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Anastasy [175]
3 years ago
7

How do I find the distance between (-2,5) and (22,12)

Mathematics
2 answers:
Andrew [12]3 years ago
8 0

Answer:

The distance between the two points is 25.

Step-by-step explanation:

In order to find the distance between two points in the coordinate plane, you must first make a right triangle and use the Pythagorean Theorem to solve for the missing side, which is the distance between the points.  Sometimes, it is best to graph the points to get a visual of the triangle, however, you can also find the lengths of the legs ('a' and 'b') by finding the distance between your x and y values.  In this case, there are 24 points between the x-values and 7 points between the y-values.  These represent our legs in the Pythagorean Theorem:  a² + b² = c².  Filling in the values for 'a' and 'b' gives us: 24² + 7² = c² or 576 + 49 = 625.  In order to find c, we need to take the √625, which is 25.  So, the distance between the points given is 25.

jok3333 [9.3K]3 years ago
6 0

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{22}~,~\stackrel{y_2}{12})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[22-(-2)]^2+[12-5]^2}\implies d=\sqrt{(22+2)^2+(12-5)^2} \\\\\\ d=\sqrt{24^2+7^2}\implies d=\sqrt{625}\implies d=25

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Answer:

(2,-5) and (2,1).

Step-by-step explanation:

We need to find the find all the points having an x coordinate of 2 whose distance from the point (-2,-4) is 5.

A circle with center (-2,-4) and radius 5 represents all the points whose distance from the point (-2,-4) is 5.

Standard form of a circle is

(x-h)^2+(y-k)^2=r^2

where, (h,k) is center and r is the radius.

(x-(-2))^2+(y-(-4))^2=(5)^2

(x+2)^2+(y+2)^2=25

Now, put x=2.

(2+2)^2+(y+2)^2=25

(4)^2+(y+2)^2=25

(y+2)^2=25-16

(y+2)^2=9

Taking square root on both sides.

y+2=\pm\sqrt{9}

y+2=\pm3

y+2=-3\text{ or }y+2=3

y=-5\text{ or }y=1

So, the y-coordinates of the points are -5 and 1.

Therefore, the required points are (2,-5) and (2,1).

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