Question

Find the point on X-axis which is equidistant from the points (2,-2) and (-4,2)
pls give step by step explanation​

Answer 1

To find:-

The point on X-axis which is equidistant from the points (2,-2) and (-4,2).

Answer:-

We know that, Point on X-Axis will be in the form of (0,x). So, let's consider this point as P(0,x).

Considering,

For point A,

$\sf{x_{1} = 0}$

$\sf{y_{1} = x}$

$\sf{x_{2} = 2}$

$\sf{y_{2} = -2}$

For point B,

$\sf{x_{1} = 0}$

$\sf{y_{1} = x}$

$\sf{x_{2} = -4}$

$\sf{y_{2} = 2}$

$\sf\red {Substitution:}$

Using distance formula,

$\bf \sqrt{({x_{2} - x_{1} })^{2} + ( {y_{2} - y_{1}})^{2} }$

➪$\sf {(2 - 0)}^{2} + {( - 2 - x})^{2} = {( - 4 - 0 ) }^{2} + {(2 - x)}^{2}$ [∵Square root on both sides gets cancelled]

➪$\sf {(2)}^{2} + ({ - 2 - x})^{2} = {( - 4)}^{2} + {(2 - x)}^{2}$

➪$\sf{4 + 4 + {x}^{2} + 4x = 16 + 4 + {x}^{2} - 4x}$

➪$\sf{8 + {x}^{2} + 4x = 20 + {x}^{2} - 4x}$

➪$\sf{8 - 20 = - 4x - 4x}$

➪$\sf{ - 12 = - 8x}$

➪$\sf \frac{ - 12}{ - 8} = x$

$\huge\pink\bigstar$$\huge\bf\red {x = \frac{3}{2}}$

∴P(0,x) = P(0,$\sf\frac {3}{2})$

$\sf\red{Verification:}$

➙$\sf {(2 - 0)}^{2} + {( - 2 - \frac{3}{2} })^{2} = {( - 4 - 0)}^{2} + {(2 - \frac{3}{2} )}^{2}$

➙$\sf {2}^{2} + {( - 4 - \frac{3}{2} )}^{2} = {( - 4)}^{2} + {(4 - \frac{3}{2} )}^{2}$

➙$\sf{4 + \frac{49}{4} = 16 + \frac{1}{4}}$

➙$\sf \frac{16 + 49}{4} = \frac{64 + 1}{4}$

$\red\bigstar$$\bf\pink {\frac{65}{4} = \frac{65}{4}}$

$\tt\large{∴LHS = RHS}$

$\sf {Proved✔}$

So,

Fomulae used are,

➠$\sf{(x+y)}^{2} = x^{2}+y^{2}+2xy$

➠$\sf{(x-y)}^{2} = x^{2}+y^{2}-2xy$

➠$\sf \sqrt{({x_{2} - x_{1} })^{2} + ( {y_{2} - y_{1}})^{2} }$

Answer 2

SOLUTION

TO DETERMINE

The point on X-axis which is equidistant from the points (2,-2) and ( - 4,2)

EVALUATION

The required point lies on x axis

Let the coordinates of the point is (h, 0)

Now by the given condition

Distance between (h, 0) & (2, - 2) = Distance between (h, 0) & ( - 4, 2)

$\displaystyle \sf{ \implies \sqrt{ {(h - 2)}^{2} + {(0 + 2)}^{2} } = \sqrt{ {(h + 4)}^{2} + {(0 - 2)}^{2} } }$

$\displaystyle \sf{ \implies \sqrt{ {(h - 2)}^{2} + 4 } = \sqrt{ {(h + 4)}^{2} + 4} }$

$\displaystyle \sf{ \implies {(h - 2)}^{2} + 4 = {(h + 4)}^{2} + 4}$

$\displaystyle \sf{ \implies {(h - 2)}^{2} = {(h + 4)}^{2} }$

$\displaystyle \sf{ \implies {h}^{2} - 4h + 4 = {h}^{2} + 8h + 16}$

$\displaystyle \sf{ \implies - 12h= 12}$

$\displaystyle \sf{ \implies h= - 1}$

Hence the required coordinate of the point is ( - 1 , 0 )

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