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3 years ago

9. Which pair of equations below represent perpendicular lines?

Mathematics

2 answers:

natima [27]3 years ago

8 0

A is correct. ...................................................

yulyashka [42]3 years ago

5 0

Answer:

$y =\frac{7}{8}x + 12$ and $y = -\frac{8}{7}x - 8$

Step-by-step explanation:

The product of the slopes of the perpendicular lines is -1

Option 1) $y =\frac{7}{8}x + 12$ and $y = -\frac{8}{7}x - 8$

General equation of line : $y=mx+c$

Comparing with general equation

Slope of line 1 = $\frac{7}{8}$

Slope of line 2= $-\frac{8}{7}$

Now product of slopes = $\frac{7}{8} \times \frac{-8}{7}$

= $-1$

Since The product of the slopes of the perpendicular lines is -1

So, $y =\frac{7}{8}x + 12$ and $y = -\frac{8}{7}x - 8$ represent perpendicular lines

Option 2) $y = 5x + 15$ and $y = -5x + 15$

General equation of line : $y=mx+c$

Comparing with general equation

Slope of line 1 = 5

Slope of line 2= -5

Now product of slopes = $5 \times -5$

= $-25$

Since The product of the slopes of the perpendicular lines is not -1

So, $y = 5x + 15$ and $y = -5x + 15$ does not represents the perpendicular lines.

Option 3) $y = 4x + 9$ and $y = 4x -9$

General equation of line : $y=mx+c$

Comparing with general equation

Slope of line 1 = 4

Slope of line 2= 4

Now product of slopes = $4 \times 4$

= $16$

Since The product of the slopes of the perpendicular lines is not -1

So, $y = 4x + 9$ and $y = 4x -9$ does not represents the perpendicular lines.

Option 4) $y = 9$ and $y = 18$

General equation of line : $y=mx+c$

Comparing with general equation

Slope of line 1 = 0

Slope of line 2=0

Now product of slopes = $0 \times 0$

= $0$

Since The product of the slopes of the perpendicular lines is not -1

So, $y = 9$ and $y = 18$ does not represents the perpendicular lines.

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Step-by-step explanation:

Since this question is lacking the matrix A, we will solve the question with the matrix

$\left[\begin{matrix}4 & -2 \\ 1 & 1 \end{matrix}\right]$

so we can illustrate how to solve the problem step by step.

a) The characteristic polynomial is defined by the equation $det(A-\lambdaI)=0$ where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,

$\left|\left[\begin{matrix}4-\lamda & -2 \\ 1 & 1-\lambda \end{matrix}\right]\right|= 0 = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+4+2 = \lambda^2-5\lambda+6$

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c) To find the bases of each eigenspace, we replace the value of lambda and solve the homogeneus system(equalized to zero) of the resultant matrix. We will illustrate the process with one eigen value and the other one is left as an exercise.

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$\left[\begin{matrix}1 & -2 \\ 1 & -2 \end{matrix}\right]$ .

Since both rows are equal, we have the equation

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For the case $\lambda=2$ , using the same process, we get the vector (1,1).

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3 years ago