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8. GEOMETRIC LOCI I |
| Block :Geometry | |
| 8.1. BISECTOR OF A SEGMENT | ||
A set of points which satisfy a certain
property is called a geometric locus |
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| Check in
the figure: 1.- That if you move the point A and/or the point B with the mouse, or you change their coordinates using the buttons at the bottom of the figure, you will see that the X points of the bisector, remain permanently the same distance from A as from B. 2.-That the bisector cuts the segment AB at its mid point M, and is perpendicular to it. 3.-That the vectors MX and AB are perpendicular, or their scalar product is zero.
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| 4- Check and copy
into your workbook the process for finding the equation
of the bisector of the segment which appears at the
beginning. A(-3,4), B(1,0) and X(x,y),
In the figure you can also verify that the line represented is y = x + 3, it has a gradient of 1 and cuts the y-axis at 3. 5.- Find the equation of the bisector of the segment between the points A(0,5) and B(4,3). 6.- Verify your calculations in the previous figure and in your exercise book write down the gradient of the line you have obtained, and its y- intercept. 7.- Calculate the scalar product of the vectors MX·AB, if you have completed everything correctly this will be equal to zero, then they are perpendicular. 8.- Calculate the lengths AX and BX, you can check this in the figure. 9.- Calculate the coordinates of the mid point M of the segment AB, and check them in the figure. |
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| 8.2. BISECTOR OF AN ANGLE | ||
| The bisector of an angle of sides
r1, r2
is
the geometric locus of the points X
which
are equidistant from r1 and from r2: dist(X,r1)
= dist(X,r2). The parameter a which appears in the button at the bottom of the figure, corresponds with the x value of the point X. If you change the value of a, you will see the different points X of the red line, which is the bisector of the angle formed by the other two blue lines. |
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1.-Verify in the figure that for
any point X on the bisector, the distance
of X from the two lines which form the
angle is the same. Move the point X and you
can prove this.
2.-Check
that in our case we are left with: But if |A|=|B| two cases could occur: A=B or A=-B therefore it could be that: 11x+2y-20=2x+11y+7 or that 11x+2y-20=-2x-11y-7. |
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| 3.-Observe that the second equality
results in: x+y-1=0, which is the bisector (red
line) as shown. This bisects the angle formed by
the two blue lines, which we have shown in yellow. 4.- Observe that the first equality results in: x-y-3=0, which is the bisector (grey line) as shown. This bisects the other angle formed by the two blue lines, that is the supplementary angle of the one shown in yellow. The two bisectors cut eachother at the same point as the lines and are perpendicular to eachother. |
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| 8.2.1. BISECTOR OF THE ANGLE BETWEEN TWO LINES | ||
| In the figure we have two lines, r1 and r2, which cut at a
point forming an angle for which we want to find the
bisector. The equations of r1 and r2 can be seen in
the figure, as can the distances from a point X on the red line
to r1
and
r2. The parameter m which appears in this figure is the gradient of the red line. |
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1.-In your exercise book write down the equations of r1 and r2 and find the equation of the bisector. Find its gradient, m. 2.- Introduce the value of m obtained in the figure, and show that it complies with: dist(X,r1) = dist(X,r2)
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| Ángela Núñez Castaín | ||
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| Ministry of Education , Social Afairs and Sport. Year 2001 | ||
Except where otherwise noted, this work is licensed under a Creative Common License
and completing the process
we are left with the equation of the line y = x + 3,
which is the bisector of the segment AB. 