If is equidistant from and , find the values of . Also find the distances and .

Understand the Question A point is equidistant from two points and if the distance is equal to the distance ( or ). The distance formula between two points and is: We first equate to solve for , which yields two values ( and ). Then, we substitute each value of into the distance formula to compute the distances and . Step 1 · Find the values of Since is equidistant from and , .Using the distance formula Squaring both sides Therefore, the coordinates of can be or . Step 2 · Calculate distances and Case 1: When () Distance Distance Case 2: When () Distance Distance Answer > When , When , Common Mistakes Missing the Negative Root: Writing only, forgetting that is also a valid coordinate on the Cartesian plane. Sign Errors in Distances: In calculating , forgetting that subtracting a negative number results in addition (). Incomplete Evaluation: Calculating for only one value of instead of checking both cases ( and ).

Coordinate Geometry | Exercise 7.1

Question 9

If $Q(0, 1)$ is equidistant from $P(5, -3)$ and $R(x, 6)$ , find the values of $x$ . Also find the distances $QR$ and $PR$ .

Solution

Understand the Question

A point $Q$ is equidistant from two points $P$ and $R$ if the distance $QP$ is equal to the distance $QR$ ( $QP = QR$ or $QP^2 = QR^2$ ).
The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
We first equate $QP = QR$ to solve for $x$ , which yields two values ( $x = 4$ and $x = -4$ ).
Then, we substitute each value of $x$ into the distance formula to compute the distances $QR$ and $PR$ .

Step 1 · Find the values of $x$

Since $Q(0, 1)$ is equidistant from $P(5, -3)$ and $R(x, 6)$ , $QP = QR$ .Using the distance formula

\begin{aligned} QP &= QR \\[0.6em] \sqrt{(5 - 0)^2 + (-3 - 1)^2} &= \sqrt{(x - 0)^2 + (6 - 1)^2} \\[0.6em] \sqrt{(5)^2 + (-4)^2} &= \sqrt{x^2 + (5)^2} \\[0.6em] \sqrt{25 + 16} &= \sqrt{x^2 + 25} \\[0.6em] \sqrt{41} &= \sqrt{x^2 + 25} \end{aligned}

Squaring both sides

\begin{aligned} 41 &= x^2 + 25 \\[0.6em] x^2 &= 41 - 25 \\[0.6em] x^2 &= 16 \\[0.6em] x &= \pm 4 \end{aligned}

Therefore, the coordinates of $R$ can be $(4, 6)$ or $(-4, 6)$ .

Step 2 · Calculate distances $QR$ and $PR$

Case 1: When $x = 4$ ( $R(4, 6)$ )

Distance $QR$

\begin{aligned} QR &= \sqrt{(4 - 0)^2 + (6 - 1)^2} \\[0.6em] &= \sqrt{4^2 + 5^2} \\[0.6em] &= \sqrt{16 + 25} = \sqrt{41} \end{aligned}

Distance $PR$

\begin{aligned} PR &= \sqrt{(4 - 5)^2 + (6 - (-3))^2} \\[0.6em] &= \sqrt{(-1)^2 + (6 + 3)^2} \\[0.6em] &= \sqrt{1 + 81} = \sqrt{82} \end{aligned}

Case 2: When $x = -4$ ( $R(-4, 6)$ )

Distance $QR$

\begin{aligned} QR &= \sqrt{(-4 - 0)^2 + (6 - 1)^2} \\[0.6em] &= \sqrt{(-4)^2 + 5^2} \\[0.6em] &= \sqrt{16 + 25} = \sqrt{41} \end{aligned}

Distance $PR$

\begin{aligned} PR &= \sqrt{(-4 - 5)^2 + (6 - (-3))^2} \\[0.6em] &= \sqrt{(-9)^2 + (6 + 3)^2} \\[0.6em] &= \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2} \end{aligned}

Answer

$x = \pm 4$

$QR = \sqrt{41}$

When $x = 4$ , $PR = \sqrt{82}$

When $x = -4$ , $PR = 9\sqrt{2}$

Common Mistakes

Missing the Negative Root: Writing $x^2 = 16 \implies x = 4$ only, forgetting that $x = -4$ is also a valid coordinate on the Cartesian plane.
Sign Errors in Distances: In calculating $y_2 - y_1 = 6 - (-3)$ , forgetting that subtracting a negative number results in addition ( $6 + 3 = 9$ ).
Incomplete Evaluation: Calculating $PR$ for only one value of $x$ instead of checking both cases ( $x = 4$ and $x = -4$ ).

Question 9

More questions in Exercise 7.1