Question

The slope of a straight line passing through $A(-2,3)$ is $-\frac{4}{3}$. The point(s) on the line that are $10$ units away from $A$ is/are

• A. $(8,0)$
• B. $(-8,11)$
• C. $(4,-5)$
• D. $(4,11)$

Answer 1

Answer: (B), (C)

The correct options are
B $(-8,11)$
C $(4,-5)$

Slope $=-\frac{4}{3}=\tan \theta \Rightarrow \theta \in 2^{nd}$ quadrant
$\Rightarrow \cos \theta =-\frac{3}{5},\sin \theta =\frac{4}{5}$
$A=(-2,3)=(x_1,y_1),r=10$
$\Rightarrow x=x_1\pm r\cos \theta =-2\pm 10(-\frac{3}{5})=-8\text{or}4$
$\Rightarrow y=y_1\pm r\sin \theta =3\pm 10\left(\frac{4}{5}\right)=11\text{or}-5$
$\therefore$ Required points are $(-8,11),(4,-5)$