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Question: The line L given by \[\dfrac{x}{5} + \dfrac{y}{b} = 1\] passes through the point (13, 32). The line ...

The line L given by x5+yb=1\dfrac{x}{5} + \dfrac{y}{b} = 1 passes through the point (13, 32). The line K is parallel to L and has the equation xc+y3=1\dfrac{x}{c} + \dfrac{y}{3} = 1. Then, what is the distance between L and K?
(a). 2315\dfrac{{23}}{{\sqrt {15} }}
(b). 17\sqrt {17}
(c). 1715\dfrac{{17}}{{\sqrt {15} }}
(d). 2317\dfrac{{23}}{{\sqrt {17} }}

Explanation

Solution

Hint: Substitute the point (13, 32) in the equation x5+yb=1\dfrac{x}{5} + \dfrac{y}{b} = 1 and find the value of b. Then find the value of c. Then use the distance between two parallel lines formula which is given as c1c2a2+b2\dfrac{{|{c_1} - {c_2}|}}{{\sqrt {{a^2} + {b^2}} }} to find the distance between them.

Complete step-by-step answer:

It is given that the line x5+yb=1\dfrac{x}{5} + \dfrac{y}{b} = 1 passes through the point (13, 32). Hence, this point should satisfy the equation of the line. Hence, we have:

135+32b=1\dfrac{{13}}{5} + \dfrac{{32}}{b} = 1

Simplifying, we have:

32b=1135\dfrac{{32}}{b} = 1 - \dfrac{{13}}{5}

32b=5135\dfrac{{32}}{b} = \dfrac{{5 - 13}}{5}

32b=85\dfrac{{32}}{b} = \dfrac{{ - 8}}{5}

Solving for b, we have:

b=32×58b = - 32 \times \dfrac{5}{8}

b=20..........(1)b = - 20..........(1)

Hence, the equation of the line is given as follows:

x5y20=1............(2)\dfrac{x}{5} - \dfrac{y}{{20}} = 1............(2)

It is given that the line xc+y3=1\dfrac{x}{c} + \dfrac{y}{3} = 1 is paralle to the line in equation (2). For parallel lines, the coeficients of x and y should be proportional, hence, we have:

5c=203\dfrac{5}{c} = \dfrac{{ - 20}}{3}

Solving for c, we have:

c=5×320c = - 5 \times \dfrac{3}{{20}}

c=34c = - \dfrac{3}{4}

Hence, we have the equation as follows:

4x3+y3=1............(3) - \dfrac{{4x}}{3} + \dfrac{y}{3} = 1............(3)

We multiply equation (2) by 20 and equation (3) by – 3 so that they have common coefficients.

4xy=204x - y = 20

4xy=34x - y = - 3

The distance between two parallel lines is given as follows:

d=c1c2a2+b2d = \dfrac{{|{c_1} - {c_2}|}}{{\sqrt {{a^2} + {b^2}} }}

Hence, we have:

d=20(3)42+(1)2d = \dfrac{{|20 - ( - 3)|}}{{\sqrt {{4^2} + {{( - 1)}^2}} }}

d=20+316+1d = \dfrac{{|20 + 3|}}{{\sqrt {16 + 1} }}

d=2317d = \dfrac{{23}}{{\sqrt {17} }}

Hence, the correct answer is option (d).

Note: The equations of the two parallel lines must have the same coefficients before using the formula to calculate the distance between them. If they don’t have the same coefficients, multiply the equation by a factor to make them equal, otherwise, the answer will be wrong.