Questions on Geometry: Length, distance, coordinates, metric length answered by real tutors!

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Question 166143: the point(x,4) is a square root of 162 units from (-6,-5). What is the value of abscissa x?: the point(x,4) is a square root of 162 units from (-6,-5). What is the value of abscissa x?
Answer by jojo14344(1023) About Me  (Show Source):
You can put this solution on YOUR website!

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