Answer :


Let, the equidistant point is C(0, y)     (on y axis, x coordinate is 0)


If it is equidistant from (5, -2) and (-3, 2) then AC = BC

Using the distance formula  

We will get





&




As, AC = BC


Squaring both sides we get,
⇒ y2 + 4y + 29 = y2 - 4y + 13

⇒  4y + 4y + 29 -13  = 0

⇒ 8y + 16 = 0

⇒ 8y = -16

y = -2

So, the equidistant point is (0, -2)

OR

Consider the points A(2, 1) and B(5, -8) .

If P is one of the trisected points then the ratio of AB at P = 1: 2

Let, coordinates of P is (x, y).
By section formula which states that if a point P(x,y) divides the line with endpoints A(x1,y1) and B(x2,y2) in the ratio m:n, the coordinates of x and y are:

The coordinates of P are:



x = 3 and y = -2
So, the coordinates of P is (3, -2).

P satisfies the equation of the given line 2x – y + k = 0.

⇒ 2(3) – (-2) + k = 0

⇒ 6 + 2 + k = 0
⇒ 8 + k = 0

⇒ k = - 8

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