If the lines 3x + y - 4 = 0,x - ay - 10 = 0,a | vedclass

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(D) Let the rectangle be $ABCD$. The lines are $AB: 3x + y - 4 = 0$,$BC: x - ay - 10 = 0$,and $CD: bx + 2y + 9 = 0$.Since $AB \parallel CD$,their slop

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$\frac{51}{4}$

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(D) Let the rectangle be $ABCD$. The lines are $AB: 3x + y - 4 = 0$,$BC: x - ay - 10 = 0$,and $CD: bx + 2y + 9 = 0$.Since $AB \parallel CD$,their slopes must be equal. Slope of $AB$ is $-3$. Slope of $CD$ is $-\frac{b}{2}$. Thus,$-\frac{b}{2} = -3 \Rightarrow b = 6$.Since $AB \perp BC$,the product of their slopes is $-1$. Slope of $BC$ is $\frac{1}{a}$. Thus,$(-3) 	imes (\frac{1}{a}) = -1 \Rightarrow a = 3$.Now,$AB: 3x + y - 4 = 0$ and $CD: 6x + 2y + 9 = 0$,which can be written as $3x + y + \frac{9}{2} = 0$.The distance between parallel lines $AB$ and $CD$ is $BC = \frac{|-4 - 9/2|}{\sqrt{3^2 + 1^2}} = \frac{17/2}{\sqrt{10}} = \frac{17}{2\sqrt{10}}$.The line $BC$ is $x - 3y - 10 = 0$. The side $CD$ is a line parallel to $AB$ and $AD$ is a line parallel to $BC$. The distance $CD$ is the distance between the parallel lines $AD$ and $BC$. Since $AD$ passes through $(1, 2)$,the distance $CD$ is the perpendicular distance from $(1, 2)$ to the line $BC: x - 3y - 10 = 0$.$CD = \frac{|1 - 3(2) - 10|}{\sqrt{1^2 + (-3)^2}} = \frac{|1 - 6 - 10|}{\sqrt{10}} = \frac{15}{\sqrt{10}}$.Area of rectangle $ABCD = BC 	imes CD = \frac{17}{2\sqrt{10}} 	imes \frac{15}{\sqrt{10}} = \frac{17 	imes 15}{2 	imes 10} = \frac{255}{20} = \frac{51}{4}$.

If the lines $3x + y - 4 = 0$,$x - ay - 10 = 0$,and $bx + 2y + 9 = 0$ 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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