Q67 / 160 · TS EAMCET 2017 · Mathematics

If the perpendicular distance between the point $(1,1)$ to the line $3 x+4 y+c=0$ is 7 , then the possible values of $c$ are

  • A
    $-35,42$
  • B
    35,28
  • C
    $42,-28$
  • Correct answer
    D
    $28,-42$

Solution

We know that, distance from point $\left(x_1, y_1\right)$ to the line $a x+b y+c=0$ is $$ \left|\frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}\right| $$ $\therefore$ Distance from $(1,1)$ to $3 x+4 y+c=0$ is 7 $$ \begin{array}{rlr|} & \quad 7 & =\left|\frac{3(1)+4(1)+c}{\sqrt{3^2+4^2}}\right| \\ \Rightarrow \quad 7 & =\left|\frac{7+c}{5}\right| \\ \Rightarrow \quad 35 & =|7+c| \\ \Rightarrow & 7+c & = \pm 35 \\ \Rightarrow \quad c & =-7 \pm 35 \\ & c & =28,-42 \end{array} $$

Hence, the correct answer is option: 'D'.

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