Question

find the distance between the points
( $\frac{-7}{5}$ , -3 and (-4 , $\frac{-6}{5}$ )
Co-Ordinate Geometry!!
Class: 10

Just need a perfect answer... for immediate work... Hope U guys help me

Answer 1

Answer:

Let the point be (x, 0) on x-axis which is equidistant from (2, -5) and (-2, 9).

Using Distance Formula and according to given conditions we have:

(x−2)2+((0−(−5))2−−−−−−−−−−−−−−−−−−√=(x−(−2))2+(0−9)2−−−−−−−−−−−−−−−−−−√

⇒(x2+4−4x+25)−−−−−−−−−−−−−−−√=x2+4+4x+81−−−−−−−−−−−−−√

Squaring both sides, we get

x2+4−4x+25=x2+4+4x+81

⇒−4x+29=4x+85

⇒8x=−56

⇒x=−568=−7

Therefore, point on the x-axis which is equidistant from (2, -5) and (-2, 9) is (-7, 0)

Answer 2

❥︎ Question :-

Find the distance between the points

$\bf \:( \dfrac{-7}{5} , -3 ) \: and \: (- 4, -\dfrac{6}{5} )$

❥︎ Answer: -

❥︎ Given :-

Two coordinates A and B

$\bf \:A( \dfrac{-7}{5} , -3 ) \: and \: B(- 4, -\dfrac{6}{5} )$

❥︎ To find :-

• Distance between A and B.

❥︎ Formula used :-

Let us consider two points A and B, then distance between

$\bf \:A(x_1,y_1) \: and \: B(x_2,y_2) \: is \: given \: by$

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

❥︎ Solution :-

Two coordinates A and B are

$\bf \:A( \dfrac{-7}{5} , -3 ) \: and \: B(- 4, -\dfrac{6}{5} )$

❥︎ Using distance formula

$\bf\implies \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

On substituting the values of

$\bf \:x_1 = - \dfrac{7}{5} ,y_1 = - 3,x_2= - 4 ,y_2= - \dfrac{6}{5}$

$\bf\implies \:AB = \sqrt{ {( - 4 + \dfrac{7}{5} )}^{2} + {( - \dfrac{6}{5} + 3 )}^{2} }$

$\bf\implies \:AB = \sqrt{ {( - \dfrac{13}{5} )}^{2} + {(\dfrac{9}{5} )}^{2} }$

$\bf\implies \:AB = \sqrt{\dfrac{169}{25} + \dfrac{81}{25} }$

$\bf\implies \:AB = \sqrt{\dfrac{169 + 81}{25} }$

$\bf\implies \:AB = \sqrt{\dfrac{250}{25} }$

$\bf\implies \:AB = \sqrt{10} \:$

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