Question

Fid the valus of y for which the distance btwn the pts P (2, -3) and Q (10, y) is 10 units. Maths.​

Answer 1

Answer:

Correct Question :

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

Answer :

The distance between two points (PQ) = –9 OR 3

Given :

Two points :-

P (2, -3).

Q (10, y).

PQ = 10 units.

To Find :

The distance between two points (PQ).

Solution :

We have to find the distance between both the points.

By using identity,

(a + b) = a² + 2ab + b²

$\begin{gathered}\begin{gathered}\implies \bf 64 + {y}^{2} + 2(y)(3) + {(3)}^{2} = {(10)}^{2} \\ \\ \\ \implies \bf 64 + {y}^{2} + 6y + 9 = 100 \\ \\ \\ \implies \bf {y}^{2} + 6y + 64 + 9 - 100 = 0 \\ \\ \\ \implies \bf {y}^{2} + 6y + 73 - 100 = 0 \\ \\ \\ \implies \bf {y}^{2} + 6y - 27 = 0 \\ \\ \\ \implies \bf {y}^{2} + 9y - 3y - 27 = 0 \\ \\ \\ \implies \bf y(y + 9) - 3(y + 9) = 0 \\ \\ \\ \implies \bf (y + 9)(y - 3) = 0 \\ \\ \\ \implies \bf y + 9 = 0 \: \: OR \: \: y - 3 = 0 \\ \\ \\ \implies \bf y = - 9 \: \: OR \: \: y = 3 \\ \\ \\ \: \: \: \therefore \bf \: \: y = - 9 \: \: OR \: \: 3\end{gathered}\end{gathered}⟹64+y2+2(y)(3)+(3)2=(10)2⟹64+y2+6y+9=100⟹y2+6y+64+9−100=0⟹y2+6y+73−100=0⟹y2+6y−27=0⟹y2+9y−3y−27=0⟹y(y+9)−3(y+9)=0⟹(y+9)(y−3)=0⟹y+9=0ORy−3=0⟹y=−9ORy=3∴y=−9OR3$

Hence,

The distance between two points (PQ) is –9 OR 3.