If a straight line through the point P(1, 2), | vedclass

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(A) The equation of the line passing through $P(1, 2)$ with slope $m = 	an(45^{\circ}) = 1$ is given by $(y - 2) = 1(x - 1)$,which simplifies to $x -

Vedclass · Accepted Answer

$\frac{16\sqrt{2}}{7}$

Vedclass · Answer

(A) The equation of the line passing through $P(1, 2)$ with slope $m = 	an(45^{\circ}) = 1$ is given by $(y - 2) = 1(x - 1)$,which simplifies to $x - y + 1 = 0$.To find the point $Q$,we solve the system of equations:$x - y + 1 = 0 \Rightarrow y = x + 1$$3x + 4y + 5 = 0$Substituting $y = x + 1$ into the second equation:$3x + 4(x + 1) + 5 = 0$$3x + 4x + 4 + 5 = 0$$7x + 9 = 0 \Rightarrow x = -\frac{9}{7}$Then $y = -\frac{9}{7} + 1 = -\frac{2}{7}$.So,$Q = \left(-\frac{9}{7}, -\frac{2}{7}ight)$.The distance $PQ$ is calculated using the distance formula:$PQ = \sqrt{\left(1 - (-\frac{9}{7})ight)^2 + \left(2 - (-\frac{2}{7})ight)^2}$$PQ = \sqrt{\left(\frac{16}{7}ight)^2 + \left(\frac{16}{7}ight)^2}$$PQ = \sqrt{2 	imes \left(\frac{16}{7}ight)^2} = \frac{16\sqrt{2}}{7}$ units.

If a straight line through the point $P(1, 2)$,which makes an angle $45^{\circ}$ with the $X$-axis,meets the line $3x + 4y + 5 = 0$ at $Q$,then the length of $PQ$ equals ......... units.

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