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sergejj [24]
3 years ago
11

which of the following pairs of lines are perpendicular? select all that apply A y=2/3x+4 and y=2/3x-8 B y=2/3x-8 and y=-3/2x-8

C y=x+3 and y=-x=3 D y=3x+n and y=3x-2 E y=3 and x=4 F y=4/5x-8 and y=-5/4x=3
Mathematics
2 answers:
jasenka [17]3 years ago
6 0

Its B, C, E, F because for B the neg reciprocal of 1 is -1

grin007 [14]3 years ago
6 0

Answer:

Hence, B, C and F are pairs of perpendicular line.

Step-by-step explanation:

We have to find the pairs of line that are perpendicular to each other.

Two lines are said to be perpendicular if the product of their slope is -1 that is:

m_1\times m_2 = -1

The slope of each line can be calculated with the help of slope intercept form:

y = mx + c

1)

y=\frac{2}{3}x+4, y=\frac{2}{3}x-8\\\\m_1 = \frac{2}{3}\\\\m_2 = \frac{2}{3}\\\\m_1\times m_2 \neq -1

2)

y=\frac{2}{3}x-8, y=\frac{-3}{2}x-8\\\\m_1 = \frac{2}{3}\\\\m_2 = \frac{-3}{2}\\\\m_1\times m_2 = -1

3)

y=x+3, y=-x-3\\\\m_1 = 1\\\\m_2 = -1\\\\m_1\times m_2 = -1

4)

y=3x+n, y=3x-2\\\\m_1 =3\\\\m_2 =3\\\\m_1\times m_2 \neq -1

5)

y=3, x=4\\\\m_1 =0\\\\m_1\times m_2 \neq -1

6)

y=\frac{4}{5}x-8, y=\frac{-5}{4}x-3\\\\m_1 = \frac{4}{5}\\\\m_2 = \frac{-5}{4}\\\\m_1\times m_2 = -1

Hence, B, C and F are pairs of perpendicular line.

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——————————

Solve the trigonometric equation:

\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}


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\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}


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So the equation becomes

\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}


Solving the quadratic equation:

\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.


\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}


\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}


You can discard the negative value for  t. So the solution for  (ii)  is

\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}


Substitute back for  t = sin x.  Remember the restriction for  x:

\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}

where  k  is an integer.


I hope this helps. =)

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