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SOVA2 [1]
3 years ago
12

Given two points M & N on the coordinate plane, find the slope of MN , and state the slope of the line perpendicular to MN .

(there's two questions)
1) M(9,6), N(1,4)

2) M(-2,2), N(4,-4)
Mathematics
1 answer:
telo118 [61]3 years ago
4 0

Answer:

Problem 1)       m = \dfrac{1}{4}     slope_{perpendicular} = -4

Problem 2)      m = \dfrac{1}{3}     slope_{perpendicular} = -3

Step-by-step explanation:

slope = m = \dfrac{y_2 - y_1}{x_2 - x_1}

slope_{perpendicular} = \dfrac{-1}{m}

Problem 1) M(9,6), N(1,4)

slope = m = \dfrac{6 - 4}{9 - 1} = \dfrac{2}{8} = \dfrac{1}{4}

slope_{perpendicular} = \dfrac{-1}{\frac{1}{4}} = -4

Problem 2) M(-2,2), N(4,-4)

slope = m = \dfrac{4 - 2}{4 - (-2)} = \dfrac{2}{6} = \dfrac{1}{3}

slope_{perpendicular} = \dfrac{-1}{\frac{1}{3}} = -3

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3 years ago
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Answer:

(a)x_1=-2,x_2=1

(b)x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix A and the matrix B_j for each variable. The matrix A is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get A more easily.

2x_1+5x_2=1\\x_1+4x_2=2

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To get B_1, replace in the matrix A the 1st column with the results of the equations:

B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]

To get B_2, replace in the matrix A the 2nd column with the results of the equations:

B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]

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x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2

In the case of B2,  the determinant is going to be zero. Instead of using the rule, substitute the values ​​of the variable x_1 in one of the equations and solve for x_2:

2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1

(b) In this system, follow the same steps,ust remember B_3 is formed by replacing the 3rd column of A with the results of the equations:

2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0

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B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]

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x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}

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