Question

23.Find the coordinates of point on X-axis which are at a distance of 5 units from the point (6,-3).​

Answer 1

Answer:

The point on X-axis is A(10, 0) or (2, 0).

Step-by-step explanation:

Let A be the point on X-axis.

We know that-

y is always 0 on X-axis.

So, coordinate of A = (x, 0)

Given that-

Distance between A(x, 0) and B(6, -3) = 5

x =?

Using formula-

Distance between M(a, b) and N(c, d) is-

$MN = \sqrt{(a-c)^{2}+(b-d)^{2} }$

So, distance between A(x, 0) and B(6, -3) is-

$AB = \sqrt{(x-6)^{2}+(0+3)^{2} }$

$5 = \sqrt{(x^{2}+36 -12x)+(3)^{2}}$

Squaring on both side

$(5)^{2} = x^{2} -12x + 36+9$

$25 = x^{2} -12x+45$

$x^{2} - 12x + 45 - 25 = 0$

$x^{2} -12x + 20 = 0$

$x^{2} -(10+2)x + 20 = 0$

$x^{2} - 10x -2x + 20 = 0$

$x(x-10) -2(x-10) = 0$

$(x-10)(x-2) = 0$

If (x - 10) = 0

Then x = 10

If (x - 2) = 0

Then x = 2

Hence, the point on X-axis is A(10, 0) or (2, 0).

Answer 2

Answer:

Step-by-step explanation:

On x axis.... point=x, 0

According to distance formula.,

D=root (x-a)^2+(y-b)^2

ie, root(6-x)^2+(-3-0)^2=5

We will get a quadratic eqn. After solving by factorisation method we will get two values for x

ie, x=10

or x=2

Coordinates may be (10, 0) or (2, 0)