On the portion of the straight line #x+2y=4# intercepted between the axes , a square is constructed on the side of the line away from the origin. The the point of intersection of its diagonals has co-ordinates is?

A) (2,3)
B)(3,2)
C)(3,3)
D)(2,2)

2 Answers
CW
Sep 28, 2017

#C(3,3)#

Explanation:

enter image source here
Given equation of line : #x+2y=4#,
when #x=0, y=2, => A(0,2) ---- y-#intercept.
when #y=0, x=4, => B(4,0) ---- x-#intercept.
#B# is #4# units right and #2# units down from #A#.
length of #AB=sqrt(2^2+4^2)=sqrt20=2sqrt5#,
Let #ABCD# be the square, as shown in the figure.
As #AD# is perpendicular to #AB#, and #AD=AB=2sqrt5#,
#=> D# is #2# units right and #4# units up from #A#,
#=> D(x_d, y_d)= (0+2, 2+4)=(2,6)#
Similarly, #C# is #2# units right and #4# units up from #B#,
#=> C(x_c, y_c)= (4+2,0+4) = (6,4)#
let #M# be the point of intersection of the diagonals #AC and BD#,
recall that the two diagonals of a square bisect each other,
#=> M# is the mid-point of #AC# or #BD#
#=> M(x_m, y_m)=((0+6)/2, (2+4)/2)=(3,3)#

CW
Oct 2, 2017

C) #(3,3)#

Explanation:

Solution 2:
enter image source here
#A(0,2), B(4,0), => AB=BC=2sqrt5#
Let #E# be the midpoint of #AB, => E(2,1)#
#=> EM=1/2(BC)=sqrt5#
slope of #AB=(0-2)/(4-0)=-1/2#,
#=># slope of #BC=2#
#=> tantheta=2#
#=> sintheta=2/sqrt5, "and" costheta=1/sqrt5#
#=> M(x_m,y_m)=(x_e+EMcostheta,y_e+EMsintheta)=(2+EMcostheta,1+EMsintheta)#
#=(2+sqrt5*1/sqrt5,1+sqrt5*2/sqrt5)#
#=(2+1,1+2)=(3,3)#

Solutin 3:
enter image source here
#AM=(2sqrt5*sqrt2)/2=sqrt10#
recall that the formula for angle between two slopes is given by :
#tanalpha=|m_1-m_2|/|1+m_1m_2|#
#angleEAM=45^@#, and slope of #AE=-1/2#,
let slope of #AM=m#
#=> tan45=1=(m+1/2)/(1-m/2)#
#=> m=1/3#
#=> tanbeta=1/3#,
#=> sinbeta=1/sqrt10#,
#=> cosbeta=3/sqrt10#,
#=> M(x_m,y_m)=(x_a+AMcosbeta, y_a+AMsinbeta)#
#=(0+sqrt10*3/sqrt10, 2+sqrt10*1/sqrt10)#
#=(3,3)#

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