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Answer:

Coordinates of #B# are #(-1.6,-1.2)#

Explanation:

Let the coordinates of B be #B(p,q)#

and coordinates of #A# being #(-1,-3)# and slope of #BA# is #(q+3)/(p+1)# and mid point of #BA# is #((p-1)/2,(q-3)/2)#. Note that this will be point of intersection of #BA# with its perpendicular bisector.

as #BA# is perpendicular to its perpendicular bisector #x-3y-5=0# and hence product of their slope should be #-1#

as #x-3y-5=0# in slope-intercept form is #y=1/3x-5/3# (1), its slope is #1/3# and hence slope of #BA# is #(-1)/(1/3)=-3#

and if #(x,y)# is a point on #BA# then #(y+3)/(x+1)=-3# i.e. #-3x-3=y+3# i.e. #3x+y+6=0# (2)

and its intersection with #x-3y-5-0# will be given by solving these equations. Putting value from (1) into (B), we have

#3x+1/3x-5/3+6=0# or #10/3x=-13/3# and #x=-13/10=-1.3#

and #y=-13/10xx1/3-5/3=-63/30=-2.1#

Hence, coordinates of point of intersection are #(-1.3,-2.1)#

and hence #(p-1)/2=-1.3# i.e. #p=-1.6# and

#(q-3)/2=-2.1# i.e #q=-1.2#

and coordinates of #B# are #(-1.6,-1.2)#

graph{(3x+y+6)(x-3y-5)((x+1)^2+(y+3)^2-0.01)((x+1.6)^2+(y+1.2)^2-0.01)=0 [-9.92, 10.08, -5.88, 4.12]}

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Let the equation of the circle which cuts the given parabola #y^2=4ax# be #(x-b)^2+(y-c)^2=r^2.#,where #(b,c)# is the coordinates of the center(D) of the circle and #r# is its radius.

Replacing #x=y^2/(4a)# in the equation of the circle we get the following equation of y of degree 4

#(x-b)^2+(y-c)^2=r^2.#

#=>x^2-2bx+b^2+y^2-2cy+c^2-r^2=0#

#=>(y^2/(4a))^2-2b(y^2/(4a))+b^2+y^2-2cy+c^2-r^2=0#

#=>y^4/(16a^2)+(1-(2b)/(4a))y^2-2cy+b^2+c^2-r^2=0color(red)(.....[1])#

Here #f(y)=0# is an equation of y of degree 4. So it will have 4 roots of y and those roots will be given by #2at_i" for " i=1,2,3,4#

Since the coefficient of #y^3# of equation [1] is zero , the sum of all 4 roots of y must be zero.

Hence #sum_(i=1)^(i=4)(2at_1)=0#

#=>sum_(i=1)^(i=4)t_1=0#

#=>t_1+t_2+t_3+t_4=0#

Answer:

#A = 238 " "units^2#

Explanation:

Area of a regular polygon #= 1/2 xx perimeter xx apothem#

#A = (Pa)/2#

Perimeter #= "number of sides" xx "side length"#

#P = ns = 8s#

#A = (8s*8.5)/2 = 34s#

Since we know that the octagon is regular and we know the apothem (height from a side to the center of the polygon, we can calculate the side length):

Each angle of the central angle: #(360^@)/8 = 45^@#

and each side of the polygon forms an isoceles triangle when joined to center. Further, a perpendicular from center to the side of the polygon divides the side in two equal parts and also cuts this angle at the center (#45^@#) in half i.e. #22.5^@#.

Using trigonometry, we can find half the side length (#x#):

#tan 22.5^@ = x/a = x/8.5#

#x = 8.5*tan 22.5^@ ~~ 3.52#

side length: #s = 2x ~~ 7#

#A = 34 * s ~~ 34 * 7 ~~238 " units"^2 #

enter image source here

Answer:

length#=2/sqrt15# unit

Explanation:

enter image source here
We know that the perpendicular distance #d# between two parallel lines #y=mx+c1, and y=mx+c2# is given by:
#d=|c1-c2|/sqrt(1+m^2)#

Given that the two parallel lines are #y=2x+1, and y=2x+2#, as shown in the figure.
#=> c1=1, c2=2# and slope of the two lines #m=2#
#=> d=PM=|1-2|/sqrt(1+2^2)=1/sqrt5#

Given that #DeltaOPQ# is an equilateral triangle,
#=> OP=PQ=OQ, and angleOPQ=angleOQP=angleQOP=60^@#
#=> POsin60=PM#
#=> PO*sqrt3/2=1/sqrt5#
#=> PO=1/sqrt5*2/sqrt3=2/sqrt15# unit

Hence, length of each side of the equilateral triangle #=2/sqrt15# unit

Answer:

#r=2# units

Explanation:

As #13^2=5^2+12^2#, the triangle is a right triangle.
Formula for the inradius (#r#) of a right triangle :
#r=(a*b)/(a+b+c)#, or #r= (a+b-c)/2#
where #a and b# are the legs of the right traingle and #c# is the hypotenuse.
#=> r=(a*b)/(a+b+c)=(5*12)/(5+12+13)=60/30=2# units

or # r=(a+b-c)/2=(5+12-13)/2=4/2=2# units

Proof 1 : #r=(a*b)/(a+b+c)#
enter image source here
Area #DeltaABC=1/2*(BC)*(AC)=1/2*r*(BC+AC+AB)#
#=> (r*(a+b+c))/2=(a*b)/2#
#=> r=(a*b)/(a+b+c)#

Proof 2 : #r=(a+b-c)/2#
enter image source here
Recall that the tangents to a cicle from an external point are equal,
#=> CD=CF=r, BE=BF=a-r#,
and #AD=AE=b-r#
#=> c=a-r+b-r#
#=> c=a+b-2r#
#=> 2r=a+b-c#
#=> r=(a+b-c)/2#

Answer:

Please see below.

Explanation:

Let us consider the following figure, where we have the median #AM# and #AP_|_BC#.

enter image source here

Now in #DeltaABM#, we have

#AB^2=AP^2+PB^2=AP^2+(BM-PM)^2#

= #color(red)(AP^2)+BM^2+color(red)(PM^2)-2BMxxPM#

or #AB^2=color(red)(AM^2)+BM^2-2BMxxPM# .....................(1)

and in #DeltaACM#, we have

#AC^2=AP^2+PC^2=AP^2+(MC+PM)^2#

= #color(red)(AP^2)+MC^2+color(red)(PM^2)-2MCxxPM#

or #AC^2=color(red)(AM^2)+MC^2-2MCxxPM#

But as #BM=MC#, we can write it as

#AC^2=color(red)(AM^2)+BM^2-2BMxxPM# .....................(2)

Adding (1) and (2), we get

#AB^2+AC^2=2AM^2+2BM^2# and as #BM=(BC)/2#

#AB^2+AC^2=2AM^2+(2BC^2)/4#

or #AB^2+AC^2=2AM^2+1/2BC^2#

This is Apollonius theorem.

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