Question

A rod of length 5 units is sliding along co-ordinate axis as shown in the figure. Find the locus of the mid point of the rod.​

Answer 1

Answer:

25 is the answer

Step-by-step explanation:

25 ok this the the answer

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

A rod of length 5 units is sliding along co-ordinate axis as shown in the figure.

Let assume that, the coordinates of the end - points of the rod of length 5 units be A(a, 0) and B(0, b) respectively.

So, AB = 5 units.

Thus, OA = a units and OB = b units

Let assume that the midpoint of rod AB be C and Let assume that Coordinates of C be (x, y).

We know, Midpoint Formula

Coordinates of mid - point C (x, y) of the line segment joining the points A and B is

$\boxed{ \tt{ \: (x,y) \: = \: \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg) \: }}$

So, using Midpoint Formula, we have

↝ Coordinates of A = (a, 0)

↝ Coordinates of B = (0, b)

↝ Coordinates of C = (x, y)

Thus,

$\rm :\longmapsto\: \: (x,y) \: = \: \bigg(\dfrac{a + 0}{2}, \dfrac{0 + b}{2}\bigg) \:$

$\rm :\longmapsto\: \: (x,y) \: = \: \bigg(\dfrac{a}{2}, \dfrac{ b}{2}\bigg) \:$

$\bf\implies \:a = 2x \: \: \: and \: \: \: b = 2y$

Now, In right triangle OAB

Using, Pythagoras Theorem, we have

$\rm :\longmapsto\: {OA}^{2} + {OB}^{2} = {AB}^{2}$

On substituting the values of OA, OB, AB, we get

$\rm :\longmapsto\: {a}^{2} + {b}^{2} = {5}^{2}$

On substituting the values of a and b, we get

$\rm :\longmapsto\: {(2x)}^{2} + {(2y)}^{2} = 25$

$\rm :\longmapsto\: {4x}^{2} + {4y}^{2} = 25$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = \dfrac{25}{4}$

$\rm \implies\:\boxed{ \tt{ \: {x}^{2} + {y}^{2} = {\bigg[\dfrac{5}{2} \bigg]}^{2} \: }}$

which is the equation of circle having center (0, 0) and radius 5/2 units.

Hence,

• Locus of the midpoint of the rod is circle.