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Find all values of k so that the given points are\sqrt {29} units apart. (-5, 5), (k, 0)

Choose the right answer

A3, 7
B7
C-3, -7
D-7 ...?

Verified Answer :

MissPhiladelphia

The solution for this would be 

29 = (k +5)^2 + (0 - 5)^2 
29 = k^2 + 10k + 25 + 25 
k^2 + 10k + 21 = 0 
(k + 3)(k + 7) = 0 

k+ 3 = 0 
k = -3 
k + 7 = 0 
k = -7 

So the correct answer to this question is c. -3, -7.

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 

Using the distance between two points, it is found that the values of k are given by:

c. -3, -7

What is the distance between two points?

Suppose that we have two points, \((x_1,y_1)\) and \((x_2,y_2)\). The distance between them is given by:

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

In this problem, the points are: (-5, 5) and (k,0).

They are a distance of \(D = \sqrt{29}\) apart, hence:

\(\sqrt{29} = \sqrt{(k + 5)^2+ (0 - 5)^2}\)

\((\sqrt{29})^2 = (\sqrt{(k + 5)^2+ (0 - 5)^2})^2\)

\((k + 5)^2 + (-5)^2 = 29\)

\(k^2 + 10k + 21 = 0\)

Which is quadratic equation with coefficients \(a = 1, b = 10, c = 21\).

Hence:

\(\Delta = b^2 - 4ac = 10^2 - 4(1)(21) = 16\)

\(x_1 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-10 - 4}{2} = -7\)

\(x_2 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-10 + 4}{2} = -3\)

Hence option c is correct.

You can learn more about distance between two points at https://brainly.com/question/18345417

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