Q113 / 155 · WBJEE 2016 · Mathematics

The coordinates of a point on the line $x+y+1=0,$ which is at a distance $\frac{1}{5}$ unit from the line $3 x+4 y+2=0,$ are

  • A
    (2,-3)
  • Correct answer
    B
    (-3,2)
  • C
    (0,-1)
  • Correct answer
    D
    (-1,0)

Solution

Let $(h, k)$ be the point on line $x+y+1=0$ So. $$ h+k+1=0 $$ $$ h=-1-k $$ Given, distance between $(h, k)$ and line $3 x+4 y+2=0$ is $\frac{1}{5}$ $\therefore$ $\frac{1}{5}=\left|\frac{(3 h)+(4 k)+2}{\sqrt{3^{2}+4^{2}}}\right|$ $\Rightarrow \quad \frac{1}{5}=\left|\frac{3(-1-k)+4 k+2}{\sqrt{25}}\right|$ $\Rightarrow \quad \frac{1}{5}=\pm\left(\frac{-3-3 k+4 k+2}{5}\right)$ $\Rightarrow$ $\frac{1}{5}=\pm\left(\frac{k-1}{5}\right)$ $\Rightarrow \quad \pm(k-1)=1$ $\Rightarrow \quad k=0$ and 2 $\therefore \quad h=-1, -3$ Hence, the required points are (-1,0) and (-3,2)

Hence, the correct answer is options: 'B', 'D'.

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