Question

find a point which is equidistant from (-3,6) and (2,4). given that the abscissa and ordinate are equal

Answer 1

Step-by-step explanation:

Given :-

(-3,6) and (2,4) and given that the abscissa and ordinate are equal.

To find :-

Find a point which is equidistant from (-3,6) and (2,4).

Solution :-

Let the required point be (x,x)

Since ,given that the abscissa and ordinate are equal.

Given points are (-3,6) and (2,4)

Let A = (-3,6) and B = (2,4)

A__________P__________B

According to the given problem

the point P is the equidistant from A and B

=> AP = PB

We know that

The distance between two points (x1, y1) and (x2,y2) is √[(x2-x1)²+(y2-y1)²] units

The distance between A and P:-

Let (x1,y1) = A(-3,6) => x1 = -3 and y1 = 6

Let (x2, y2) = P(x,x) => x2 = x and y2 = x

Now,

AP = √[(x-(-3))²+(x-6)²]

=> AP = √[(x+3)²+(x-6)²]

=> AP = √(x²+6x+9+x²-12x+36)

=> AP = √(2x²-6x+45) ------------------(1)

The distance between P and B:-

Let (x1,y1) = P(x,x) => x1 = x and y1 = x

Let (x2, y2) = B(2,4) => x2 = 2 and y2 = 4

Now,

PB= √[(2-x)²+(4-x)²]

=> PB = √(x²-4x+4+x²-8x+16)

=> PB = √(2x²-12x+20) ------------------(2)

Now,

(1) = (2)

=> √(2x²-6x+45) = √(2x²-12x+20)

On squaring both sides then

=> [√(2x²-6x+45)]² = [√(2x²-12x+20)]²

=> 2x²-6x+45 = 2x²-12x+20

=> 2x²-6x-2x²+12x = 20-45

=> 6x = -25

=> x = -25/6

The point = (-25/6,-25/6)

Answer:-

The required point for the given problem is (-25/6,-25/6)

Used formulae:-

The distance between two points (x1, y1) and (x2,y2) is √[(x2-x1)²+(y2-y1)²] units