What is the distance between $(15, 24)$ $(15,24)$ and $(42, 4)$ $(42,4)$ ?

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What is the distance between $(15, 24)$ $(15,24)$ and $(42, 4)$ $(42,4)$ ?

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Answer 1 · 2015-11-23T04:49:48

The distance between $(15, 24)$ $(15, 24)$ and $(42, 4)$ $(42, 4)$ is approximately $33.6$ $33.6$ units.

Explanation:

The formula for the distance between $2$ $2$ points is:

$d = \sqrt{({(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2})}$ $d=sqrt(((x_"2"-x_"1")^2+(y_"2"-y_"1")^2))$

$1^{s t}$ $1^(st)$ point: $(x_{1}, y_{1})$ $(x_"1",y_"1")$ = $(15, 24)$ $(15, 24)$
$2^{n d}$ $2^(nd)$ point: $(x_{2}, y_{2})$ $(x_"2",y_"2")$ = $(42, 4)$ $(42, 4)$

Substitute the points into the distance formula:

$d = \sqrt{({(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2})}$ $d=sqrt(((x_"2"-x_"1")^2+(y_"2"-y_"1")^2))$
$d = \sqrt{{((42) - (15))}^{2} + {((4) - (24))}^{2}}$ $d=sqrt(((42)-(15))^2+((4)-(24))^2)$
$d = \sqrt{{(27)}^{2} + {(- 20)}^{2}}$ $d=sqrt((27)^2+(-20)^2)$
$d = \sqrt{(729) + (400)}$ $d=sqrt((729)+(400)$
$d = \sqrt{1129}$ $d=sqrt(1129)$
$d \approx 33.6$ $d~~33.6$

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What is the distance between $(15, 24)$ $(15,24)$ and $(42, 4)$ $(42,4)$ ?

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Explanation:

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