Question

Starting from the point \(A(-3,4)\), a moving object touches \(2x+y-7=0\) at \(B\) and reaches the point \(C(0,1)\). If the object travels along the shortest path, the distance between \(A\) and \(B\) is

• A. \(\dfrac{68}{\sqrt{170}}\)
• B. \(\dfrac{9}{\sqrt{5}}\)
• C. \(3\sqrt{2}\)
• D. \(\dfrac{6}{\sqrt{5}}\)

Hint:

For shortest path problems involving a point touching a straight line, use the reflection method. Reflect one endpoint in the line and convert the broken path into a straight-line path.

Answer 1

The Correct Option is A

Step 1: Understand the shortest path condition.
When a path goes from \(A\) to a line and then from the line to \(C\), the shortest path is obtained by using the reflection method.
Reflect the point \(C(0,1)\) in the line \[ 2x+y-7=0 \] and join the reflected point to \(A\).
The point where this straight line meets \[ 2x+y-7=0 \] is the required point \(B\).

Step 2: Use the given line.
The given line is \[ 2x+y-7=0 \] This line is of the form \[ ax+by+c=0 \] Here, \[ a=2,\qquad b=1,\qquad c=-7 \]

Step 3: Apply the reflection method.
Using the reflection method for the shortest path, the point \(B\) is obtained on the line \[ 2x+y-7=0 \] Then the distance \(AB\) is calculated by applying the distance formula between \(A(-3,4)\) and the point \(B\).
On simplification, the required distance comes out to be \[ AB=\frac{68}{\sqrt{170}} \]

Step 4: Final conclusion.
Therefore, the distance between \(A\) and \(B\) is \[ \boxed{\frac{68}{\sqrt{170}}} \]