Question

If the distance between the points $A(4,k)$ and $B(1,0)$ is $5$ units, then what can be the possible values of k?

Answer 1

Hint : Here we will use the distance formula for the given points A and B, where
AB $= \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$ then substitutes the value and simplify for the possible values of k.

Complete step-by-step answer :
Let us assume that –
$A(4,k) = ({x_1},{y_1})$
And $B(1,0) = ({x_2},{y_2})$
We are also given distance AB $= 5$ units
Place the values –
$5 = \sqrt {{{(1 - 4)}^2} + {{(0 - k)}^2}}$
Simplify the above equation-
$\Rightarrow 5 = \sqrt {{{( - 3)}^2} + {{( - k)}^2}}$
Remembering squares of negative terms also gives you the positive term.
$\Rightarrow 5 = \sqrt {9 + {k^2}}$
Take square on both the sides of the equation-
$\Rightarrow {\left( 5 \right)^2} = {\left( {\sqrt {9 + {k^2}} } \right)^2}$
Square and square-root cancel each other on the right hand side of the equation.
$\Rightarrow 25 = 9 + {k^2}$
The above equation can be re-written as –
$\Rightarrow 9 + {k^2} = 25$
Make required “k” the subject and take other terms on the opposite side. Remember when you move any term from one side to another, the sign of the term also changes. Positive terms become negative and vice versa.
$\Rightarrow {k^2} = 25 - 9$
Simplify –
$\Rightarrow {k^2} = 16$
Take square-root on both the sides of the equation –
$\Rightarrow \sqrt {{k^2}} = \sqrt {16}$
Square and square-root cancel each other on the left hand side of the equation.
$\Rightarrow k = \sqrt {16}$
Also, we know that squares of negative and positive terms give positive terms.
$\Rightarrow k = \pm 4$
Hence, possible values for k is $( - 4, + 4)$
So, the correct answer is “ $( - 4, + 4)$ ”.

Note : Be careful while simplification and dealing with positive and negative terms. Know the difference between the square and square-roots and apply accordingly. Perfect square number is the square of an integer, simply it is the product of the same integer with itself. For example - $25{\text{ = 5 }} \times {\text{ 5, 25 = }}{{\text{5}}^2}$ , generally it is denoted by n to the power two i.e. ${n^2}$ . Whereas square-root is defined as $\sqrt n$ , for example $25{\text{ = 5 }} \times {\text{ 5, }}\sqrt {{\text{25}}} {\text{ = }}\sqrt {{{\text{5}}^2}} = 5$