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In order to prove this i am assuming that s is the semiperimeter of the triangle #

# s= (a+b+c)/2 #

ok so first let

#s-a = x#
#s-b =y#
#s-c= z#

on solving for #a,b,c# we get

#a=y+z#
#b=x+z#
#c=x+y#

Now , #(abc)/8 = ((x+y)(y+z)(z+x))/8#

and we know that
# (x+y)/2 >= sqrt(xy) #
As Arithmetic mean of #x,y # is greater than their geometric mean

so ,
# ((y+z)(x+y)(z+x))/8 >=( (2 sqrt(x y) ) (2 sqrt(yz) )(2 sqrt(zx)))/8 = xyz #
# and xyz=(s-a)(s-b)(s-c) #

Hence

# (abc)/8 >= (s-a)(s-b)(s-c) #

Answer:

#1 + 5 X + 10 Y + 5Y^2=0#

Explanation:

Rotating the coordinate system around the origin by an angle #theta# so that

#x = X cos(theta)-Ysin(theta)#
#y = X sin(theta)+Y cos(theta)#

and substituting we have in the conic now in the coordinates #X,Y#

the coefficient of #XY# is #3sin(2theta)-4cos(2theta)# so choosing #theta# such that #3sin(2theta)-4cos(2theta)=0# or #tan(2theta)=4/3# or #theta = 1/2 arctan(4/3)# we will have

#1 + 5 X + 10 Y + 5Y^2=0# which is a parabola.

Answer:

see explanation.

Explanation:

enter image source here

Given #AB=AC, and angleCAB=20^@#,
#=> angleABC=angleACB=(180-20)/2=80^@#
Given #angleBDC=30^@, => angleBCD=70^@#
#=> angleACD=80-70=10^@#

In #DeltaADC, (AD)/sin10=(CD)/sin20#
#=> AD=CD*sin10/sin20=0.50771xxCD#

In #DeltaBDC, (BC)/sin30=(CD)/sin80#
#=> BC=CD*sin30/sin80=0.50771xxCD#

Hence, #AD=BC#

proof of #sin10/sin20=sin30/sin80#

cross multiplying :
#sin10sin80=sin20sin30#
#LHS = sin10cos(90-80)=sin10cos10#
#=1/2*sin20# ....(as #sinxcosx=1/2sin2x#)
#RHS=sin20sin30=sin20*1/2=1/2*sin20#

#LHS=RHS# (proved)

Answer:

Longest diagonal: #color(green)(sqrt(657+90sqrt(23)))~~color(green)(33.0)#

Explanation:

enter image source here
Since the area is given as #9# (sq.units), if we use a side with length #9# as the base,
the height of the parallelogram must be #color(brown)1#

Taking the extension of the base to a point perpendicularly below the opposite vertex of the parallelogram and denoting the length of this extension as #color(blue)a#,
by the Pythagorean Theorem
#color(white)("XXX")color(blue)a=sqrt(24^2-1^2) =5sqrt(23)#

Denoting the longest diagonal as #color(red)b#
and re-applying the Pythagorean Theorem using the base plus its extension (#9+color(red)a#) and the height #color(brown)1#
#color(white)("XXX")color(red)b=sqrt((9+color(blue)(5sqrt(23)))^2+color(brown)1^2)=sqrt(81+90sqrt(23)+575+1)=sqrt(657+90sqrt(23))#

We can use a calculator to derive the approximation
#color(white)("XXX")sqrt(657+90sqrt(23))~~32.99431522#

Answer:

Any point on the line #color(purple)(2y-3x=23)#
or
if the given line #color(red)(4y-6x=8)# is to be a perpendicular bisector then #color(brown)(""(-23/15,46/5))#

Explanation:

Apology:
Even omitting some details, this explanation is quite long.

Consider the vertical line #x=7# which passes through the given point #(7,3)#
This vertical line will intersect #color(red)(4y-6x=8)# at #(7,25/2)#, a point which is #25/2-3=19/2# above the given point #color(green)(""(7,3))#
A point twice as far above #color(green)(""(7,3))# would be at #(7,3+2xx19/2)=(7,22)#
That is the line segment from #color(green)(""(7,3))# to #(7,22)# is bisected by #color(red)("4y-6x=8)#
enter image source here

Furthermore, as we can see from similar triangles any line parallel to #color(red)(4y-6x=8)# and through #(7,22)# gives an infinite collection of points, any one of which would serve as an endpoint with #color(green)(""(7,3))# for a line segment bisected by #color(red)(4y-6x=8)#

enter image source here
#color(red)(4y-6x=8)# has a slope of #3/2#
so any line parallel to it must also have a slope of #3/2#
and
if such a line passes through #(7,22)# then we can write its equation using the slope-point form:
#color(white)("XXX")y-22=3/2(x-7)#
or simplified as
#color(white)("XXX")color(purple)(2y-3x=23)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

It is possible, since this question was asked under the heading "Perpendicular Bisectors" that it was intended that #color(red)(4y-6x=8)# should be the perpendicular bisector of a derived line segment.

In this case the perpendicular line to #color(red)(4y-6x=8)# passing through #color(green)(""(7,3))#
would have a slope of #-2/3# (the negative inverse of #color(red)(4y-6x=8)# and (again, working through the slope-point form)
an equation of #color(brown)(3y+2x=23)#
enter image source here
The system of equations:
#color(purple)(2y-3x=23)#
#color(brown)(3y+2x=23)#
can be solved for the point of intersection: #color(brown)(""(-23/15,46/5))#

Answer:

#x=5/4#
minimum value of #(PA+PB)=sqrt41#

Explanation:

enter image source here

See Fig 1.
Reflect #B# over #"X-axis"#,
#=> B(5,3) -> B'(5,-3)#
#=> PB=PB' "( reflection symmetry)"#
As long as #P# moves along the X-axis, #PB=PB'#

See Fig 2.
When #A,P, and B'# lie in a straight line, minimum value of #(PA+PB')#can be achieved.
When #A,P and B'# are collinear, #AP and PB'# have the same slope #m#
#=> m=(0-1)/(x-0)=(-3-0)/(5-x)#
#=> -3x=-5+x#
#=> x=5/4#

Hence, min. value of #(PA+PB)=PA+PB'=AB'#
#=sqrt((5-0)^2+(-3-1)^2)#
#= sqrt(25+16)#
#=sqrt41#

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