Question

A person standing at the junction (crossing) of two straight paths represented by the equations 2 x – 3 y + 4 = 0 and 3 x + 4 y – 5 = 0 wants to reach the path whose equation is 6 x – 7 y + 8 = 0 in the least time. Find equation of the path that he should follow.

Answer 1

The given equations of lines are,

2x−3y+4=0(1)

3x+4y−5=0 (2)

6x−7y+8=0(3)

The person is standing at the junction of equations (1) and (2) of lines.

Solve equations (1) and (2) to get,

x=− 1 17 y= 22 17

The person is standing at point ( − 1 17 , 22 17 ).

The person will reach the path of equation (3) in least time if he walks in a perpendicular direction from point ( − 1 17 , 22 17 ) to the line 6x−7y+8=0.

Let the slope of the line 6x−7y+8=0 be m.

Rewrite the equation of line (3).

7y=6x+8 y= 6 7 x+ 8 7

Compare the equation (1) with equation of line y=mx+c, we get

m 1 = 6 7

The condition for perpendicularity is,

m 1 ⋅ m 2 =−1

Substitute the value of m 1 to obtain the slope of the line perpendicular to the line 6x−7y+8=0.

m 2 = −1 6 7 =− 7 6

The formula for the equation of line having slope m and passing through the points ( x 0 , y 0 ) is given by,

( y− y 0 )=m⋅( x− x 0 )(4)

Substitute the values of ( x 0 , y 0 ) as ( − 1 17 , 22 17 ) and m as − 7 6 in equation (4).

( y− 22 17 )=− 7 6 ⋅( x+ 1 17 ) 6( 17y−22 )=−7( 17x+1 ) 102y−132=−119x−7 119x+102y−125=0

Thus, the person should follow the path of equation, 119x+102y−125=0.