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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.
Solution
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.
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