Question

The line $\frac { x } { 3 } + \frac { y } { 4 }$ = 1 meets the axis of y and axis of x at A and B respectively. A square ABCD is constructed on the line segment AB away from the origin, the coordinates of the vertex of the square farthest from the origin are –

• A. (7, 3)
• B. (4, 7)
• C. (6, 4)
• D. (3, 8)

Answer 1

Answer: (B)

(4, 7)

Answer 2

The coordinates of A are (0, 4) and that of B are (3, 0).

Let CL and DM be perpendiculars on x-axis and

y-axis respectively then if ŠOBA = q.

ŠCBL = ŠADM = 90° – q [See figure]

also, BC = AB = $\sqrt { 3 ^ { 2 } + 4 ^ { 2 } }$ = 5

Ž BL = BC sin q and CL = BC cos q

Ž BL = 5 × $\frac { 4 } { 5 }$ = 4 and CL = 5 × $\frac { 3 } { 5 }$ = 3

Similarly, MD = 4 and AM = 3.

So the co-ordinates of C are (OB + BL, CL) = (7, 3) and of D are (MD, OA + AM) = (4, 7)

The co-ordinates of the vertex farthest from the origin are therefore (4, 7).