Question

Show that point (5,3) is equidistant
from the point (1,1) and (3,-1)​

Answer 1

Question :–

▪︎Show that point (5,3) is equidistant from the points (1,1) and (3,-1).

ANSWER :–

▪︎ Let the point are A(1,1) , B(3,-1) and P(5,3).

▪︎ Now we have to prove – AP = BP

☞ If two points are (a,b) and (c,d) , then distance between them is –

$\\ \implies \: { \boxed{ \bold{Distance = \sqrt{ {(c - a)}^{2} + {(d - b)}^{2} } }}}$

▪︎Now , let's find AP –

$\\ \implies \: { \bold{AP = \sqrt{ {(5 - 1)}^{2} + {(3 - 1)}^{2} } }}$

$\\ \implies \: { \bold{AP = \sqrt{ {(4)}^{2} + {(2)}^{2} } }}$

$\\ \implies \: { \bold{AP = \sqrt{16+ 4} }}$

$\\ \implies \: { \bold{AP = \sqrt{20} \: \: \: \: \: \: - - - -eq.(1) }}$

▪︎ Now , let's find BP –

$\\ \implies \: { \bold{BP = \sqrt{ {(5 - 3)}^{2} + {(3 - ( - 1))}^{2} } }}$

$\\ \implies \: { \bold{BP = \sqrt{ {(2)}^{2} + {(4)}^{2} } }}$

$\\ \implies \: { \bold{BP = \sqrt{ 4 + 16 } }}$

$\\ \implies \: { \bold{BP = \sqrt{20 } \: \: \: \: \: \: \: \: - - - -eq.(2) }}$

By eq.(1) and eq.(2) –

$\\ \implies \: { \boxed{\bold{AP =BP }}}$

Hence proved , Point (5,3) is equidistant

from the point (1,1) and (3,-1).

Answer 2

Given: Points (5, 3), (1, 1) and (3, -1).

To show: That the point (5, 3) is equidistant from points (1, 1) and (3, -1).

Answer:

We'll be using the distance formula to check and see if the distances of both points are equal.

Distance formula:

$\tt Distance\ =\ \sqrt{\Bigg(x_2\ -\ x_1\Bigg)^2\ +\ \Bigg(y_2\ -\ y_1\Bigg)^2}$

Let's first find the distance between points (5, 3) and (1, 1).

From those points,

$\tt x_1\ =\ 5\\\\x_2\ =\ 1\\\\y_1\ =\ 3\\\\y_2\ =\ 1$

Using them in the formula,

$\tt Distance\ =\ \sqrt{\Bigg(1\ -\ 5\Bigg)^2\ +\ \Bigg(1\ -\ 3\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(-4\Bigg)^2\ +\ \Bigg(-2\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(16\ +\ 4\Bigg)}\\\\\\\\\bf Distance\ =\ 4.47\ units\ \tt [approx.]$

Now, let's find the distance between points (5, 3) and (3, -1).

From those points,

$\tt x_1\ =\ 5\\\\x_2\ =\ 3\\\\y_1\ =\ 3\\\\y_2\ =\ -1$

Using them in the formula,

$\tt Distance\ =\ \sqrt{\Bigg(3\ -\ 5\Bigg)^2\ +\ \Bigg(-1\ -\ 3\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(-2\Bigg)^2\ +\ \Bigg(-4\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(4\ +\ 16\Bigg)}\\\\\\\\\bf Distance\ =\ 4.47\ units\ \tt [approx.]$

• Distance between (5, 3) and (1, 1) = 4.47 units.
• Distance between (5, 3) and (3, -1) = 4.47 units.

Therefore, the point (5, 3) is equidistant from points (1, 1) and (3, -1).