Question

Find a relation between x and y such that the point (x,y) is equidistant from the points (-4,-1) and (5,-2).

Please do answer with an explanation.​

Answer 1

Step-by-step explanation:

(X+4)^2 - (y+1)^2 = (X-5)^2 - (y+2)^2

X2+16+8x-y2-1-2y=X2+25-10x-y2-4-4y

16-1-25+4+8x+10x-2y+4y=0

-6+18x+2y=0

18x+2y=6

6x+y=3

Answer 2

$\underline{\boxed{\bold{ \bigstar \; Question \; \bigstar }}}$  

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➠ Find a relation between x and y such that the point (x,y) is equidistant from the points (-4,-1) and (5,-2)

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$\underline{\boxed{\bold{ \bigstar \; Important \; Information \; \bigstar }}}$  

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❏ Midpoint of a line segment is such a point that divides line segment in 2 parts of equal measure.

❏ Here is a formula that is unique relation between co-ordinates of midpoints and endpoints of line segment.

• Mid - Point : (x , y)

Let the given co-ordinates of endpoints be :

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• End Point 1 : (x₁ , y₁)

• End Point 2 : (x₂ , y₂)

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$\star \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \boxed { :\longmapsto \; x = \dfrac{x_1 + x_2}{2} }$

$\star \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \boxed { :\longmapsto \; y = \dfrac{y_1 + y_2}{2} }$

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$\underline{\boxed{\bold{ \bigstar \; Given \; \bigstar }}}$  

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➠ End Point 1 : (x₁,y₁) = (-4,-1)

➠ End Point 2 : (x₂,y₂) = (5,-2)

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$\underline{\boxed{\bold{ \bigstar \; Answer \; \bigstar }}}$  

Let's Start !

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❖ Here we go with relation of x-coordinates.

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$\star \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \boxed { :\longmapsto \; x = \dfrac{x_1 + x_2}{2} }$

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$:\implies \; x = \dfrac{(-4)+ 5}{2}$

$:\implies \; x = \dfrac{1}{2}$

$\boxed {\red{:\implies \; 2x = 1 } } \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \mathtt { \{ Equation \;: 1 \} }$

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❖ Here we go with relation of y-coordinates.

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$\star \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \boxed { :\longmapsto \; y = \dfrac{y_1 + y_2}{2} }$

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$:\implies \; y = \dfrac{(-1)+ (-2)}{2}$

$:\implies \; y = \dfrac{-3}{2}$

$\boxed {\red{:\implies \; 2y = -3 } } \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \mathtt { \{ Equation \;: 2 \} }$

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Now by adding Equation : 1 and Equation : 2 ,

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$\boxed {\red{:\leadsto \; 2x = 1 } } \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \mathtt { \{ Equation \;: 1 \} }$

$\boxed {\red{:\leadsto \; 2y = -3 } } \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \mathtt { \{ Equation \;: 2 \} }$

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$:\implies 2x + 2y = 1 + (-3)$

$:\implies 2x + 2y = 1 -3$

$:\implies 2x + 2y = -2$

$:\implies 2( x + y)= -2$

$\boxed {\orange{:\leadsto x + y= \dfrac{-2}{2} = -1 } }$

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$\boxed{\boxed{\bold{\therefore Unique \; Relation : x + y = -1}}}$

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Hope this helps u.../

【Brainly Advisor】