Question

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

Answer 1

Answer:

y = 1, -11

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Step-by-step explanation:

According to the question;

• P = (2, -5)

• Q = (10, y)

• PQ = 10 units.

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For any two points with the co-ordinates (x₁, y₁) and (x₂, y₂), the formula used to calculate the distance is;

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$\sf Distance \ Formula = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

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We'll substitute the values given in the question in the formula to find out the value of 'y'.

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$\sf \dashrightarrow \ PQ = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

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$\sf \dashrightarrow \ 10 = \sqrt{(2 - 10)^2 + (-5 - y)^2}$

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$\sf \dashrightarrow \ 10 = \sqrt{(-8)^2 + (-5 - y)^2}$

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Square on both sides;

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$\sf \dashrightarrow \ \Big(10\Big)^2 = \Big(\sqrt{(-8)^2 + (-5 - y)^2} \ \Big)^2$

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$\sf \dashrightarrow \ 100 = (-8)^2 + (-5 - y)^2$

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$\sf \dashrightarrow \ 100 = 64 + (-5)^2 + (y)^2 - 2(-5)(y)$

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$\sf \dashrightarrow \ 100 - 64 = 25 + y^2 + 10y$

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$\sf \dashrightarrow \ 36 = 25 + y^2 + 10y$

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$\sf \dashrightarrow \ y^2 + 10y = 36 - 25$

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$\sf \dashrightarrow \ y^2 + 10y = 11$

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$\sf \dashrightarrow \ y^2 + 10y - 11 = 0$

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Splitting the middle term;

• Sum ➝ +10

• Product ➝ -11

• Split ➝ 11 × -1

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$\sf \dashrightarrow \ y^2 + 11y - y - 11 = 0$

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$\sf \dashrightarrow \ y(y + 11) - 1(y + 11) = 0$

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$\sf \dashrightarrow \ (y - 1) (y + 11) = 0$

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Therefore y = 1, or y = -11, we'll substitute both values in the distance formula to see if both cases are true.

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Case I:

y = 1

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$\sf \dashrightarrow \ PQ = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$\sf \dashrightarrow \ 10 = \sqrt{(2 - 10)^2 + (-5 - 1)^2}$

$\sf \dashrightarrow \ 10 = \sqrt{(-8)^2 + (-6)^2}$

$\sf \dashrightarrow \ 10 = \sqrt{64 + 36}$

$\sf \dashrightarrow \ 10 = \sqrt{100}$

$\sf \dashrightarrow \ 10 = 10$

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LHS = RHS

∴ y = 1 is one of the answers.

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Case II:

y = -11

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$\sf \dashrightarrow \ PQ = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$\sf \dashrightarrow \ 10 = \sqrt{(2 - 10)^2 + (-5 - (-11))^2}$

$\sf \dashrightarrow \ 10 = \sqrt{(-8)^2 + (-5 + 11)^2}$

$\sf \dashrightarrow \ 10 = \sqrt{64 + (6)^2}$

$\sf \dashrightarrow \ 10 = \sqrt{64 + 36}$

$\sf \dashrightarrow \ 10 = \sqrt{100}$

$\sf \dashrightarrow \ 10 = 10$

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LHS = RHS

∴ y = -11 is also one of the answers.

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Therefore, the values of 'y' are 1 and -11.

Answer 2

Answer:

Given :-

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

To Find :-

• What is the value of y.

Formula Used :-

$\diamondsuit \longmapsto \sf\boxed{\bold{\pink{Distance\: Formula =\: \sqrt{{(x_1 - x_2)}^{2} + {(y_1 - y_2)}^{2}}}}}$

where,

• (x₁ - x₂) = Coordinates for the first point
• (y₁ - y₂) = Coordinates for the second point

Solution :-

Given :

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

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

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

Here,

• PQ = 10 units
• x₁ = 2
• x₂ = 10
• y₁ = - 5
• y₂ = y

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

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

Now, by squaring both sides we get :

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

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

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

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

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

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

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

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

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

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

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

$\implies \sf y(y + 11) - 1(y + 11) =\: 0$

$\implies \sf (y + 11)(y - 1) =\: 0$

$\implies \sf (y + 11) =\: 0$

$\implies \sf\bold{\red{y =\: - 11}}$

Either,

$\implies \sf (y - 1) =\: 0$

$\implies \sf\bold{\red{y =\: 1}}$

$\therefore$ The value of y is - 11 or 1 .