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Dimas [21]
3 years ago
8

Is it possible for a line segment to have more than one bisector?

Mathematics
1 answer:
ki77a [65]3 years ago
8 0

Yes, it is possible to have more than one bisector in a line segment.

Bisector is a line that divides a line or an angle in to two equivalent parts. There are two types of Bisectors based on what geometrical shape it bisects.

<span> Bisector of a Line Angle Bisector </span>

 <span>In general 'to bisect' something means to cut it into two equal parts. The bisector  is the one that doing the cutting process.</span>

 

With a line bisector, we cut a line segment into two equal parts with another line - the bisector. Just imagine the line PQ is being cut into two equal lengths (PF and FQ) by the bisector line AB.

 

Whenever AB intersects at a right angle, it is called the "perpendicular bisector" of PQ. If it crosses at any other angle it is simply called a bisector. Drag the points A or B and see both types.

 

<span>For obvious reasons, the point F is called the midpoint of the line PQ,</span>

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A store bought an oil painting and marked it up 25% from the original cost of $1,000. Later
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1275

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The fromula of a distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

We have the points J(-5, 6), K(3, 4) and L(-2, 1). Substitute:

|JK|=\sqrt{(3-(-5))^2+(4-6)^2}=\sqrt{8^2+(-2)^2}=\sqrt{64+4}=\sqrt{68}\\\\=\sqrt{4\cdot17}=\sqrt4\cdot\sqrt{17}=2\sqrt{17}\\\\|JL|=\sqrt{(-2-(-5))^2+(1-6)^2}=\sqrt{3^2+(-4)^2}=\sqrt{9+25}=\sqrt{34}\\\\|KL|=\sqrt{(-2-3)^2+(1-4)^2}=\sqrt{(-5)^2+(-3)^2}=\sqrt{25+9}=\sqrt{34}

The perimeter of ΔJKL:

P_{\Delta JKL}=|JK|+|JL|+|KL|\\\\P_{\Delta JKL}=2\sqrt{17}+\sqrt{34}+\sqrt{34}=2\sqrt{17}+2\sqrt{34}=2(\sqrt{17}+\sqrt{34})

The area of ΔJKL:

A_{\Delta JKL}=\dfrac{1}{2}|JL||KL|\\\\A_{\Delta JKL}=\dfrac{1}{2}\cdot\sqrt{34}\cdot\sqrt{34}=\dfrac{1}{2}(\sqrt{34})^2=\dfrac{1}{2}(34)=17

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3 years ago
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