Question

n
find the point on y: axis which is
equidistance from (-5,-2) and (3,4)​

Answer 1

Question

Find the point on y-axis which is equidistant from (-5,-2) & (3,4)

ANSWER

Given : -

A point on y-axis which is equidistant from (-5,-2) & (3,4)

Required to find : -

• co-ordinate of the point ?

Formula used : -

>>> Distance formula <<<

The distance between 2 points is given as √[(x2-x1)²+(y2-y1)²] units

Solution : -

A point on y-axis is equidistant from (-5,-2) & (3,4)

We need to find the co-ordinates of the point.

Since, it is mentioned that the point lies on y-axis we can clearly say that the x co-ordinate of the the above point is zero.

If, it is not zero (x co-ordinate ≠0) then the point won't lie on the y-axis.

So, we need to find the value of y co-ordinate as the value of x co-ordinate is zero (0).

However,

Let the point which is equidistant from the 2 points be A(0,y)

On the other hand,let the other 2 points be B(-5,-2) & C(3,4)

This implies;

Since, the point A(0,y) is equidistant from the points B & C.

So,

• AB = AC

Using the distance formula;

Distance between 2 sides is √[(x2-x1)²+(y2-y1)²] units

A(x1,y1) = (0,x)

B(x2,y2) = (-5,-2)

[ Here, I am taking A(x1,y1) because I want to show how I am comparing and substituting the value in the formula]

AB = √[(-5-0)²+(-2-x)²]

AB = √[(-5)²+(-2)²+(x)²-2(-2)(x)]

AB = √[25+4+x²+4x]

AB = √[29+x²+4x] units .......[1]

Consider this as Equation-1

Similarly,

A(x1,y1) = (0,x)

C(x2,y2) = (3,4)

AC = √[(3-0)²+(4-x)²]

AC = √[(3)²+(4)²+(x)²-2(4)(x)]

AC = √[9+16+x²-8x]

AC = √[25+x²-8x] units ......[2]

Consider the as Equation-2

According to the problem;

AB = AC

√[29+x²+4x] = √[25+x²-8x]

squaring on both sides

(√[29+x²+4x]) = (√[25+x²-8x])

29+x²+4x = 25+x²-8x

x² get's cancelled on both sides

29+4x = 25-8x

4x+8x = 25-29

12x = -4

x = (-4)/(12)

x = (-1)/(3)

Therefore,

The co-ordinates of the point A is (0,[-1]/[3])