Question

The distance between two points is √39 . One endpoint is (4,−3) . It is known that the y-coordinate of the other endpoint is 2.

What is the x-coordinate of this endpoint? (more than one answer)
Choices:
A). 2√7
B). 2
C). 4-√14
D). 4+√14
E). 10
F). -6

Answer 1

Answer:

C & D

Step-by-step explanation:

$\sqrt{(x-4)^2+(2+3)^2} =(\sqrt{39} )^2\\(x-4)^2+25=39\\(x-4)^2=39-25=14\\x-4=\pm \sqrt{(14)}\\x=4+\sqrt{14} \\x=4-\sqrt{14}$

Answer 2

Using the formula for the distance between two points, it is found that the x-coordinate of the endpoint can be of:

$x = 4 \pm sqrt{14}$, options C and D.

The distance between two points, $(x_0,y_0)$ and $(x_1,y_1)$, is given by:

$D = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$

In this problem, points (4, -3) and (x, 2), with a distance of $\sqrt{39}$ units. Thus:

$\sqrt{(x - 4)^2 + (2 - (-3))^2} = \sqrt{39}$

$\sqrt{(x - 4)^2 + 25} = \sqrt{39}$

$(\sqrt{(x - 4)^2 + 25})^2 = (\sqrt{39})^2$

$(x - 4)^2 + 25 = 39$

$x^2 - 8x + 16 + 25 - 39 = 0$

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

Quadratic equation with coefficients $a = 1, b = -8, c = 2$, thus:

$\Delta = (-8)^{2} - 4(1)(2) = 56$

$x_{1} = \frac{-(-8) + \sqrt{56}}{2} = \frac{8 + 2\sqrt{14}}{2} = 4 + \sqrt{14}$

$x_{2} = \frac{-(-8) - \sqrt{56}}{2} = \frac{8 - 2\sqrt{14}}{2} = 4 - \sqrt{14}$  

Options C and D.

A similar problem is given at https://brainly.com/question/14849255