x}^{2} - 2(a + b)x + \cancel{(b - a)}^{2} + \cancel{y}^{2} - 2(b - a)y = \\ \rm \: \cancel{(a - b)}^{2} + \cancel{x}^{2} - 2(a - b)x +\cancel{(a + b)}^{2} + \cancel {y}^{2} - 2(a + b)y \: \: \:$

$\rm \: - 2(a + b)x - 2(b - a)y = - 2(a - b)x - 2(a + b)y \: \: \:$

$\rm \: - 2[(a + b)x + (b - a)y] = - 2[(a - b)x + (a + b)y] \: \: \:$

$\rm \:(a + b)x + (b - a)y =(a - b)x + (a + b)y \: \: \:$

$\rm \:(a + b)x - (a - b)x = (a + b)y - (b - a)y \: \: \:$

$\rm \:[\cancel{a} + b- \cancel{a} + b]x = [a + \cancel{b} - \cancel{b} + a]y \: \: \:$

$\rm \: 2bx \: = \: 2ay$

$\rm\implies \:\rm \: bx \: = \: ay$

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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

Refer the given attachment.