Question

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)

Answer 1

`C(3,3)`

Explanation:

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)`

Answer 2

C) `(3,3)`

Explanation:

Solution 2:

`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:

`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)`