Find the values of for which the distance between the points and is 10 units.

Understand the Question The distance between points and is given by the distance formula: Given , , and distance units. Substituting the given values and squaring both sides gives a quadratic equation in , which can be solved to find all possible values of . Step 1 · Set up the Quadratic Equation By the distance formula Substitute , , and Squaring both sides Step 2 · Solve for Factorising the quadratic equation by splitting the middle term Setting each factor to zero Answer > or Common Mistakes Sign Error in Coordinates: Writing instead of . Discarding Negative Values: Forgetting that coordinates can be negative. While distance is always positive, the coordinate is fully valid. Expansion Mistakes: Expanding incorrectly as instead of .

Coordinate Geometry | Exercise 7.1

Question 8

Find the values of $y$ for which the distance between the points $P(2, -3)$ and $Q(10, y)$ is 10 units.

Solution

Understand the Question

The distance $d$ between points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Given $P(2, -3)$ , $Q(10, y)$ , and distance $PQ = 10$ units.
Substituting the given values and squaring both sides gives a quadratic equation in $y$ , which can be solved to find all possible values of $y$ .

Step 1 · Set up the Quadratic Equation

By the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substitute $P(2, -3)$ , $Q(10, y)$ , and $d = 10$

\begin{aligned} 10 &= \sqrt{(10 - 2)^2 + (y - (-3))^2} \\[0.6em] 10 &= \sqrt{8^2 + (y + 3)^2} \\[0.6em] 10 &= \sqrt{64 + (y^2 + 6y + 9)} \\[0.6em] 10 &= \sqrt{y^2 + 6y + 73} \end{aligned}

Squaring both sides

\begin{aligned} 10^2 &= y^2 + 6y + 73 \\[0.6em] 100 &= y^2 + 6y + 73 \end{aligned}

$y^2 + 6y + 73 - 100 = 0$

$y^2 + 6y - 27 = 0$

Step 2 · Solve for $y$

Factorising the quadratic equation by splitting the middle term $y^2 + 9y - 3y - 27 = 0$

\begin{aligned} y(y + 9) - 3(y + 9) &= 0 \\[0.6em] (y + 9)(y - 3) &= 0 \end{aligned}

Setting each factor to zero $y + 9 = 0 \quad \text{or} \quad y - 3 = 0$

$y = -9 \quad \text{or} \quad y = 3$

Answer

$y = 3$ or $y = -9$

Common Mistakes

Sign Error in Coordinates: Writing $(y - 3)^2$ instead of $(y - (-3))^2 = (y + 3)^2$ .
Discarding Negative Values: Forgetting that coordinates can be negative. While distance is always positive, the coordinate $y = -9$ is fully valid.
Expansion Mistakes: Expanding $(y + 3)^2$ incorrectly as $y^2 + 9$ instead of $y^2 + 6y + 9$ .

Question 8

More questions in Exercise 7.1