A segment in the complex plane has a midpoint at 3 – 2i. If one endpoint of the segment is at 7 + i, what is the other endpoint?

Respuesta :

well, we know the midpoint is 3 - 2i.. well, skipping for a second the 'i', is just (3, - 2) then

one endpoint is at 7 + i, or 7 + 1i or , ( 7, 1)

let's say the other endpoint is at x,y

[tex]\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 7}}\quad ,&{{ 1}})\quad % (c,d) &({{ x}}\quad ,&{{ y}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right) \\\\\\ \left(\cfrac{{{ x}} + {{ 7}}}{2}\quad ,\quad \cfrac{{{ y}} + {{ 1}}}{2} \right)=(3,-2)\implies \begin{cases} \frac{x+7}{2}=3\\ x+7=6\\ x=6-7\\ ------\\ \frac{y+1}{2}=-2\\ y+1=-4\\ y=-4-1 \end{cases}[/tex]

so, x = -1 and y = -5, that gives us ( -1, -5)   or   -1 - 5i on the imaginary grid

the other endpoint is at (-1,-5)

Endpoint is at -1-5i

Given:

A segment in the complex plane has a midpoint at 3 – 2i.

one endpoint of the segment is at 7 + i

Midpoint is at 3-2i , the coordinates of midpoint is (3,-2)

Like the one endpoint of the segment is (7,1)

Let the other end point is at (x,y)

Lets apply midpoint formula

[tex](\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2} )= midpoint \\[/tex]

Two points are (7,1) and (x,y)  and midpoint is (3,-2)

Substitute all the  values

[tex](\frac{7+x}{2} , \frac{1+y}{2} )= (3,-2)\\\frac{7+x}{2}=3\\7+x=6\\x=-1\\\\ \frac{1+y}{2}=-2\\\\1+y=-4\\y=-5[/tex]

so the other endpoint is at (-1,-5)

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