Question

Find all possible values of $y$ for which the distance between the points $A(2, - 3)$ and $B(10,y)$ is $10$ units.

Answer 1

Hint : We have to find the possible values of $y$ for which the distance between the points $A(2, - 3)$ and $B(10,y)$ is $10$ units. For this, we will use the distance formula to find the distance between $A(2, - 3)$ and $B(10,y)$, then we will equate this distance with the given distance of $10$ units to find the value of $y$ by solving the quadratic equation obtained.

Complete step-by-step answer :
We have to find the possible values of $y$ for which the distance between the points $A(2, - 3)$ and $B(10,y)$ is $10$ units i.e., $AB = 10$.
We will solve this question using the distance formula. As we know, the distance between two given points $({x_1},{y_1})$ and $({x_2},{y_2})$which is given by $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$.
Using the distance formula we can write, distance between the points $A(2, - 3)$ and $B(10,y)$ as
$\Rightarrow AB = \sqrt {{{\left( {\left( {10} \right) - \left( 2 \right)} \right)}^2} + {{\left( {\left( y \right) - \left( { - 3} \right)} \right)}^2}}$
Subtracting the terms in the bracket, we get
$\Rightarrow AB = \sqrt {{{\left( 8 \right)}^2} + {{\left( {y + 3} \right)}^2}}$
Applying exponents on terms using formula ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$, we get
$\Rightarrow AB = \sqrt {64 + {y^2} + 6y + 9}$
On simplification we get
$\Rightarrow AB = \sqrt {73 + {y^2} + 6y}$
On squaring both the sides we get
$\Rightarrow {\left( {AB} \right)^2} = 73 + {y^2} + 6y - - - (1)$
Now, as we know $AB = 10$, so on squaring both the sides we get ${\left( {AB} \right)^2} = {\left( {10} \right)^2}$ i.e., ${\left( {AB} \right)^2} = 100$.
Putting this in equation $(1)$ we get
$\Rightarrow 100 = 73 + {y^2} + 6y$
Taking $100$ from left hand side to the right hand side of the above equation and on rewriting, we get
$\Rightarrow 73 + {y^2} + 6y - 100 = 0$
On simplification,
$\Rightarrow {y^2} + 6y - 27 = 0$
On splitting the middle term, we can write
$\Rightarrow {y^2} + 9y - 3y - 27 = 0$
On taking common we get
$\Rightarrow y\left( {y + 9} \right) - 3\left( {y + 9} \right) = 0$
Taking $\left( {y + 9} \right)$ common we get
$\Rightarrow \left( {y + 9} \right)\left( {y - 3} \right) = 0$
On equating we get
$\Rightarrow \left( {y + 9} \right) = 0$ or $\left( {y - 3} \right) = 0$
On solving we get
$\Rightarrow y = - 9$ or $y = 3$
Therefore, the possible values of $y$ are $- 9$ and $3$.
So, the correct answer is “$- 9$ and $3$”.

Note : Here, we have used the distance formula to find the distance between two points by taking $A(2, - 3)$ as $({x_1},{y_1})$ and $B(10,y)$ as $({x_2},{y_2})$. We can also solve this by taking $B(10,y)$ as $({x_1},{y_1})$ and $A(2, - 3)$ as $({x_2},{y_2})$ to find $BA$ and then equating it to the given value of distance between $A$ and $B$ because distance $AB$ and $BA$ are equal.