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1 year ago

Find anequation for the perpendicular bisector of the line segment whose endpoints are(3, -1) and (-9,5).

Mathematics

1 answer:

Mariulka [41]1 year ago

4 0

The perpendicular bisector goes through the midpoint.

What is the midpoint of the 2 points?

We simply add up the x points and divide by 2. Then we add up the y points and divide by 2.

So,

Midpoint is:

$(\frac{3-9}{2},\frac{-1+5}{2})=(-3,2)$

Also, the perpendicular bisector's slope is the negative reciprocal of the line's slope.

The slope of the line is change in y points divided by change in x points.

Slope =

$\frac{5--1}{-9-3}=\frac{5+1}{-12}=\frac{6}{-12}=-\frac{1}{2}$

The slope (m) of the perpendicular line (negative reciprocal) is basically:

$2$

Equation of line is:

$y-y_1=m(x-x_1)$

m is the slope (what we got "2")

x1 and y1 are the respective point where it passes through (which is the midpoint, which is (-3,2)

So, equation of perpendicular bisector is:

$\begin{gathered} y-y_1=m(x-x_1) \\ y-2=2(x+3) \\ y=2x+6+2 \\ y=2x+8 \end{gathered}$

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In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$