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defon
3 years ago
8

The equation of four lines are given. Identify which lines are perpendicular.

Mathematics
2 answers:
PilotLPTM [1.2K]3 years ago
7 0

Answer: A.) Lines 2 and 4 are perpendicular

Step-by-step explanation:

Two lines are said to be perpendicular if the product of their slope gives - 1 . That is , if m_{1} is the slope of the first line and  m_{2} is the slope of the second line , if the two lines are perpendicular , then :

m_{1} m_{2} = -1  , which can also be written as ;

m_{1} = -1 /  m_{2}.

The equation of line in slope intercept form is given as :

y = mx + c , where m is the slope and c is the y - intercept.

Line one : y = - 1 , comparing with the equation of line in slope intercept form , it means line one has no slope

Line two : y = -1/2x - 2 , it has a slope of - 1/2

Line 3 : x = -4 , has no slope

Line 4 : y + 3 = 2 ( x - 2 )

Writing this in slope intercept form , we will make y the subject of the formula.

y + 3 = 2x - 4

y  = 2x -4 - 3

y = 2x - 7

This means that the slope of line 4 is 2.

Comparing line two and line four to check if they are perpendicular

- 1/2 x 2 = - 1 , surely , it obeys the perpendicularity rule , this means that line 2 and 4 are perpendicular

Bumek [7]3 years ago
6 0
Lines 2 and four r perpendicular I think
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The equation of the circle is (x+3)^{2}+(y-3)^{2}=37

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<u>Center:</u>

The center of the circle can be determined using the midpoint formula,

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The radius of the circle can be determined using the distance formula,

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Substituting the center (-3,3) and the endpoint (-4,9), we get,

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Substituting the values, we have,

(x+3)^{2}+(y-3)^{2}=(\sqrt{37})^{2}

(x+3)^{2}+(y-3)^{2}=37

Thus, the equation of the circle is (x+3)^{2}+(y-3)^{2}=37

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