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

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 the distance formula,

$\boxed{ \bf \sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2}} - y_{1})^{2}}}$

$\implies \bf \sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2}} - y_{1})^{2}} = PQ \\ \\ \\ \implies \bf \sqrt{ {(10 - 2)}^{2} + { \bigg(y - ( - 3) \bigg)}^{2}} = 10 \\ \\ \\ \implies \bf \sqrt{ {(8)}^{2} + {(y + 3)}^{2} } = 10$

By using identity,

• (a + b) = a² + 2ab + b²

$\implies \bf 64 + {y}^{2} + 2(y)(3) + {(3)}^{2} = {(10)}^{2} \\ \\ \\ \implies \bf 64 + {y}^{2} + 6y + 9 = 100 \\ \\ \\ \implies \bf {y}^{2} + 6y + 64 + 9 - 100 = 0 \\ \\ \\ \implies \bf {y}^{2} + 6y + 73 - 100 = 0 \\ \\ \\ \implies \bf {y}^{2} + 6y - 27 = 0 \\ \\ \\ \implies \bf {y}^{2} + 9y - 3y - 27 = 0 \\ \\ \\ \implies \bf y(y + 9) - 3(y + 9) = 0 \\ \\ \\ \implies \bf (y + 9)(y - 3) = 0 \\ \\ \\ \implies \bf y + 9 = 0 \: \: OR \: \: y - 3 = 0 \\ \\ \\ \implies \bf y = - 9 \: \: OR \: \: y = 3 \\ \\ \\ \: \: \: \therefore \bf \: \: y = - 9 \: \: OR \: \: 3$

Hence,

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

Answer 2

ɢɪᴠᴇɴ:-

• The points P (2, -3) and Q (10, y).
• PQ = 10 units.

ᴛᴏ ꜰɪɴᴅ :-

• The distance between two points (PQ).

ꜱᴏʟᴜᴛɪᴏɴ :-

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak { \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2}} - y_{1})^{2}} = PQ} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {\sqrt {{(10 - 2)}^{2} + { \bigg(y - ( - 3) \bigg)}^{2}} = 10} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {\sqrt{{(8)}^{2} + {(y + 3)}^{2} } = 10} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {64 + {y}^{2} + 2(y)(3) + {(3)}^{2} = {(10)}^{2}} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {64 + {y}^{2} + 6y + 9 = 100} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {{y}^{2} + 6y + 64 + 9 - 100 = 0} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {{y}^{2} + 6y + 73 - 100 = 0} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {{y}^{2} + 6y - 27 = 0} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {{y}^{2} + 9y - 3y - 27 = 0 } \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak { (y + 9)(y - 3) = 0} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {2y + 9 = 0 \: \: OR \: \: y - 3 = 0} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

Therefore, the distance between two points (PQ) is –9 or 3.

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Identity used:-

• (a + b) = a² + b²+ 2ab