Question

COORDINATE GEOMETRY
CLASS 10
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•Please answer the question in the picture.
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THANKS...​ ​​

Answer 1

Solution :

$\underline{\bf{Given\::}}}}$

If P(2,4) is equidistant from Q(7,0) & R(x,9).

$\underline{\bf{To\:find\::}}}}$

The value of x & the distance of PR .

$\underline{\bf{Explanation\::}}}}$

We know that formula of the distance;

$\boxed{\bf{D=\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} }}}$

A/q

PQ = PR

$\longrightarrow\sf{PQ=\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} } }\\\\\longrightarrow\sf{PQ=\sqrt{(7 - 2)^{2} + (0-4)^{2} } }\\\\\longrightarrow\sf{PQ=\sqrt{(5)^{2} + (4)^{2} } }\\\\\longrightarrow\sf{PQ=\sqrt{25 + 16} }\\\\\longrightarrow\bf{PQ=\sqrt{41} }$

&

$\longrightarrow\sf{PR=\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} } }\\\\\longrightarrow\sf{PR=\sqrt{(x - 2)^{2} + (9-4)^{2} } }\\\\\longrightarrow\sf{PR=\sqrt{(x-2)^{2} + (5)^{2} } }\\\\\longrightarrow\sf{PR=\sqrt{(x-2)^{2}+ 25} }$

$\therefore \longrightarrow\sf{PQ=PR}\\\\\longrightarrow\sf{\sqrt{41} =\sqrt{(x-2)^{2} + 25} }$

Squaring both the side;

$\longrightarrow\sf{(\sqrt{41} )^{2} = \big(\sqrt{(x-2)^{2} + 25} \big)^{2} }\\\\\longrightarrow\sf{41 = (x-2)^{2} + 25}\\\\\longrightarrow\sf{(x-2)^{2} = 41 - 25}\\\\\longrightarrow\sf{(x-2)^{2} = 16}\\\\\longrightarrow\sf{x-2 = \sqrt{16} }\\\\\longrightarrow\sf{x-2=4}\\\\\longrightarrow\sf{x=4+2}\\\\\longrightarrow\bf{x = 6}$

Now;

$\longrightarrow\sf{PR=\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} } }\\\\\longrightarrow\sf{PR=\sqrt{(6 - 2)^{2} + (9-4)^{2} } }\\\\\longrightarrow\sf{PR=\sqrt{(4)^{2} + (5)^{2} } }\\\\\longrightarrow\sf{PR=\sqrt{16 + 25} }\\\\\longrightarrow\bf{PR=\sqrt{41} }$

Thus;

The distance of PR will be √41 .

Answer 2

Given:

P(2, 4)

Q(7, 0)

R(x, 9)

P is equidistant from Q & R.

In simple words, PQ = PR

To Find:

Value of x.

Distance of PQ.

Solution:

ATQ:

$\sf \Longrightarrow PQ = PR$

Squaring on both sides we get:

$\sf \Longrightarrow PQ^{2} = PR^{2}$

Applying Distance Formula we get:

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

For PQ:

x₁ = 2

x₂ = 7

y₁ = 4

y₂ = 0

For PR

x₁ = 2

x₂ = x

y₁ = 4

y₂ = 9

$\sf \Longrightarrow \Big\{\sqrt{(2 - 7)^2 + (4 - 0)^2}\Big\}^{2} = \Big\{\sqrt{(2-x)^2 + (4 - 9)^2}\Big\}^{2}^$

Squares and roots get cancelled.

$\sf \Longrightarrow (-5)^2 + (4)^2 = (2-x)^2 + (-5)^2$

Using (a - b)² = a² + b² - 2ab we get:

$\sf \Longrightarrow 25 + 16 = (2)^2 + (x)^2 - 2(2)(x) + 25$

$\sf \Longrightarrow 25 + 16 - 25 = 4 + x^2 - 4x$

$\sf \Longrightarrow 16 - 4 = x^2 - 4x$

$\sf \Longrightarrow x^2 - 4x = 12$

$\sf \Longrightarrow x^2 - 4x - 12 = 0$

$\sf \Longrightarrow x^2 - 6x + 2x - 12 = 0$

$\sf \Longrightarrow x(x - 6) + 2(x - 6) = 0$

$\sf \Longrightarrow (x - 6) (x + 2) = 0$

∴ x = 6 or x = - 2.

But, one of the values might not be the correct value, so let's cross-check with this equation that we've got from one of the steps of the previous part:

$\sf \Longrightarrow (-5)^2 + (4)^2 = (2-x)^2 + (-5)^2$

Case I

When x = 6

$\sf \Longrightarrow (-5)^2 + (4)^2 = (2-x)^2 + (-5)^2$

$\sf \Longrightarrow (-5)^2 + (4)^2 = (2-6)^2 + (-5)^2$

$\sf \Longrightarrow 25 + 16 = (-4)^2 + (-5)^2$

$\sf \Longrightarrow 25 + 16 = 16 + 25$

$\sf \Longrightarrow 41 = 41$

$\sf \Longrightarrow LHS = RHS$

Therefore x = 6 is correct.

Case II

When x = -2

$\sf \Longrightarrow (-5)^2 + (4)^2 = (2-x)^2 + (-5)^2$

$\sf \Longrightarrow (-5)^2 + (4)^2 = (2-(-2))^2 + (-5)^2$

$\sf \Longrightarrow (-5)^2 + (4)^2 = (2+2)^2 + (-5)^2$

$\sf \Longrightarrow 25 + 16 = (4)^2 + 25$

$\sf \Longrightarrow 25 + 16 = 16 + 25$

$\sf \Longrightarrow 41 = 41$

$\sf \Longrightarrow LHS = RHS$

Therefore x = -2 is also correct.

Hence, the values of x are 6 & -2.

Now, we have to find the distance of PQ.

Using the Distance Formula we get:

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

$\sf \Longrightarrow PQ = \sqrt{(2 - 7)^2 + (4 - 0)^2}$

$\sf \Longrightarrow PQ = \sqrt{(-5)^2 + (4)^2}$

$\sf \Longrightarrow PQ = \sqrt{25+ 16}$

$\sf \Longrightarrow PQ = \sqrt{41} \ sq.units$

Final answers:

⇔ The values of x are 6 and -2.

⇔ The distance of PQ is $\sf \sqrt{41}$ sq.units.