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In the figure above, what is the slope of line segment BC?
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24 Nov 2024, 23:10
24 Nov 2024, 23:10
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In the figure above, what is the slope of line segment BC?
A. \( - \dfrac{1}{\sqrt{3}}\)
B. \(- \dfrac{1}{\sqrt{2}}\)
C. \(-1\)
D. \(- \sqrt{2}\)
E. \(- \sqrt{3}\)
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In the figure above, what is the slope of line segment BC?
[#permalink]
28 Nov 2024, 02:49
28 Nov 2024, 02:49
1
to find the slope of the line BC we need the coordinates of C
and it is clear from the given data that the given triangle is an equilateral triangle
coordinate A is (0,0) and it is clear from the figure that the coordinate B is (5,0)
when you draw a straight line perpendicular to the line AB which is also passing through the middle of AB, you can see it that passes through the coordinate C, since it is an equilateral triangle
from this we can conclude that the x coordinate of C is 5/2 = 2.5
now the area of any triangle is 0.5 * B * H where B - base length of a triangle which is 5 from the given data and H - height of the triangle which is unknown and the y coordinate of C
area of an equilateral triangle is √3/4 × a^2 and a is the side of an equilateral triangle which is 5 from the given data
equating these 2 equations and substituting the values, we get 0.5 * 5 * H = √3/4 x 5^2
solving for H we get H = 4.33
therefore the coordinates of C are (2.5 , 4.33)
now the formula for the slope is (y2 -y1)/(x2 -x1)
slope of BC = (4.33 - 0)/(2.5-5) = 1.732 which is √3
option E is the right answer
please give me kudos if you find this solution helpful
and it is clear from the given data that the given triangle is an equilateral triangle
coordinate A is (0,0) and it is clear from the figure that the coordinate B is (5,0)
when you draw a straight line perpendicular to the line AB which is also passing through the middle of AB, you can see it that passes through the coordinate C, since it is an equilateral triangle
from this we can conclude that the x coordinate of C is 5/2 = 2.5
now the area of any triangle is 0.5 * B * H where B - base length of a triangle which is 5 from the given data and H - height of the triangle which is unknown and the y coordinate of C
area of an equilateral triangle is √3/4 × a^2 and a is the side of an equilateral triangle which is 5 from the given data
equating these 2 equations and substituting the values, we get 0.5 * 5 * H = √3/4 x 5^2
solving for H we get H = 4.33
therefore the coordinates of C are (2.5 , 4.33)
now the formula for the slope is (y2 -y1)/(x2 -x1)
slope of BC = (4.33 - 0)/(2.5-5) = 1.732 which is √3
option E is the right answer
please give me kudos if you find this solution helpful
gmatclubot
In the figure above, what is the slope of line segment BC? [#permalink]
28 Nov 2024, 02:49
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