Mathematics, 07.01.2020 13:31 alex6496

Which of the following pairs of lines are perpendicular? select all that apply.

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y = $\frac{2}{3}$ x - 8 and y = - $\frac{3}{2}$ x - 8y = x + 2 and y = -x + 3y = 3 and x = 4y = $\frac{4}{5}$ x - 8 and y = - $\frac{5}{4}$ x + 3

Step-by-step explanation:

Two lines are perpendicular to each other if the product of their slopes is -1

In the case of lines y = 3 and x = 4 , their slopes are zero and infinite respectively but this two lines are by construction perpendicular to each other.

Answer from: richisolano14

The correct options are: B, C, E and F

Step-by-step explanation:

Consider the provided information.

Two lines are said to be perpendicular if the slopes are opposite reciprocals.

$m_1\times m_2=-1$

The slope intercept form is: y=mx+c

Where m is the slope of line.

Now consider the provided options:

Option A) $y=\frac{2}{3}x+4\ and\ y= \frac{2}{3}x-8$

Both the equation have same slope i.e 2/3.

Thus, the pair of line is not perpendicular.

Option B) $y=\frac{2}{3}x-8\ and\ y= -\frac{3}{2}x-8$

$\frac{2}{3}\times -\frac{3}{2}=-1$

Hence, the pair of line is perpendicular.

Option C) $y=x+2\ and\ y= -x+3$

$1\times \frac{1}{-1}=-1$

Hence, the pair of line is perpendicular.

Option D) $y=3x+2\ and\ y=3x-2$

Both the equation have same slope i.e 3.

Thus, the pair of line is not perpendicular.

Option E) $y=3\ and\ x=4$

y=3 is a horizontal line parallel to x-axis and x=4 is a vertical line parallel to y axis. We know that x and y axis are perpendicular to each other and the provided lines follows the same property.

Thus, the pair of line is perpendicular.

Option F) $y=\frac{4}{5}x-8\ and\ y= -\frac{5}{4}x+3$

$\frac{4}{5}\times -\frac{5}{4}=-1$

Thus, the pair of line is perpendicular.

Hence, the correct options are: B, C, E and F

Answer from: Quest

in if you ask questions, the points are taken from your total and given to the one who answered it..

so if i have 30 points and i ask a question for 5 points, then when someone asks a question i will lose 5 points and the answerer earns 5

happy to

Answer from: Quest

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

$x^2-2x+y^2-6y=26$

we now create space between the x and y terms with parenthesis.

$(x^2-2x)+(y^2-6y)=26$

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

$\frac{-2}{2} =(-1)^{2} =1$

$\frac{-6}{2} =(-3)^{2} =9$

we add the squares to both sides.

$(x^2-2x+1)+(y^2-6y+9)=26+1+9$

simplify.

$(x^2-2x+1)+(y^2-6y+9)=36$

and write the quadratics in factored form.

$(x-1)^{2} +(y-3)^{2} =36$

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.

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Which of the following pairs of lines are perpendicular? select all that apply.
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