Question

find all the following value for x for which distance between point A(x,-1) and B(5,3) is 5units

Answer 1

Answer:

x = 2, x = 8

Step-by-step explanation:

Calculate the distance using the distance formula and equate to 5

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = A (x, - 1) and (x₂, y₂ ) = B(5, 3)

d = $\sqrt{(5-x)^2+(3+1)^2}$

   = $\sqrt{(5-x)^2+4^2}$

   = $\sqrt{(5-x)^2+16}$ , thus

$\sqrt{(5-x)^2+16}$ = 5 ( square both sides )

(5 - x)² + 16 = 25 ( subtract 16 from both sides )

(5 - x)² = 9 ( take the square root of both sides )

5 - x = ± $\sqrt{9}$ = ± 3 ( subtract 5 from both sides )

- x = - 5 ± 3 , thus

- x = - 5 + 3 = - 2 ( multiply both sides by - 1 )

x = 2

or

x = - 5 - 3 = - 8 ( multiply both sides by - 1 )

x = 8

Answer 2

Answer:

Step-by-step explanation:

The formula to find the distance between 2 points in the coordinate plane is

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

We have our distance, and we also have all the coordinates but the first x.  Fillling in with what we have gives us this:

$5=\sqrt{(5-x_1)^2+(3-(-1))^2}$

which simplifies to

$5=\sqrt{(5-x_1)^2+(4)^2}$ . Expanding that binomial gives us:

$5=\sqrt{25-10x+x^2+16}$ .  Combining like terms gives us:

$5=\sqrt{x^2-10x+41}$ which is the same thing as above, only in standard form for polynomials. Now we need to get that x out from under that square root sign. We do that by squaring both sides to get:

$25=x^2-10x+41$ . Now we have to factor to solve for x. We'll put everything on one side of the equals sign, set the polynomial equal to 0, then factor.

$0=x^2-10x+16$ is our polynomial now. a = 1, b = -10, c = 16. The product ac is 1 * 16 which is 16. Some combination of the factors of 16 will result in a -10. So we need the factors of 16.

16: {1, 16}, {2, 8}, {4, 4}

The only combination of those factors that will result in a -10 is the second pair, {2, 8}. If we add 2 and 8 we get 10, but in order for our 10 to be negative, both 2 and 8 have to be negative. So we rewrite the polynomial in terms of -2 and -8:

$0=x^2-8x-2x+16$

Now we can factor by grouping. Group the first 2 terms together and the second 2 terms together without moving any of their positions:

$0=(x^2-8x)-(2x+16)$

From each set of parenthesis we will now factor out what's common.  x is common in the first set of ( ), and 2 is common in the second set of ( ):

$0=x(x-8)-2(x-8)$

What's common now is the binomial (x - 8). So we'll factor that out now:

$0=(x-8)(x-2)$

By the Zero Product Property, either

x - 8 = 0 or x - 2 = 0.

If x - 8 = 0, then x = 8. If x - 2 = 0, then x = 2.

It looks like we have 2 solutions. Let's try them both and see if, when we stick an 8 and then a 2 into our distance formula, the distance is 5:

$d=\sqrt{(5-8)^2+(4^2)}$ is

$d=\sqrt{(-3)^2+(4)^2}$ is

$d=\sqrt{9+16}$ is

$d=\sqrt{25}$ which does in fact equal 5. Now let's try the 2:

$d=\sqrt{(5-2)^2+(4)^2}$ which is

$d=\sqrt{(3)^2+(4)^2}$ is

$d=\sqrt{9+16}$ is

$d=\sqrt{25}$ which also comes out to equal 5.

So the 2 values of x which will work here are 2 and 8.