Question

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

Answer 1

Answer:

1 and 17

Step-by-step explanation:

Using distance formula:

Distance between any two points ( x$_1$, y$_1$ ) and ( x$_2$, y$_2$ ) is given by:

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

Here, points are ( x, 4 ) and ( 9, 10 ), and distance between them is 10.

= > $\sqrt{(x-9)^2+(4-10)^2}$ = 10

= > ( x - 9 )^2 + ( - 6 )^2 = 10^2

= > ( x - 9 )^2 = 100 - 36

= > ( x - 9 )^2 = 64 = 8^2

= > x - 9 = ± √8^2 = ± 8

= > x = ± 8 + 9

So,

x = - 8 + 9 or 8 + 9

x = 1 or 17

Answer 2

Solution :

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

The distance between the points P(x,4) & Q(9,10) is 10 units .

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

As we know that formula of the distance;

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

Where,

• x1 = x
• x2 = 9
• y1 = 4
• y2 = 10

$\longrightarrow\sf{PQ=\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}}\\\\\longrightarrow\sf{10=\sqrt{(9-x)^{2} + (10-4)^{2} } }\\\\\longrightarrow\sf{10=\sqrt{(9-x)^{2} + (6)^{2} }} \\\\\longrightarrow\sf{10=\sqrt{(9-x)^{2} + 36}}$

∴ Squaring both the sides;

$\longrightarrow\sf{(10)^{2} = \big(\sqrt{(9-x)^{2} + 36}\big)^{2}}\\\\\longrightarrow\sf{100 = (9-x)^{2} + 36}\\\\\longrightarrow\sf{(9-x)^{2} = 100 - 36}\\\\\longrightarrow\sf{(9-x)^{2} = 64}\\\\\longrightarrow\sf{(9)^{2} + (x)^{2} - 2\times 9\times x = 64}\\\\\longrightarrow\sf{81 + x^{2} - 18x = 64}\\\\\longrightarrow\sf{x^{2} -18x +81 - 64 = 0}\\\\\longrightarrow\sf{x^{2} -18x + 17 = 0}\\\\\longrightarrow\sf{x^{2} - 17x - x +17 = 0}\\\\\longrightarrow\sf{x(x-17) -1(x-17)=0}$

$\longrightarrow\sf{(x-17)(x-1)=0}\\\\\longrightarrow\sf{x-17=0\:\:\:Or\:\:\:x-1=0}\\\\\longrightarrow\bf{x=17\:\:\:Or\:\:\:x=1}$

Thus;

The value of x will be 1 or 17 .