Find all the points having an x-coordinate of 4 whose distance from the point (-2,-1) is 10

1 answer:

4 0

$A(x_1;\ y_1);\ B(x_2;\ y_2)\\\\the\ distance\ between\ A\ and\ B:\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\A(-2;-1);\ B(4;\ y);\ d=10\\\\subtitute\\\\\sqrt{(4-(-2))^2+(y-(-1))^2}=10\\\\\sqrt{(4+2)^2+(y+1)^2}=10\\\\\sqrt{6^2+(y+1)^2}=10\\\\\sqrt{36+(y+1)^2}=10\ \ \ |square\ both\ sides\\\\36+(y+1)^2=10^2\\\\36+(y+1)^2=100\ \ \ |subtract\ 36\ from\ both\ sides\\\\(y+1)^2=64\iff y+1=-\sqrt{64}\ or\ y+1=\sqrt{64}\\\\y+1=-8\ or\ y+1=8\ \ \ \ |subtract\ 1\ from\ both\ sides\\\\y=-9\ or\ y=7$

$Answer:{\boxed{(4;-9)\ or\ (4;\ 7)}$

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How do you find x? and how do you round to the nearest tenth?

expeople1 [14]

Hey there!

In order to solve this problem, we can use the Pythagorean Theorem, which is $c^2 = a^2 + b^2$ . Since the Pythagorean Theorem would find only half of the length of x, we'll need to double it in the end.

First, let's plug in what we know to the Pythagorean Theorem: $2.1^2 = 1.4^2 + b^2$

Next, we'll simplify to that $b^2$ is by itself:

$b^2 = 2.45$

Then, we'll find the square root of 2.45, multiply that by 2 and round to the nearest tenth:

$2(\sqrt{2.45} ) = 3.1$

Therefore, x = 3.1 ft.

How to round to the nearest tenth:

In a decimal, the tenth is the first number after the decimal point. To round to the tenth, if the number in the hundredth place is less than 5, you keep the number as it is, but if the number in the hundredth place is more than 5, you increase the number in the tenth place by 1.

Hope this helps!! :)

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irina1246 [14]

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