Simple math! please...easy points?

Question

1. The Locus of points equidistant from two fixed points is
one circle?
one straight line?
two circles?
two straight lines?

2. What is the equation of the locus of points equidistant from the points (4,2) and (-2,2)
y=1?
x=1?
x=-1?

3.
there are two boys in a lake. A scuba diver swims so that he is always equidistant from both boys. Decribe his path.

4. true or false: The locus of points equidistant from the points (-1,-3) and (-1,4) is a line whose equation is y=1/2
true?
false?

Answer 1

1.) Whenever you want to be equidistant between two points or two parallel lines, then your locus will always be one line down the center. When it is two points, you can connect them with a line and the locus of points equidistant between them will be a line that is perpendicular to the connecting line.

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2.) This is the same idea as the first question, but they give you specific points on the coordinate grid. If you plotted those two points on graph paper, you would notice that they have the same y (same vertical - up and down - position) but different x values (horizontal - side to side - locations). The locus of points equidistant between them is going to be a vertical line the is halfway between the two x values given. Since the x's are 4 and -2, they are 6 spaces away from each other (4 - -2 = 4 + +2 = 6). Split that in halfway and then subtract from 4 to find that your line is at 1, and since it is a vertical line, it is x = 1.

3.) Same as the first two.

4.) This answer is true. This time, since the x's match, find the difference between the y's and then subtract from the larger y. 4 - -3 = 7 and 7/2 = 3.5. 4 - 3.5 = 1/2. Since the two points will be lined up vertically (because they have the same horizontal value), the answer line will be horizontal and have the equation y = 1/2, which is the line where all points on the line have the coordinates (x, 1/2) such as (-1, 1/2) which is the midpoint between a line connecting the points (-1,-3) and (-1,4).

Answer 2

1. A straight line: the height of an equilateral triangel having the line between the fixed points for base will always bisect it, and at a point on a straight line there is only one line perpendicular.

2. If (x-4)^2 + (y -2)^2 = (x + 2)^2 + (y - 2)^2
then subtracting (y - 2)^2 from each side:

(x - 4)^2 = (x + 2)^2

x^2 - 8x + 16 = x^2 + 4x + 4 // -x^2

-8x + 16 = 4x + 4 // +8x - 4

12 = 12x

x=1

3. The straight line from question 1 with the two boys for fixed points.

4. True:

(x + 1)^2 + (y + 3)^2 = (x + 1)^2 + (y - 4)^2

(y + 3)^2 = (y - 4)^2

y^2 + 6y + 9 = y^2 - 8y + 16

14y = 7

y=1/2

Answer 3

ok, you won't be able to be intense, it is undemanding! ok, unique $30. complete of scholars: 3; each and each and every make contributions with 10 funds. fee of the watch: $25.00 Paid to the shopkeeper: $25 Trainee's earnings: $2 ok, 25+2= 27 Plus, 3 (each and each and every student gained a million greenback decrease back) answer 27+3=30 Do the issue backwards, it provides the reply. Hesabi turned right into a exceedingly wise guy huh?

Answer 4

1. One straight line

2. x = 1

3. Plane

4. True

Answer 5

If it is so simple then do your own homework.

Answer 6

ir it is so easy why cant you do it