Question

Coordinates of a point which is at $3$ units distance from the point $(1,-3)$ on the line $2x+3y+7=0$ is/are

• A. $(1+\frac{9}{\sqrt{13}},3-\frac{6}{\sqrt{13}})$
• B. $(1-\frac{9}{\sqrt{13}},-3+\frac{6}{\sqrt{13}})$
• C. $(1+\frac{9}{\sqrt{13}},-3-\frac{6}{\sqrt{13}})$
• D. $(1-\frac{9}{\sqrt{13}},3-\frac{6}{\sqrt{13}})$

Answer 1

Answer: (B), (C)

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

Let the coordinate of point be $(h,k)$

As this point lies on $2x+3y+7=0$

Hence $2h+3k+7=0$ ...(1)

And as this point is at distance $3$ from point $(1,-3)$

$\sqrt{{(h-1)}^2+{(k+3)}^2}=3\Rightarrow h^2-2h+1+k^2+6k+9=9$

$\Rightarrow h^2+k^2-2h+6k+1=0$ ...(2)
From (1) and (2)

$(h,k)=(1\pm \frac{9}{\sqrt{13}},-3\mp \frac{6}{\sqrt{13}})$