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.