Question

Find k so that the distance from (–3, –1) to (k, 5) is 10 units.

Answer 1

Answer:

k = - 11, 5

Step-by-step explanation:

By distance formula:

$\sqrt{ \{ {k - ( - 3) \}}^{2} + \{5 - ( - 1) \}^{2} } = 10 \\ \\ \sqrt{ {(k + 3)}^{2} + {(5 + 1)}^{2} } = 10 \\ \\ \sqrt{ {(k + 3)}^{2} + {(6)}^{2} } = 10 \\ \\ \sqrt{ {(k + 3)}^{2} + 36 } = 10 \\ \\ (k + 3)^{2} + 36 = {10}^{2} \\ ..(squaring \: both \: sides) \\ {k}^{2} + 6k + 9 + 36 = 100 \\ {k}^{2} + 6k + 45 - 100 = 0 \\ {k}^{2} + 6k - 55 = 0 \\ {k}^{2} + 11k - 5k - 55 = 0 \\ k(k + 11) - 5(k + 11) = 0 \\ (k + 11)(k - 5) = 0 \\ k + 11 = 0 \: or \: k - 5 = 0 \\ k = - 11 \: or \: k = 5 \\ \\ \huge \red { \boxed{k = \{ - 11, \: 5\}}}$

Answer 2

There are two solutions for $k$: 5 and -11.

In this case, we must use the Length Equation for Line Segment to solve for $k$, whose formula is described below:

$l = \sqrt{(x_{B}-x_{A})^{2}+ (y_{B}-y_{A})^{2}}$ (1)

Where:

$(x_{A}, y_{A})$ - Coordinates of the initial endpoint.

$(x_{B}, y_{B})$ - Coordinates of the final endpoint.

$l$ - Length of the line segment.

If we know that $(x_{A}, y_{A}) = (-3, -1)$, $(x_{B}, y_{B}) = (k, 5)$ and $l = 10$, then the equation is expanded into this form:

$(k+3)^{2}+[5-(-1)]^{2} = 100$

$k^{2}+6\cdot k + 9 + 36 = 100$

$k^{2} + 6\cdot k -55 = 0$

The roots of this Second Order Polynomial are determined by Quadratic Formula:

$k = 5$ or $k = -11$.

There are two solutions for $k$: 5 and -11.

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