Question

If the distance between the points $(4,p)$ and $(1,0)$ is $5$, then find the value of p.
A.$4$ only
B.$ \pm 4$
C.$ - 4$ only
D.$0$

Answer 1

Hint: The distance between two points in the plane is the length of the line segment joining the points. If $P({x_1},{y_1})$ and $Q({x_2},{y_2})$ are two points then the distance between these two points is given by $PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
i.e., \[PQ = \sqrt {(Difference{\text{ }}of{\text{ }}abscissa) + (Difference{\text{ }}of{\text{ }}ordinates)} \]
In the given question the distance between two points is already given, we have to find the value of p.
Let $P(4,p)$ and $Q(1,0)$ be the given points and distance between them is $PQ = 5$ unit.
Formula: $PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $

Complete step by step solution:
Distance formula: $PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
 Let $P(4,p)$ and $Q(1,0)$ be the given points. Then,
Here, ${x_1} = 4$ , ${y_1} = p$ and ${x_2} = 1$, ${y_2} = 0$.
Distance between $P(4,p)$ and $Q(1,0)$ is $PQ = 5$unit.
Substitute the value of ${x_1}$ , ${x_2}$ , ${y_1}$ , ${y_2}$ and $PQ$ in the distance formula.
 $ \Rightarrow \sqrt {{{(1 - 4)}^2} + {{(0 - p)}^2}} = 5$
On simplifying the brackets, we get
$ \Rightarrow \sqrt {{{(- 3)}^2} + {{( - p)}^2}} = 5$
$ \Rightarrow \sqrt {9 + {p^2}} = 5$
On squaring both the sides, we get
$ \Rightarrow \left( {\sqrt {9 + {p^2}} } \right) = {\left( 5 \right)^2}$
$ \Rightarrow 9 + {p^2} = 25$
Shift $9$ to the Left hand side.
$ \Rightarrow {p^2} = 25 - 9$
$ \Rightarrow {p^2} = 16$ or ${p^2} = {4^2}$
The Square of the positive and negative numbers is the same.
$\therefore p = \pm 4$
Hence, the value of p is $ \pm 4$.
So, the correct option is B.
So, the correct answer is “Option B”.

Note: In the given question we are given the points in pairs, what if the point is given as: the abscissa is this and ordinate is this, in that condition remember that the abscissa of a point is its perpendicular distance from the y-axis whereas the ordinate of a point is its perpendicular distance from x-axis. Remember to write the value of p as $ \pm 4$ instead of just writing $4$ because the value of square root can be positive and negative.