Question

Conordinates of a point which is at 3 units distane from the point (1, –3) on the line 2x + 3y + 7 = 0 is/are.

(A) $\left(1+\frac{9}{\sqrt{13}},3-\frac{6}{\sqrt{13}}\right)$ (B) $\left(1-\frac{9}{\sqrt{13}},-3+\frac{6}{\sqrt{13}}\right)$

(C) $\left(1+\frac{9}{\sqrt{13}},-3-\frac{6}{\sqrt{13}}\right)$ (D) $\left(1-\frac{9}{\sqrt{13}},3-\frac{6}{\sqrt{13}}\right)$

Answer 1

dear student

$$\begin{gathered} co-ordinatesofpointatadis\tan ce"r"fromapoint(x,y)alongalinewithslope\tan \theta is \\ (x\pm r\cos \theta ,y\pm r\sin \theta ) \\ inthiscaseslopeofline2x+3y+7=0 \\ slope=-\frac{2}{3} \\ \tan \theta =-\frac{2}{3} \\ \cos \theta =\frac{-3}{\sqrt{13}} \\ \sin \theta =\frac{2}{\sqrt{13}} \\ requiredpoint \\ =(x\pm r\cos \theta ,y\pm r\sin \theta ) \\ =(1\pm 3\frac{-3}{\sqrt{13}},-3\pm 2\frac{2}{\sqrt{13}}) \\ =(1+\frac{9}{\sqrt{13}},-3-\frac{6}{\sqrt{13}}),(1-\frac{9}{\sqrt{13}},-3+\frac{6}{\sqrt{13}}) \end{gathered}$$

regards