Question

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

Answer 1

Answer:

- 1 or - 3

Step-by-step explanation:

Distance between P and A = distance between P and ( - 2k, 3 )

Using distance formula:

= > √[(-2-2)² + (k-2)²] = √[(-2k-2)²+(-3-2)²]

= > (-2-2)² + (k-2)² = (-2k-2)²+(-3-2)²

= > (-4)² + (k-2)² = (-2k-2)² + (-5)²

= > 16 + k² + 4 - 4k = 4k² + 4 + 8k + 25

= > k² - 4k + 20 = 4k² + 8k + 29

= > 0 = 4k² - k² + 8k + 4k + 29 - 20 = 0

= > 0 = 3k² + 12k + 9

= > 0 = k² + 4k + 3

= > 0 = k² + 3k + k + 3

= > 0 = k( k + 3 ) + ( k + 3 )

= > ( k + 1 )( k + 3) = 0

= > k = - 1 or - 3

Thus,

AP = √(-2-2)²+(k-2)²

AP = √(-4)²+(-1-2)² or √(-4)²+(-3-2)²

AP = √16+9 or √16+25

AP = √25 = 5 or √41

Answer 2

Solution :

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

If the point P(2,2) is equidistant from the points A(-2,k) & B(-2k,-3).

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

As we know that formula of the distance;

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

$\underline{\boldsymbol{According\:to\:the\:question\::}}}$

PA = PB

• P(2,2)
• A(-2,k)

$\longrightarrow\sf{PA=\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} } }\\\\\longrightarrow\sf{PA=\sqrt{(-2 - 2)^{2} + (k-2)^{2} } }\\\\\longrightarrow\sf{PA=\sqrt{(-4)^{2} + (k-2)^{2} } }\\\\\longrightarrow\sf{PA=\sqrt{16 + (k-2)^{2}} }$

&

• P(2,2)
• B(-2k-3)

$\longrightarrow\sf{PB=\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} } }\\\\\longrightarrow\sf{PB=\sqrt{(-2k - 2)^{2} + (-3-2)^{2} } }\\\\\longrightarrow\sf{PB=\sqrt{(-2k-2)^{2} + (-5)^{2} } }\\\\\longrightarrow\sf{PB=\sqrt{(-2k-2)^{2} + 25} }$

Now;

$\longrightarrow\sf{PA=PB}$

Squaring both the side;

$\longrightarrow\sf{\big(\sqrt{16 + (k-2)^{2} } \big)^{2} = \big(\sqrt{(-2k-2)^{2} + 25} \big)^{2}}\\\\\longrightarrow\sf{16 + (k-2)^{2} = (-2k-2)^{2} + 25}\\\\\longrightarrow\sf{16 + k^{2} +(2)^{2} -2\times k\times 2 = (-2k)^{2} + (2)^{2} +2\times (-2k) \times (-2)+25}\\\\\longrightarrow\sf{16 + k^{2} + 4 -4k = 4k^{2} + 4 + 8k + 25}\\\\\longrightarrow\sf{20 + k^{2} - 4k = 4k^{2} +8k + 29}\\\\\longrightarrow\sf{4k^{2} -k^{2} +8k+4k + 29 -20 = 0}\\\\\longrightarrow\sf{3k^{2} +12k + 9 =0 }$

$\longrightarrow\sf{3k^{2} +3k+9k + 9=0}\\\\\longrightarrow\sf{3k(k +1) +9(k+1)=0}\\\\\longrightarrow\sf{(k+1)(3k+9)=0}\\\\\longrightarrow\sf{k+1=0\:\:Or\:\:3k+9=0}\\\\\longrightarrow\sf{k=-1\:\:\:Or\:\:\:3k=-9}\\\\\longrightarrow\sf{k=-1\:\:\:Or\:\:\:k=-\cancel{9/3}} \\\\\longrightarrow\bf{k=-1\:\:\:Or\:\:\:k=-3}$

∴Putting the value of k = -1 in length of AP;

$\longrightarrow\sf{AP=\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} } }\\\\\longrightarrow\sf{AP=\sqrt{(-2 - 2)^{2} + (-1-2)^{2} } }\\\\\longrightarrow\sf{AP=\sqrt{(-4)^{2} + (-3)^{2} } }\\\\\longrightarrow\sf{AP=\sqrt{16 + 9 }}\\\\\longrightarrow\sf{AP=\sqrt{25} }\\\\\longrightarrow\bf{AP=5\:unit}$

∴Putting the value of k = -3 in length of AP;

$\longrightarrow\sf{AP=\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} } }\\\\\longrightarrow\sf{AP=\sqrt{(-2 - 2)^{2} + (-3-2)^{2} } }\\\\\longrightarrow\sf{AP=\sqrt{(-4)^{2} + (-5)^{2} } }\\\\\longrightarrow\sf{AP=\sqrt{16 + 25 }}\\\\\longrightarrow\bf{AP=\sqrt{41}\:unit}$

Thus;

The length of the AP will be 5 & √41 units .