Question

Find a point on the y axis which is equidistant from the point A(6,5) and B(-4,3)

Answer 1

Given: Points A(6, 5) and B(-4, 3).

To find: A point on the y - axis that is equidistant from the two points.

Answer:

Since it's given that the point lies on the y - axis, the point is (0, y).

Let's label it as P(0, y).

Distance formula:

$\tt \sqrt{\bigg(x_2\ -\ x_1\bigg)\ +\ \bigg(y_2\ -\ y_1\bigg)}$

Now, as per the question, AP = PB.

For AP [A(6, 5); P(0, y)], we have:

$\tt x_1\ =\ 6\\\\x_2\ =\ 0\\\\y_1\ =\ 5\\\\y_2\ =\ y$

For BP [B(-4, 3); P(0, y)], we have:

$\tt x_1\ =\ -4\\\\x_2\ =\ 0\\\\y_1\ =\ 3\\\\y_2\ =\ y$

Using them in the distance formula for AP = BP,

$\tt \sqrt{\bigg(0\ -\ 6\bigg)^2\ +\ \bigg(y\ -\ 5\bigg)^2}\ =\ \sqrt{\bigg(0\ +\ 4\bigg)^2\ +\ \bigg(y\ -\ 3\bigg)^2}\\\\\\\\\sqrt{\bigg(-6\bigg)^2\ +\ y^2\ -\ 10y\ +\ 25}\ =\ \sqrt{\bigg(4\bigg)^2\ +\ y^2\ -\ 6y\ +\ 9}\\\\\\\\\sqrt{\bigg[36\ +\ y^2\ -\ 10y\ +\ 25\bigg]}\ =\ \sqrt{\bigg[16\ +\ y^2\ -\ 6y\ +\ 9\bigg]}$

Squaring both sides,

$\tt 36\ +\ y^2\ -\ 10y\ +\ 25\ =\ 16\ +\ y^2\ -\ 6y\ +\ 9\\\\\\61\ -\ 10y\ =\ 25\ -\ 6y\\\\\\61\ -\ 25\ =\ -6y\ +\ 10y\\\\\\36\ =\ 4y\\\\\\y\ =\ 9$

Therefore, the point on the y - axis that is equidistant from the points A(6, 5) and B(-4, 3) is P(0, 9).

Answer 2

Question: To find a point on the y-axis that is equidistant from the points A(6,5) and B(-4,3)

Solution:

ATQ, we have to find a point on the y-axis that is equidistant to two points. Let this point be P.

If the point 'P' lies on the y-axis, its 'x' coordinate will be 0. Therefore, the coordinates for this point will be - P(0, y)

Since the 'x' coordinate is 0, We'll find the 'y' co-ordiante by equating the equal sides AP and BP using the distance formula.

$\underline{\boxed{\sf \ Distance \ Formula = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 \ }}}$

A(6, 5)

x₁ → 6

y₁ → 5

B(-4, 3)

x₂ → -4

y₂ → 3

P(0, y)

x₃ → 0

y₃ → y

$\sf ATQ,\\ \\ \\\sf \Longrightarrow AP=BP\\ \\ \\\sf \Longrightarrow \sqrt{(x_1-x_3)^2+(y_1-y_3)^2}=\sqrt{(x_2-x_3)^2+(y_2-y_3)^2}\\ \\ \\\sf \Longrightarrow \sqrt{(6-0)^2+(5-y)^2}=\sqrt{(-4+0)^2+(3-y)^2}\\ \\ \\\underline{\boxed{\sf Squaring \ on \ both \ sides,}}\\ \\ \\\sf \Longrightarrow \left(\sqrt{(6)^2+(5-y)^2}\right)^2=\left(\sqrt{(-4)^2+(3-y)^2}\right)^2\\ \\ \\\sf \Longrightarrow (6)^2+(5-y)^2=(-4)^2+(3-y)^2\\ \\ \\\underline{\boxed{\sf Using \ (a-b)^2 = a^2 - 2ab + b^2}}$

$\sf {\Longrightarrow 36+25-2(5)(y)+y^2=16+9-2(3)(y)+y^2}\\ \\ \\\sf \Longrightarrow 61-10y+\bcancel{y^2}=25-6y+\bcancel{y^2}\\ \\ \\\sf \Longrightarrow 61-10y=25-6y\\ \\ \\\sf \Longrightarrow 36=-6y+10y\\ \\ \\\sf \Longrightarrow 36=4y\\ \\ \\\sf \Longrightarrow 9=y\\ \\ \\\sf \Longrightarrow \underline{\underline{\bold{y=9}}}$

Therefore the y coordinate is '9', making the point that lies on the y-axis equidistant from the points A(6,5) and B(-4,3) as P(0,9).

$\boxed{\boxed{\sf \underline{ANSWER: P(0, 9)}}}$