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waigaK

Answered question

2021-08-13

Find all of the points on the x-axis which are 4 units from the point (-1, 1).

Answer & Explanation

Corben Pittman

Skilled2021-08-14Added 83 answers

$Let {({x}_{{1}}, {y}_{{1}})} = {({x}, {0})} be the point on the x-axis. We also let {({x}_{{2}}, {y}_{{2}})} = {(- {1}, {1})} . Use the distance formula to find x where d=4:$
${d} = \sqrt{{{({x}_{{2}} - {x}_{{1}})}^{2} + {({y}_{{2}} - {y}_{{1}})}^{2}}}$
${4} = \sqrt{{{(- {1} - {x})}^{2} + {({1} - {0})}^{2}}}$
${4} = \sqrt{{{1} + {2} {x} + {x}}} + {1}^{2}$
${4} = {\sqrt{{y}}}^{2} + {6} {y} + {45}$
$Square both sides:$
${16} = {x}^{2} + {2} {x} + {2}$
$Subtract 36 from both sides:$
${0} = {x}^{2} + {2} {x} - {14}$
$Solve for x using quadratic formula:$
${x} = {\frac{{- {b} \pm \sqrt{{{b}^{2} - {4} {a} {c}}}}}{{{2} {a}}}}$
$where a=1,b=2, and c=-14$
${x} = {{- {2} \pm \sqrt{{{({2})}^{}}$

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