Math, asked by Anonymous, 1 month ago

If the distances of P(x, y) from A(5, 1) and B(- 1, 5) are equal, then prove that 3x = 2y

No spam!! ​

Answers

Answered by mathdude500
9

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

Given that,

The distances of P(x, y) from A(5, 1) and B(- 1, 5) are equal.

\rm \implies\:PA \: = \: PB

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

We know,

Distance Formula :- The distance between two points P(x₁, y₁) and Q(x₂, y₂) is evaluated as

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

\rm :\longmapsto\: {(x - 5)}^{2} +  {(y - 1)}^{2} =  {(x + 1)}^{2} +  {(y - 5)}^{2}

 \rm \:\cancel{x}^{2} + \cancel{25 }- 10x +\cancel{y}^{2} + \cancel1 - 2y = \cancel{x}^{2} + \cancel1 + 2x +\cancel{y}^{2} + \cancel{25} - 10y

\rm :\longmapsto\: - 10x - 2y = 2x - 10y

\rm :\longmapsto\: 10y - 2y = 2x + 10x

\rm :\longmapsto\: 8y = 12x

\bf\implies \:2y = 3x

Hence, Proved

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Learn More :-

1. Section formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

\boxed{\tt{ \sf\implies R = \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)}}

2. Mid-point formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be

\boxed{\tt{ \sf\implies R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}}

3. Centroid of a triangle

Centroid of a triangle is the point where the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

\boxed{\tt{ \sf\implies R = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)}}

Answered by Unni007
8

Given,

  • P = (x, y)

So A(5, 1) and B(-1, 5) can be written as (x₁, y₁) and (x₂, y₂).

It is given that PA = PB.

By Distance formula,

\boxed{\bold{\sf{(x-x_1)^2 + (y-y_1)^2 = (x-x_2)^2 + (y-y_2)^2}}}

So applying values to the equation,

\sf{\implies (x-5)^2 + (y-1)^2 = (x+1)^2 + (y-5)^2}

\sf{\implies (x^2-10x+25)+(y^2-2y+1) = (x^2+2x+1)+(y^2-10y+25)}

\sf{\implies x^2-10x+25+y^2-2y+1 = x^2+2x+1+y^2-10y+25}

\sf{\implies x^2-10x+25-x^2-2x-1=y^2-10y+25-y^2+2y-1}

\sf{\implies 10x+2x=10y-2y}

\sf{\implies 12x=8y \ [Divide \ by \ 4]}

\sf{\implies 3x=2y }

Hence Proved.

\boxed{\bold{\sf{\therefore 3x=2y}}}

Similar questions