Question

Point P(x, y) is equidistant from the points A(-5, 3) and B(7, 2). What is the Telation between x and y? (a)12x+12y-17 = 0 -- (b)24x + 2y-19 = 0 (c)24x-2y - 19 = 0 (d)12x-12y + 17 = 0

pls answer only if you know​

Answer 1

Answer:

Hello friend your answer is option number (c) 24x–2y–19=0

Please mark me as a brainlist

Answer 2

$\textsf{\large{\underline{Solution}:}}$

Given: Point P(x, y) is equidistant from point A(-5, 3) and B(7, 2)

Therefore:

$\rm \longrightarrow PA=PB$

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

Using distance formula:

$\rm \longrightarrow PA = \sqrt{ {(x + 5)}^{2} + {(y - 3)}^{2} }$

$\rm \longrightarrow PA^{2} = {(x + 5)}^{2} + {(y - 3)}^{2}$

$\rm \longrightarrow PA^{2} = {x}^{2} + {y}^{2} + 10x - 6y+ 34$

Also:

$\rm \longrightarrow PB = \sqrt{ {(x - 7)}^{2} + {(y - 2)}^{2} }$

$\rm \longrightarrow PB^{2} = {(x - 7)}^{2} + {(y - 2)}^{2}$

$\rm \longrightarrow PB^{2} = {x}^{2} - 14x + 49 + {y}^{2} - 4y + 4$

$\rm \longrightarrow PB^{2} = {x}^{2} + {y}^{2} - 14x - 4y+ 53$

Now:

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

$\rm \longrightarrow {x}^{2} + {y}^{2} + 10x - 6y + 34 = {x}^{2} + {y}^{2} - 14x - 4y + 53$

$\rm \longrightarrow 10x - 6y + 34 = - 14x - 4y + 53$

$\rm \longrightarrow 24x -2y - 19 = 0$

Which is our required answer.

$\textsf{\large{\underline{Learn More}:}}$

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:

$\rm\longrightarrow R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)$

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:

$\rm\longrightarrow R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)$

3. Centroid of a triangle formula.

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:

$\rm\longrightarrow R = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)$