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iren2701 [21]
1 year ago
6

2 - 14. What is the equation of the line perpendicular to y - 4 = (x - 6) and passes through the point - (-3,2)? 5 5 - 2 5 = O -

2= O A. y-2=-(x + 3) x OB. y + 3 = -(x - 2) (- 2 O c. y - 2 = = (x + 3) OD. y + 3 = (x - 2) 2 2 2 5

Mathematics
1 answer:
Oduvanchick [21]1 year ago
5 0

Solution:

The equation of a line that passes through a point is expressed as

\begin{gathered} y-y_1=m(x-x_1)\text{ ---- equation 1} \\ where \\ m\Rightarrow slope\text{ of the line} \\ y\Rightarrow y-intercept\text{ of the line} \\ (x_1,y_1)\Rightarrow coordinate\text{ of the point through which the line passes} \end{gathered}

Two lines A and B are said to be perpendicular if the slope of line A is equal to the negative reciprocal of the slope of line B.

Thus, lines A and B are perpendicular if

m_A=-\frac{1}{m_B}\text{ ---- equation 2}

Let the line equation

y-4=\frac{2}{5}(x-6)

represent the line A.

step 1: Evaluate the slope of line A.

Comparing the equation of line A with equation 1, we can conclude that the slope of the line A is

m_A=\frac{2}{5}

step 2: Evaluate the slope of line B.

Since lines A and B are perpendicular, we have

\begin{gathered} From\text{ equation 2,} \\ \begin{equation*} m_A=-\frac{1}{m_B} \end{equation*} \\ where \\ m_A=\frac{2}{5} \\ thus, \\ \frac{2}{5}=-\frac{1}{m_B} \\ cross-multiply \\ -2m_B=5 \\ divide\text{ both sides by -2} \\ \frac{-2m_B}{-2}=\frac{5}{-2} \\ \Rightarrow m_B=-\frac{5}{2} \end{gathered}

step 3: Evaluate the line B.

Since the line B passes through the points (-3,2), recall from equation 1

\begin{equation*} y-y_1=m(x-x_1)\text{ } \end{equation*}

where

\begin{gathered} x_1=-3 \\ y_1=2 \\ m=m_B=-\frac{5}{2} \end{gathered}

Substitute these values into equation 1.

Thus,

\begin{gathered} y-2=-\frac{5}{2}(x-(-3)) \\ \Rightarrow y-2=-\frac{5}{2}(x+3) \end{gathered}

Hence, the equation of the line is

y-2=-\frac{5}{2}(x+3)

The correct option is A

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can be represented by the matrix  

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Our goal is to transform this 3 x 4 matrix so that it ends up looking like:

1  0  0  a
0  1  0  b
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and the solution you want is the vector (a, b, c) (three numeric values).

</span>I have more or less arbitrarily chosen to start with the third row:   
2   -3  -2    3.  We want this row to begin with a 1, so we multiply each of the original four digits by (1/2), obtaining 1   -3/2   -2/2   3/2, or 1  -3/2   -1   3/2.

We can present the original matrix in any order without changing its value.  Thus, the original 

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becomes 

-6  -1    -5  -10
-5    6     4   -7
 1  -3/2  -1   3/2

We want that "1" to appear in the upper, left hand corner of the matrix.  We are free to interchange rows, so we interchange the first and 3rd rows, obtaining 

1  -3/2  -1   3/2
-5    6     4   -7
-6  -1    -5  -10

Next, we manipulate the first row (which begins with 1) so as to get the first element of the 2nd and 3rd rows to be 0.

To achieve this for the 2nd row, we multiply the 1st row by 5, obtaining

5   -15/2   -5   15/2

and then we add this to the existing 2nd row.  The result will be an "0"
in the first column:

0   (6-15/2)   ( 4-5)  (-7+15/2), or   0   -3/2   -1   1/2.

Substitute this new 2nd row for the original 2nd row.  We'll now have:

  1  -3/2  -1   3/2
  0  -3/2   -1   1/2
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Now we have to "fix" the 3rd row, so that it starts with a zero (0):
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Next steps involve transforming the 2nd column so that it looks lilke

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1
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To do this, mult. the entire 2nd row by -2/3,  Here's the expected result:

0    1     2/3    -1/3

Replace the existing 2nd row with this revised 2nd row:

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In the end we want this matrix to look like 

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and the solution you want is the vector (a, b, c) (three numeric values).

Use this new 2nd row to further fix the 2nd column, so that it looks like

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I ask that you go thru this discussion and work out each set of calculations yourself, to verify what I have done so far.  Reply with any questions that arise.  We'll find a way to finish this solution.

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