Question

Find the point on x-axis which are at a distance of 5 units from the point (5,-4)​

Answer 1

Given Question :-

• Find the point on x-axis which are at a distance of 5 units from the point (5,-4)

Answer

Let the point on x - axis be P (x, 0) and let (5, - 4) represents coordinates of Q.

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{coordintes \: of \: Q(5, - 4)} \\ &\sf{distance \: between \: PQ = 5 \: units} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find -   \begin{cases} &\sf{point \: on \: x \: - \: axis}  \end{cases}\end{gathered}\end{gathered}$

Concept Used

Distance Formula between two points

☆ We know,

$\sf \: Distance \: between \: A(x_1, y_1) \: and \: B(x_2, y_2) \: is \: given \: by$

$\boxed{ \red{ \bf \: AB = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} } }}$

$\large\underline\purple{\bold{Solution :- }}$

● According to statement, the distance between P(x, 0) and Q(5, - 4) is 5 units.

$: \implies \bf \: PQ\: = \sf \: 5$

$: \implies \sqrt{ {(5 - x)}^{2} + ( - 4 - 0)} = 5$

● On squaring both sides, we get

$: \implies \sf \: {(5 - x)}^{2} + 16 = 25$

$: \implies \sf \: {(5 - x)}^{2} = 25 - 16$

$: \implies \sf \: {(5 - x)}^{2} = 9$

$: \implies \sf \: 5 - x \: = \: \pm \: 3$

$: \implies \bf \: 5 - x \: = - 3 \: \: \: or \: \: \: 5 - x = 3$

$: \implies \bf \: x \: = 8 \: \: \: or \: \: \: x \: = 2$

$\boxed{ \red{ \sf \: Hence, \: the \: points \: on \: x - axis \: are \: (8, 0) \: or \: (2, 0)}}$