Question

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Find a point on X- Axis , which is equidistant from the point ( 7,6) and (-3,4).


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Answer 1

$\huge \red{Questions}$
Find a point on X- Axis , which is equidistant from the point ( 7,6) and (-3,4).

$\mathfrak\green{answer} \\ \\ (3 , 0 )$

$\huge \red{Solution}$

Now ,

Let A ( X , 0 ) be any point on the X- Axis , which is equidistant from B ( 7,6) and C (-3,4).

.
. . AB = BC

[ on squaring both sides]

$= > {AB }^{2} = { BC}^{2} \\ \\ = > (7 - x {})^{2} + (6 - 0) {}^{2} = ( - 3 - x ){}^{2} + (4 - 0) {}^{2} .$

$[ distance = \\ \sqrt{(x 2 - x1 {}^{2} ) + (y2 - y1 {}^{2} } ] \\ \\ = > 49 + {x}^{2} - 14x + 36 = 9 + {x}^{2} + 6x + 16 => 20x = 60 \\ \\ = > x = \frac{60}{20} \\ \\ = 3$

Therefore the required value is ( 3, 0 ) .

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Answer 2

Question :

Find a point on X- Axis , which is equidistant from the point ( 7,6) and (-3,4).

Answer:

( 3,0 )

Step-by-step explanation:

Any point on x - axis can be written as ( x , 0 ) .

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

( 7 , 6 ) ----------- ( x . 0 ) ---------- ( - 3 , 4 )

A                            B                        C

So AB = AC .

Applying distance formula we can write :

$\sf{\sqrt{(x-7)^2+(0-6)^2}=\sqrt{(-3-x)^2+(4-0)^2}}\\\\\textsf{Squaring both sides we get :}\\\\\implies \sf{(x-7)^2+(0-6)^2=(-3-x)^2+(4-0)^2}\\\\\implies \sf{x^2+49-14x+36=9+x^2+6x+16}$

$\implies \sf{49-14x+36=9+6x+16}\\\implies \sf{6x+14x=49-9+36-16}\\\implies \sf{20x=60}\\\implies \sf{x=\frac{60}{20}}$

$\implies \sf{x=3}$

Hence the point will be ( 3 , 0 ) .

Note :

You cannot apply distance formula since the line may not be co-linear .

If it was co-linear , then this could have been done by Mid-point formula .