Question

The distance between the points (5, -9) and (11, y) is 10 units. Find the values of y.
(a) -2, -17
(b) -1, -17
(c) -1, -27
(d) -1, 17

Answer 1

Assume the given points as $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ and apply the distance formula given as: - $d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ to find the distance. Here, ‘d’ is the distance. Substitute the given value of d = 10 and square both sides to get rid of the square root term. Solve the equation for the variable y and get the answer.

Complete step-by-step answer:
Here, we have been provided with two points whose coordinates are given as (5, -9) and (11, y). It is said that the distance between these two points is 10 units. We have been asked to find the values of y.
Now, we know that in coordinate geometry the distance two points having coordinates $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by the formula: -
$\Rightarrow d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$, where ‘d’ is the distance between the points given.
So, let us consider the points (5, 9) and (11, y) as $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ respectively. So, we have,

& \Rightarrow d=\sqrt{{{\left( 11-5 \right)}^{2}}+{{\left( y-\left( -9 \right) \right)}^{2}}} \\\ & \Rightarrow d=\sqrt{{{6}^{2}}+{{\left( y+9 \right)}^{2}}} \\\ \end{aligned}$$ Substituting the given values of distance (d) = 10 units, we get, $$\Rightarrow 10=\sqrt{{{6}^{2}}+{{\left( y+9 \right)}^{2}}}$$ On squaring both sides, we get, $$\begin{aligned} & \Rightarrow {{10}^{2}}={{6}^{2}}+{{\left( y+9 \right)}^{2}} \\\ & \Rightarrow {{\left( y+9 \right)}^{2}}={{10}^{2}}-{{6}^{2}} \\\ \end{aligned}$$ Using the algebraic identity, $${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$$, we get, $$\begin{aligned} & \Rightarrow {{\left( y+9 \right)}^{2}}=\left( 10+6 \right)\left( 10-6 \right) \\\ & \Rightarrow {{\left( y+9 \right)}^{2}}=16\times 4 \\\ & \Rightarrow {{\left( y+9 \right)}^{2}}=64={{8}^{2}} \\\ \end{aligned}$$ Taking square root both sides, we get, $$\Rightarrow \left( y+9 \right)=\pm 8$$ Here, we can have two cases. In the first case we will consider the positive value and in the second case we will consider the negative value. So, we have, 1\. Case (i): - When y + 9 = + 8: - $$\begin{aligned} & \Rightarrow y+9=8 \\\ & \Rightarrow y=8-9 \\\ & \Rightarrow y=-1 \\\ \end{aligned}$$ 2\. Case (ii): - When y + 9 = -8: - $$\begin{aligned} & \Rightarrow y+9=-8 \\\ & \Rightarrow y=-8-9 \\\ & \Rightarrow y=-17 \\\ \end{aligned}$$ Therefore, the two values of y can be -1 and -17. **So, the correct answer is “Option (b)”.** **Note:** One may know that without using the distance formula we will not be able to solve the question, so it must be memorized. Here, you can see that we have obtained two values of y and none of them is rejected. This is because both the values satisfy the obtained equation and there is no reason for rejection. It was necessary to square both sides of the equation so that we can get rid of the square root that contained the variable y, and the value of y can be determined.