Question

Find the values of m for which the distance between the points P (m, -9) and
Q (2, -3), is 10 units.

Answer 1

Step-by-step explanation:

PQ=10

⇒(10−2)2+(y+3)2=10

64+y2+9+6y=10

Squaring both sides,

y2+6y−27=0

⇒y2+9y−3y−37=0

⇒y(y+9)−3(y+9)=0

⇒(y+9)(y−3)=0

∴y=−9,3

Answer 2

$PQ = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } \\ 10 = \sqrt{{(2 - m) ^{2} } + {( - 3 + 9)}^{2} } \\ 100 = {m}^{2} + 4 - 4m + 36 \\ {m}^{2} - 4m + 4 - 64 = 0 \\ {m}^{2} - 4m - 60 = 0 \\ {m}^{2} - 10m + 6m - 60 = 0 \\ m(m - 10) + 6(m - 10) = 0 \\ (m + 6)(m - 10) = 0$

∴ m = -6 or, 10

Hope It Helps :)