If the lines $3 x + y - 4 = 0, x - a y - 10 = 0, b x + 2 y + 9 = 0$ form three successive sides of a rectangle in that order and the fourth side passes through (1, 2), then the area of that rectangle (in sq. units) is …….

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answer is 12.75.

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Detailed Solution

Let ABCD be a rectangle such that the equation of $\bar{A B}$ is $3 x + y - 4 = 0,$ the equation of $\bar{B C}$ is x- ay -10=0 and the equation $\bar{C D}$ is $b x + 2 y + 9 = 0 .$

Slope of $\bar{A B} = - 3,$ slope of $\bar{B C} = \frac{1}{a},$ slope of $\bar{C D} = \frac{- b}{2}$

$\bar{A B}$ is parallel to $\bar{C D} \Rightarrow - 3 = \frac{- b}{2} \Rightarrow b = 6$

$\bar{A B}$ is perpendicular of $\bar{B C} \Rightarrow (- 3) (\frac{1}{a}) = - 1 \Rightarrow a = 3.$

Equation of $\bar{A D}$ which passes through (1, 2) and parallel to $x - 3 y - 10 = 0$ is $1 (x - 1) - 3 (y - 2) = 0 \Rightarrow x - 3 y + 5 = 0$

$p_{1} =$ Distance between parallel sides $x - 3 y - 10 =, x - 3 y + 5 = 0 i s \frac{| 5 + 10 |}{\sqrt{1 + 9}} = \frac{5}{\sqrt{10}}$

$p_{2} =$ Distance between parallel sides $6 x + 2 y + 9 = 0, 6 x + 2 y - 8 = 0 i s \frac{| 9 + 8 |}{\sqrt{36 + 4}} = \frac{17}{\sqrt{40}}$

Area of rectangle $A B C D = p_{1} \times p_{2} = \frac{5}{\sqrt{10}} \times \frac{17}{\sqrt{40}} = \frac{15 \times 17}{\sqrt{10} + 2 \sqrt{10}} = \frac{15 \times 17}{2 \times 10} = \frac{3 \times 17}{4} = \frac{51}{4} = 12.75$