Question

2.The points on x-axis which are at a distance of √ 13 units from (- 2, 3) is____.​

Answer 1

Answer:0

Step-by-step explanation:

root [-2-0] ^2+ [3-y]^2= root 13

solving 4 +[3-y]^2 = 13

 3-y=3

y=0  [0,0] as x=o y=o

Answer 2

The points on x-axis which are at a distance of √13 units from (- 2, 3) are (0,0) & ( - 4,0)

Given :

The points on x-axis which are at a distance of √13 units from (- 2, 3)

To find :

The points

Solution :

Step 1 of 2 :

Form the equation to find the coordinates of the point

Let the point on x axis is (h,0)

The distance between (h, 0) and (- 2, 3) is √13 units

So by the given condition

$\displaystyle \sf \sqrt{{(h + 2) }^{2} + {(0 - 3)}^{2} } = \sqrt{13}$

Step 2 of 2 :

Find the coordinates of the point

$\displaystyle \sf \sqrt{{(h + 2) }^{2} + {(0 - 3)}^{2} } = \sqrt{13}$

$\displaystyle \sf{ \implies }{(h + 2) }^{2} + {(0 - 3)}^{2} = 13$

$\displaystyle \sf{ \implies }{(h + 2) }^{2} + {( - 3)}^{2} = 13$

$\displaystyle \sf{ \implies }{(h + 2) }^{2} + 9 = 13$

$\displaystyle \sf{ \implies }{(h + 2) }^{2} = 4$

$\displaystyle \sf{ \implies }{(h + 2) }^{2} = {2}^{2}$

$\displaystyle \sf{ \implies }h + 2 = \pm \: 2$

Now ,

h + 2 = 2 gives h = 0

h + 2 = - 2 gives h = - 4

∴ The coordinates of the points are (0,0) & ( - 4,0)

Hence the points on x-axis which are at a distance of √13 units from (- 2, 3) are (0,0) & ( - 4,0)

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