The points on the straight line 3x - 4y + 1 = | vedclass

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(D) Let the point $A(x, y)$ on the line $3x - 4y + 1 = 0$ be at a distance of $5$ units from the point $P(3, 2)$.The equation of a line passing throug

Vedclass · Accepted Answer

$(7, 5), (-1, -1)$

Vedclass · Answer

(D) Let the point $A(x, y)$ on the line $3x - 4y + 1 = 0$ be at a distance of $5$ units from the point $P(3, 2)$.The equation of a line passing through $(3, 2)$ with inclination $	heta$ is given by $\frac{x-3}{\cos 	heta} = \frac{y-2}{\sin 	heta} = r$.For $r = \pm 5$,we have $x = 3 \pm 5 \cos 	heta$ and $y = 2 \pm 5 \sin 	heta$.Since this point lies on the line $3x - 4y + 1 = 0$,we substitute these coordinates into the equation:$3(3 \pm 5 \cos 	heta) - 4(2 \pm 5 \sin 	heta) + 1 = 0$$9 \pm 15 \cos 	heta - 8 \mp 20 \sin 	heta + 1 = 0$$2 \pm 15 \cos 	heta \mp 20 \sin 	heta = 0$$\pm 15 \cos 	heta \mp 20 \sin 	heta = -2$Alternatively,using the slope of the line $3x - 4y + 1 = 0$,which is $m = \frac{3}{4}$,we have $	an 	heta = \frac{3}{4}$.This implies $\sin 	heta = \pm \frac{3}{5}$ and $\cos 	heta = \pm \frac{4}{5}$.Substituting these values into $x = 3 \pm 5 \cos 	heta$ and $y = 2 \pm 5 \sin 	heta$:For the positive sign: $x = 3 + 5(\frac{4}{5}) = 7$ and $y = 2 + 5(\frac{3}{5}) = 5$.For the negative sign: $x = 3 - 5(\frac{4}{5}) = -1$ and $y = 2 - 5(\frac{3}{5}) = -1$.Thus,the required points are $(7, 5)$ and $(-1, -1)$.

The points on the straight line $3x - 4y + 1 = 0$ which are at a distance of $5$ units from the point $(3, 2)$ are

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