Question 1203373


given:

the distance between ({{{ x}}},{{{  y}}} ) and the {{{ x}}}-axis is {{{ 6}}} units.

The nearest point to ({{{ x}}},{{{  y}}}) in the {{{ x}}}-axis is the point ({{{ x}}}, {{{ 0}}}) so we have:


 {{{ 6 = sqrt( (x - x)^2 + (y - 0)^2) }}}

{{{sqrt(y^2)=6}}}

{{{abs(y) = 6|||


so {{{ y}}} can be {{{ 6}}} or {{{ -6}}}


So we know that {{{x}}} and {{{y}}} are negative, our point will be in Q III.


so, {{{ y=-6}}} and now we can write our point as ({{{ -x}}}, {{{ -6}}})


 the distance between our point and ({{{ 5}}},{{{ 7}}}) is {{{ 15}}} units:


 {{{ sqrt( (x - 5)^2 + (y - 7)^2) = 15}}}

{{{ x = 5 - 2 sqrt(14)}}}

or

{{{ x = 5 + 2 sqrt(14)}}}


And we know that the distance from the origin, ({{{ 0}}}, {{{ 0}}}) is {{{ sqrt(n) }}}:

 {{{ sqrt(n )= sqrt(x^2 + y^2)}}}

{{{ n =x^2 + y^2}}}

{{{ n = (5 - 2sqrt(14))^2 + (-6)^2}}}

{{{ n = 42.2}}} (approximately)