Question 166143
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{{{Distance}}} from 2 set of points:
{{{highlight(D^2=(x[2]-x[1])^2+(y[2]-y[1])^2)}}}
Given: points ------> (x,4)(-6,-5)
{{{Distance=sqrt(162)}}}
Subst.
{{{(sqrt(162))^2=(-6-x)^2+(-5-4)^2}}}, cancels out "square root" on the left term
{{{162=x^2+12x+36+(-9)^2}}}
{{{x^2+12x+36+81-162=0}}}
{{{x^2+12x-45=0}}}, SOLVE BY PYTH. THEOREM
where----{{{system(a=1,b=12,c=-45)}}}
Then, {{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}
{{{x=(-12+-sqrt(12^2-4*1*-45))/(2*1)}}}
{{{x=(-12sqrt(144+180))/2}}}
{{{x=(-12+-sqrt(324))/2=(-12+-18)/2}}}
2 Values:
{{{x=(-12+18)/2=6/2=cross(6)3/cross(2)1}}} ---> {{{highlight(x=3)}}}, abscissa
Also, {{{x=(-12-18)/2=-30/2=cross(-30)15/cross(2)}}} --> {{{x=-15}}}
We'll use {{{highlighted}}} ---> x=3, for the graph below:
{{{drawing(300,300,-7,7,-7,7,grid(1),graph(300,300,-7,7,-7,7),red(circle(-6,-5,.20)),red(circle(3,4,.20)),blue(line(-6,-5,3,4)))}}}---> See the BLUE Line @ points (3,4) & (-6,-5) that has distance of {{{sqrt(162)}}}.
.
With points (3,4) & (-6,-5), will it equate to {{{Distance=sqrt(162)}}}? Let's see:
With our {{{Distance}}} formula, we'll find 1st {{{x[2]}}} & {{{x[1]}}}:
{{{drawing(300,300,-7,7,-7,7,grid(1),graph(300,300,-7,7,-7,7),blue(circle(-6,-5,.20)),blue(circle(3,-5,.20)),blue(line(-6,-5,3,4)),red(line(3,-5,-6,-5)))}}}
---> As you see on the graph, {{{x[2]=3}}} & {{{x[1]=-6}}}, good!
Next we find {{{y[2]}}} & {{{y[1]}}}, and as we see the graph:
{{{drawing(300,300,-7,7,-7,7,grid(1),graph(300,300,-7,7,-7,7),blue(circle(3,-5,.20)),blue(circle(3,4,.20)),green(line(3,4,3,-5)),blue(line(3,4,-6,-5)),red(line(3,-5,-6,-5)))}}}
---> As you can see, {{{y[2]=4}}} & {{{y[1]=-5}}}:
By then, going back to our formula:
{{{D^2=(x[2]-x[1])^2+(y[2]-y[1])^2}}}
{{{(sqrt(162))^2=(3-(-6))^2+(4-(-5))^2}}}
{{{162=(3+6)^2+(4+5)^2}}}
{{{162=9^2+9^2=81+81}}}
{{{162=162}}}, good!
*Our dimensions are right.
Thank you,
Jojo</pre>