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Find the values of y for which the distance between the points P( 2 , – 5 ) and Q ( 10 , y ) is 10 units.

Answer y=1,-11
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Answer 1

Answer:

1

Step-by-step explanation:

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Given:-

• Distance = 10 units
• x₁ = 2     x₂ = 10
• y₁ = -5    y₂ = x

To find

• y₂

The distance between two points P&Q is.

     $\sqrt{(x_2-x_1)^{2} + (y_2 - y_1)^{2} }$

      $\sqrt{(x_2-x_1)^{2} + (y_2 - y_1)^{2} } = 10\\\\\sqrt{(10-2)^{2} + (x + 5)^{2} } = 10\\\\(10 - 2)^{2} + (x + 5)^{2} = 100\\\\8^{2} + (x+5)^{2} = 100\\\\64 + (x+5)^{2} = 100\\\\(x+5)^{2} = 100-64\\\\(x+5)^{2} = 36\\\\x+5 = 6\\\\x = 1$

     

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Answer 2

Given :-

The distance between the points P(2 , - 5) and Q(10 , y) is 10 units.

To Find :-

the value of y.

By using Formula :-

$\begin{gathered}\longmapsto \sf\boxed{\bold{\red{Distance\: Formula =\: \sqrt{{(x_1 - x_2)}^{2} + {(y_1 - y_2)}^{2}}}}}\\\end{gathered}$

where,

(x₁ - x₂) = Coordinates for the first point

(y₁ - y₂) = Coordinates for the second point

Solution :-

Here Given that :

• P(2 , - 5)
• Q(10 , y)
• PQ = 10 units

Now, according to the question by using the formula we get :

$\begin{gathered}\implies \sf PQ =\: \sqrt{{(x_1 - x_2)}^{2} + {(y_1 - y_2)}^{2}}\\\end{gathered}$

• PQ = 10 units

x₁ = 2

x₂ = 10

y₁ = - 5

y₂ = y

$\begin{gathered}\implies \sf 10 =\: \sqrt{{(2 - 10)}^{2} + {(- 5 - y)}^{2}}\\\end{gathered}$

$\begin{gathered}\implies \sf 10 =\: \sqrt{{(- 8)}^{2} + {(- 5 - y)}^{2}}\\\end{gathered}$

Now, by squaring both sides we get :

$\begin{gathered}\implies \sf {\bigg \lgroup 10 \bigg \rgroup}^{2} =\: {\bigg \lgroup \sqrt{{(- 8)}^{2} + {( - 5 - y)}^{2}} \bigg \rgroup}^{2} \\\end{gathered}$

$\begin{gathered}\implies \sf 100 =\: {(- 8)}^{2} + {(- 5 - y)}^{2}\\\end{gathered}$

$\begin{gathered}\implies \sf 100 =\: (- 8)(- 8) + {(- 5)}^{2} + {(y}^{2} - 2(- 5)(y)\\\end{gathered}$

$\begin{gathered}\implies \sf 100 =\: 64 + 25 + {(y)}^{2} + 10y\\\end{gathered}$

$\begin{gathered}\implies \sf 100 - 64 =\: 25 + {(y)}^{2} + 10y\\\end{gathered}/tex]</p><p>[tex]\begin{gathered}\implies \sf 36 =\: 25 + {(y)}^{2} + 10y\\\end{gathered}$

$\begin{gathered}\implies \sf 36 - 25 =\: {(y)}^{2} + 10y\\\end{gathered}$

$\begin{gathered}\implies \sf 11 =\: {(y)}^{2} + 10y\\\end{gathered}$

$\begin{gathered}\implies \sf {(y)}^{2} + 10y - 11 =\: 0\\\end{gathered}$

$\begin{gathered}\implies \sf {(y)}^{2} + (11- 1)y - 11 =\: 0\\\end{gathered}$

$\begin{gathered}\implies \sf {(y)}^{2} + 11y - y - 11 =\: 0\\\end{gathered}$

$\begin{gathered}\implies \sf y(y + 11) - 1(y + 11) =\: 0\\\end{gathered}$

$\begin{gathered}\implies \sf y(y +11)(y-1)=\:0\\\end{gathered}$

$\begin{gathered}\implies \sf y(y + 11) =\: 0\\\end{gathered}$

$\begin{gathered}\implies \sf y=-11\\\end{gathered}$

$\sf \pink{y=-11}$

Either,

$\begin{gathered}\implies \sf (y-1) =\: 0\\\end{gathered}$

$\sf \pink{y=1}$

Therefore, the value of y is -11 or 1