Question

A vertex of a rectangle is at the origin and two of its sides lie along the lines

5 x - 12 y + 26 = 0

and

12 x + 5 y - 39 = 0

. Find the area of the rectangle.

Accepted Answer

Given lines are, $5 x - 12 y + 26 = 0 \to (1)$ and $12 x + 5 y - 39 = 0 \to (2)$ .

Now, the slope of the line $(1)$ $= \frac{- 5}{- 12} = \frac{5}{12}$

And also, the slope of the line $(2)$ $= \frac{- 12}{5}$

Therefore, product of slopes of $(1)$ and $(2)$ $= \frac{5}{12} \times (\frac{- 12}{5}) = - 1$

Thus, both lines are perpendicular to each other.

Therefore, the given lines are along the sides $A B$ and $A D$ of a rectangle $A B C D$ .

And also, clearly $(0, 0)$ does not lie on both lines $(1)$ and $(2)$ . Thus $(0, 0)$ be the coordinate of vertex $C$ rectangle $A B C D$ .

Therefore, the length of side $B C$ of rectangle $= |B C|$

$=$ length of perpendicular from $(0, 0)$ to the line $A B$

$= \frac{| 5 \times 0 - 12 \times 0 + 26 |}{\sqrt{5^{2} + (- 12)^{2}}}$

$= \frac{26}{13}$

$= 2$

Therefore, the slope of $B C = \frac{- 1}{slope of A B}$

$= - \frac{1}{\frac{- 5}{- 12}} = \frac{- 12}{5}$

Thus, the equation of $B C$ is $y - 0 = \frac{- 12}{5} (x - 0)$

$\Rightarrow 12 x + 5 y = 0 \to (3)$

Solving equations $(1)$ and $(2)$ simultaneously, we get

$\frac{x}{468 - 130} = \frac{y}{312 + 195} = \frac{1}{25 + 144}$ (Using cross-multiplication method)

$\Rightarrow \frac{x}{338} = \frac{y}{507} = \frac{1}{169}$

$\Rightarrow x = \frac{338}{169} = 2, y = \frac{507}{169} = 3$

So, the point of intersection of $(1)$ and $(2)$ is $(2, 3)$ , hence the coordinates of the vertex $A (2, 3)$ .

Thus, $|A B| =$ length of the perpendicular form $A (2, 3)$ to the line $(3)$

$= \frac{| 12 \times 2 + 5 \times 3 |}{\sqrt{12^{2} + 5^{2}}}$

$= \frac{| 24 + 15 |}{13} = 3$

Hence, required area of rectangle $= | A B | \times | B C | = 3 \times 2 = 6$ sq.units.

M L Aggarwal Solutions for Chapter: Straight Lines, Exercise 9: EXERCISE

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Straight Lines, Exercise 9: EXERCISE

Questions from M L Aggarwal Solutions for Chapter: Straight Lines, Exercise 9: EXERCISE with Hints & Solutions

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