If a line L passing through the point A(-2, 4 | vedclass

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(A) The line $L$ passes through $A(-2, 4)$ with an inclination $	heta = 60^{\circ}$.The coordinates of any point $B(p, q)$ on the line at a distance

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) The line $L$ passes through $A(-2, 4)$ with an inclination $	heta = 60^{\circ}$.The coordinates of any point $B(p, q)$ on the line at a distance $r = 6$ from $A(x_1, y_1)$ are given by $p = x_1 + r \cos 	heta$ and $q = y_1 + r \sin 	heta$.Since $B$ lies in the $3^{	ext{rd}}$ quadrant,we move in the opposite direction along the line,so $r = -6$.$p = -2 + (-6) \cos 60^{\circ} = -2 - 6(\frac{1}{2}) = -2 - 3 = -5$.$q = 4 + (-6) \sin 60^{\circ} = 4 - 6(\frac{\sqrt{3}}{2}) = 4 - 3\sqrt{3}$.We need to calculate $\sqrt{p^2 + q^2 - 8q}$.Note that $p^2 + q^2 - 8q = p^2 + (q-4)^2 - 16$.Substituting $p = -5$ and $q = 4 - 3\sqrt{3}$:$p^2 = (-5)^2 = 25$.$(q-4)^2 = (4 - 3\sqrt{3} - 4)^2 = (-3\sqrt{3})^2 = 9 	imes 3 = 27$.So,$p^2 + (q-4)^2 - 16 = 25 + 27 - 16 = 52 - 16 = 36$.Therefore,$\sqrt{p^2 + q^2 - 8q} = \sqrt{36} = 6$.

If a line $L$ passing through the point $A(-2, 4)$ makes an angle of $60^{\circ}$ with the positive direction of $X$-axis in the anti-clockwise direction and $B(p, q)$ lying in the $3^{\text{rd}}$ quadrant is a point on $L$ at a distance of $6$ units from the point $A$,then $\sqrt{p^2+q^2-8q} = $

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