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(D) The equation of the line is $3x - 4y + 5 = 0$,which can be written as $3x - 4y = -5$. Dividing by $-5$,we get $-\frac{3}{5}x + \frac{4}{5}y = 1$.

Vedclass · Accepted Answer

$19$

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(D) The equation of the line is $3x - 4y + 5 = 0$,which can be written as $3x - 4y = -5$. Dividing by $-5$,we get $-\frac{3}{5}x + \frac{4}{5}y = 1$. Comparing this with the normal form $x \cos \alpha + y \sin \alpha = p$,we have $\cos \alpha = -\frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$. Thus,$	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{4/5}{-3/5} = -\frac{4}{3}$. The line $ax + by = 1$ passes through $(1, -1)$ and has slope $m = 	an \alpha = -\frac{4}{3}$. The equation of the line is $y - (-1) = -\frac{4}{3}(x - 1)$. $3(y + 1) = -4(x - 1)$ $\Rightarrow 3y + 3 = -4x + 4$ $\Rightarrow 4x + 3y = 1$. Comparing $4x + 3y = 1$ with $ax + by = 1$,we get $a = 4$ and $b = 3$. Therefore,$a + ab + b = 4 + (4 	imes 3) + 3 = 4 + 12 + 3 = 19$.

If $\alpha$ is the angle made by the perpendicular drawn from the origin to the line $3x - 4y + 5 = 0$ with the positive $X$-axis in the positive direction,and $ax + by = 1$ is the equation of a line passing through the point $(1, -1)$ with $\tan \alpha$ as its slope,then $a + ab + b =$

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