Question

A is a point on the x-axis and B is (-7,9). Distance between the points A and B is 15 units. Find the coordinates of point A​

Answer 1

Given ,

A is a point on the x axis and B is (-7,9)

The distance between the point A and B is 15 units

Let ,

The coordinate of point A be (x , 0)

We know that , the distance formula is given by

$\boxed{ \tt{Distance = \sqrt{ {( x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} } }}$

Thus ,

$\tt \implies 15 = \sqrt{ { \{x - ( - 7) \}}^{2} + { \{0- 9\}}^{2} }$

Squaring on both sides , we get

$\tt \implies 225 = {(x)}^{2} + 49 + 14x + 81$

$\tt \implies 225 = {(x)}^{2} + 14x + 130$

$\tt \implies {(x)}^{2} + 14x - 95 = 0$

$\tt \implies {(x)}^{2} - 5x + 19x- 95 = 0$

$\tt \implies x(x - 5) + 19(x - 14) = 0$

$\tt \implies x + 19 = 0 \: \: or \: \: x - 5 = 0$

$\tt \implies x = - 19 \: \: or \: \: x = 5$

Hence , the coordinates of point A is (-19 , 0) or (5 , 0)

Answer 2

Given:

• A is a point on x - axis .
• Co - ordinate of point B is ( -7 , 9).
• Distance between A and B is 15 units.

To Find:-

• The co - ordinate of point A .

Formula Used:-

We will use distance formula which is stated as :

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

Where

• $\sf{x_1}$ and $\sf{y_1}$ are co-ordinate of first point.
• $\sf{x_2}$ and $\sf{y_2}$ are co-ordinate of second point.

We will also use a identity , which is :

$\large{\underline{\boxed{\purple{\sf{\dag (a+b)^2=a^2+b^2+2ab.}}}}}$

Answer:-

Firstly it is given that A is a co - ordinate on X - axis . This means that value of ordinate will be 0 . If we take abssicca to be p .

So , Co - ordinate of A would be something like this ( p , 0 ) .

Now here if we apply distance formula between A and B , we can find abssicca of A .

Here ,

• A ' s co-ordinate is ( p , 0 ) .
• B ' s co-ordinate is ( -7 , 9) .
• Given Distance is 15 units.

Let's apply Distance Formula now :

$\sf{\implies 15 = \sqrt{[p - (-7)]^2 + [0-9]^2} .}$

$\sf{\implies 15 = \sqrt{(p+7)^2+(-9)^2}.}$

$\sf{\implies 15 = \sqrt{p^2+49+14p+81} .}$

[Now , Squaring both sides :]

$\sf{\implies (15)^2=(\sqrt{p^2+14p+130})^2 .}$

$\sf{\implies 225 = p^2 + 14p + 130 .}$

$\sf{\implies p^2+14p+130-225=0 .}$

$\sf{\implies p^2 +14p -95 = 0.}$

$\sf{\implies p^2 + 19p - 5p - 95=0.}$

$\sf{\implies p(p+19)-5(p+19)=0 .}$

$\sf{\implies (p+19)(p-5)=0.}$

$\underline{\boxed{\red{\bf{\longmapsto p = 5,(-19)}}}}$

Hence p can have two values which are 5 and (-19) .

Hence the co-ordinate of A is ( 5,0) or (-19,0).