Question

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

Answer 1

AB and BC are two linear paths. Equations of lines AB and BC

2x – 3y + 4 = 0        ....(i)

And 3x + 4y – 5 = 0     ......(ii)

AB and BC meet at point B.

On multiplying equation (i) by 3 and equation (ii) by 2

6x – 9y = –12      ....(iii)

6x + 8y = 10       .....(iv)

By subtracting equation (iii) from equation (iv),

17y = 10 + 12 = 22

∴ y = `22/7`

Putting the value of y in equation (i),

`2"x" - 3 xx (22/17) = -4`

or `2"x" = -4 + 66/17 = (-2)/17`

∴ x = `-1/17`

Thus the coordinates of B are `((-1)/17, 22/17)`.

To reach from B to AC in minimum time the minimum distance to be covered is BD (BD ⊥ AC).

Equation of line AC, slope of 6x – 7y + 8 = 0 = `6/7`

Slope of BD = `-7/6`

BD passes through point B `((-1)/17, 22/17)`.

∴ Equation of line BD

y – y₁ = m(x – x₁)

`"y" - 22/17 = -7/6 ("x" + 1/17)`

On multiplying by 102,

102y – 132 = –119x – 7

119x + 102y – 125 = 0

Therefore, to reach AC from B, we have to take path BD whose equation is 119x + 102y – 125 = 0.