Biology, 18.10.2019 06:00, jack626259

(-. ,, -7.8) and (2. -3,4) are the vertices of a parallelogram
find the equation of the set of points are equidistant from the po
(1.2.3) and (3.2, -1).

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The correct answer was given: kumarivishakha118

Explanation:

i think this is the right answer

(-. ,, -7.8) and (2. -3,4) are the vertices of a parallelogramfind the equation of the set of points

The correct answer was given: Firdaus17f

answer in the image ☺☺☺☺

Find the equation of the set of points are equidistant from the points (1,2,3) and (3,2,-1).

The correct answer was given: ravichandrareddybuss

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

T(x,y,z)

This is the point equidistant from points P(1,2,3) and Q(3,2,-1)

$\textbf{\underline{Therefore, }}$

We have

TP = TQ

Hence,

TP² = TQ²

(x - 1)² + (y - 2)² + (z - 3)² = (x - 3)² + (y - 2)² + (z + 1)²

Using Identity

${\boxed{\sf\:{(a-b)^2=a^2+b^2-2ab}}}$

${\boxed{\sf\:{(a+b)^2=a^2+b^2+2ab}}}$

Hence,

x² - 2x + 1 + y² - 4y + 4 + z² - 6z + 9 = x² - 6x + 9 + y² - 4y + 4 + z² + 2z + 1

⇒ -2x - 4y - 6z + 14 = -6x - 4y + 2z + 14

⇒ -2x - 6z + 6x - 2z = 0

⇒ 4x - 8z = 0

⇒ 4(x - 2z) = 0

⇒ x - 2z = 0

$\textbf{\underline{Hence\;we\;get\;the\;equation, }}$

$\Large{\boxed{\sf\:{x-2z=0}}$

The correct answer was given: ashiii12

A={(x,y)∈R : (x−4)²+(y−2)²=y²}

Step-by-step explanation:

The correct answer was given: riddhima95

Here's ur answer

Hope it helps

#BAL

The correct answer was given: meandart

Hope u understand ....

answer in the attachment ....

Check it out ....

The correct answer was given: alikarki2580

Refer to the Attachment.

Hope it Helps !!!!

7.8) and (2. -3,4) are the vertices of a parallelogramfind the equation of the set of points are eq

The correct answer was given: Firdaus17f

Step-by-step explanation:

Refer to the attachment ...

The correct answer was given: amruthdeep5258

Any point on the plane (x, y, z)

(x-1)²+(y-2)²+(z-3)²=(x-3)²+(y-2)²+(z+1)²
-2x+1-4y+4+-6z+9=-6x+9-4y+4+2z+1 (cancelling x², y², z² both sides)
4x-8z=0

The correct answer was given: nikhil3810rhmschool

Concept : if Q is a point in the plane in such a way that it is equidistant from both the given points A and B , then we have to use the relation for equidistant of two points .
AQ = BQ

Let Q ≡ ( x, y, z )
now,
AQ = √{ (x - 1)² + ( y - 2)² + (z - 3)²}
BQ = √{ (x - 3)² + (y - 2)² + ( z + 1)²}

now,
AQ = BQ
take square both sides,
AQ² = BQ²
(x - 1)² + ( y-2)² + (z - 3)² = (x - 3)² + (y - 2) + (z +1)²
(x - 1)² + (z - 3)² = (x - 3)² + (z + 1)²
(x - 1)² - (x - 3)² = (z + 1)² - (z - 3)²

use formula ,
a² - b² = (a - b)(a + b)

(x - 1 -x + 3)(x - 1 + x - 3) = (z + 1 - z + 3)(z + 1 +z -3)
2(2x -4 ) = 4(2z -2)
4(x - 2) = 8(z - 1)
x - 2 = 2z - 2
x - 2z = 0

hence, required equation is , x - 2z = 0

The correct answer was given: Brain

answer:

lol, we'll pray for u buddy, ! have a nice day ✌✌

The correct answer was given: Brain

answer:

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