Maxim Rosales

## Answered question

2022-12-23

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

### Answer & Explanation

Mikayla Cox

Beginner2022-12-24Added 15 answers

The following formula is used to determine the separation between two points:
$d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
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Substituting these values and 5 for d, we can solve for ${y}_{1}$:
$5=\sqrt{{\left(-1-3\right)}^{2}+{\left(-4-{y}_{1}\right)}^{2}}$
First, square both sides of the equation:
${5}^{2}={\left(\sqrt{{\left(-1-3\right)}^{2}+{\left(-4-{y}_{1}\right)}^{2}}\right)}^{2}$
$25={\left(-1-3\right)}^{2}+{\left(-4-{y}_{1}\right)}^{2}$
$25={\left(-4\right)}^{2}+{\left(-4-{y}_{1}\right)}^{2}$
$25=16+{\left(-4-{y}_{1}\right)}^{2}$
$25=16+\left({\left(-4\right)}^{2}-\left(2\cdot -4\cdot {y}_{1}\right)+{{y}_{1}}^{2}\right)$
$25=16+\left(16-\left(-8{y}_{1}\right)+{{y}_{1}}^{2}\right)$
$25=16+16+8{y}_{1}+{{y}_{1}}^{2}$
$25=32+8{y}_{1}+{{y}_{1}}^{2}$
$25={{y}_{1}}^{2}+8{y}_{1}+32$
$25-25={{y}_{1}}^{2}+8{y}_{1}+32-25$
$0={{y}_{1}}^{2}+8{y}_{1}+7$
${{y}_{1}}^{2}+8{y}_{1}+7=0$
This can be considered as:
$\left({y}_{1}+7\right)\left({y}_{1}+1\right)=0$
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:
$\left(3,-7\right)$ and $\left(3,-1\right)$

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