First slide

Equation of line in Straight lines

Question

The line $\frac{x}{3} + \frac{y}{4} = 1$ meets the y and x-axis 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

Moderate

A

(7,3)
B

(4,7)
C

(6,4)
D

(3, 8)

Solution

The coordinates of A are (0, 4) and those of B are (3, 0).
$\begin{array}{l} B C = A B = \sqrt{3^{2} + 4^{2}} = 5 \\ I n △ B C L ∠ C B L = 180 - 90 - θ = 90 - θ \\ \sin (90 - θ) = \frac{C L}{B C} \\ cos θ= \frac{C L}{B C} \\ and C L = B C \cos θ \\ \cos (90 - θ) = \frac{B L}{B C} \\ or B L = B C \sin θ \\ or B L = 5 \times \frac{4}{5} = 4 \\ and C L = 5 \times \frac{3}{5} = 3 \end{array}$
$Similarly, M D = 4 and A M = 3 . So, the coordinates of C are (O B + B L, C L) \equiv (7, 3) and those of D are (M D, O A + A M) \equiv (4, 7)$
The coordinates of the vertex farthest from the origin are (4,7).

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