Maxim Rosales

2022-12-23

How to find all points having an x coordinate of 3 whose distance from the point (-1, -4) is 5?

Mikayla Cox

Beginner2022-12-24Added 15 answers

The following formula is used to determine the separation between two points:

$d=\sqrt{{({{x}_{2}}-{{x}_{1}})}^{2}+{({{y}_{2}}-{{y}_{1}})}^{2}}$

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Substituting these values and 5 for d, we can solve for${y}_{1}$ :

$5=\sqrt{{({-1}-{3})}^{2}+{({-4}-{{y}_{1}})}^{2}}$

First, square both sides of the equation:

$5}^{2}={\left(\sqrt{{({-1}-{3})}^{2}+{({-4}-{{y}_{1}})}^{2}}\right)}^{2$

$25={({-1}-{3})}^{2}+{({-4}-{{y}_{1}})}^{2}$

$25={(-4)}^{2}+{({-4}-{{y}_{1}})}^{2}$

$25=16+{({-4}-{{y}_{1}})}^{2}$

$25=16+({\left({-4}\right)}^{2}-(2\cdot {-4}\cdot {{y}_{1}})+{{{y}_{1}}}^{2})$

$25=16+(16-(-8{{y}_{1}})+{{{y}_{1}}}^{2})$

$25=16+16+8{{y}_{1}}+{{{y}_{1}}}^{2}$

$25=32+8{{y}_{1}}+{{{y}_{1}}}^{2}$

$25={{{y}_{1}}}^{2}+8{{y}_{1}}+32$

$25-{25}={{{y}_{1}}}^{2}+8{{y}_{1}}+32-{25}$

$0={{{y}_{1}}}^{2}+8{{y}_{1}}+7$

${{{y}_{1}}}^{2}+8{{y}_{1}}+7=0$

This can be considered as:

$({{y}_{1}}+7)({{y}_{1}}+1)=0$

Now that we have found the solutions, we can solve each term on the left for 0:

Solution 1:

${{y}_{1}}+7=0$

${y}_{1}}+7-{7}=0-{7$

${{y}_{1}}+0=-7$

${{y}_{1}}=-7$

Solution 2:

${{y}_{1}}+1=0$

${y}_{1}}+1-{1}=0-{1$

${{y}_{1}}+0=-1$

${{y}_{1}}=-1$

The two points satisfying this problem are:

$(3,-7)$ and $(3,-1)$

We can let

We can let

Substituting these values and 5 for d, we can solve for

First, square both sides of the equation:

This can be considered as:

Now that we have found the solutions, we can solve each term on the left for 0:

Solution 1:

Solution 2:

The two points satisfying this problem are:

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