Question

Find the value of ‘𝒂’, if the distance between the points
𝑷(𝟑, −𝟔) and 𝑸(−𝟑, 𝒂) is 𝟏𝟎 units.

Answer 1

Step 1: Use the distance formula and obtain the equation in terms of ‘a’.
Let the given points be:

P(3, -6) = $x_1,y_1$
Q(-3, a) = $x_2,y_2$

Given that the distance between the points P(3, -6) and Q(-3, a) is 10 units.
$\sqrt{({(-3–3)}^2+{(a+6)}^2)}=10units$


Step 2 : Find the value of ‘a’.

Squaring both sides of the equation,
$(-6{)}^2+(a+6{)}^2=100$
$(a+6{)}^2=100–36=64$
$a+6=\pm 8$

Case I: Considering +8,
$a+6=8$ ,
$a=8–6=2$

Case II: Considering –8
$a+6=–8$
$a=-8–6$
$a=–14$

Therefore, the coordinates of Q are either Q(–3, 2) or Q(–3, –14).