Question

The equation of straight line passing through the point of intersection of family of lines $x(3 + 4\lambda) + y(4 + 3\lambda) + 1 + 6\lambda = 0$ which is at a minimum distance from the point $(1, 1)$ can be expressed as $x + by - c = 0$, where $b$ and $c$ are natural numbers, then the value of $(b + c)$ is

• A. 12
• B. 13
• C. 14
• D. 15

Answer 1

Answer: (D)

15

Answer 2

Step 1 Find the fixed intersection point $P$ of the two generating lines:

$$3x+4y+1=0,\quad4x+3y+6=0$$

Solving gives $P(-3,2)$.

Step 2 Among all lines through $P$, the one at minimum distance from $Q(1,1)$ is the perpendicular to $PQ$. The slope of $PQ$ is

$$m_{PQ}=\frac{1-2}{1-(-3)}=-\tfrac14,$$

so the required line has slope $4$. Its equation through $(-3,2)$:

$$y-2=4(x+3)\;\Longrightarrow\;4x - y +14=0.$$

Step 3 Match to the form $x + by - c=0$. Up to an overall sign, one may take

$$x + (\,1\,)y - (\,14\,)=0,$$

so $b=1,\;c=14$, giving $b+c=15$.