If the distance between a point P and the point (1, 1, 1) on the line \frac{x \textrm{ } - \textrm{ } 1}{3} \textrm{ } = \textrm{ } \frac{y - \textrm{ } 1}{4} \textrm{ } = \textrm{ } \frac{z \textrm{ } - 1}{12} is 13, then the coordinates of P are

Engineering

Mathematics

Line of Greatest Slope

Question

If the distance between a point P and the point (1, 1, 1) on the line

\frac{x - 1}{3} = \frac{y - 1}{4} = \frac{z - 1}{12}

is 13, then the coordinates of P are

(3, 4, 12)

(\frac{3}{13}, \frac{4}{13}, \frac{12}{13})

(4, 5, 12)

(40, 53, 157)

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\begin{array}{l} A c c o r d i n g t o q u e s t i o n : \\ w e h a v e t h e l i n e e q u = \frac{x - 1}{3} = \frac{y - 1}{4} = \frac{z - 1}{12} a n d p o i n t (P) i s 13. \\ N o w, \\ l i n e o f e q u : \\ \Rightarrow \frac{x - 1}{3} = \frac{y - 1}{4} = \frac{z - 1}{12} = 13 \\ N o w f i n d x, y & a m p; z c o o r d i n a t e s : \\ \Rightarrow \frac{x - 1}{3} = 13 \\ ∴ x = 39 + 1 = 40 \\ \Rightarrow \frac{y - 1}{4} = 13 \\ ∴ y = 52 + 1 = 53 \\ \Rightarrow \frac{z - 1}{12} = 13 \\ ∴ z = 156 + 1 = 157 \\ N o w, w e g e t n e w c o o r d i n a t e s o f P : (40, 53, 157), \\ s o t h a t t h e c o r r e c t o p t i o n i s D . \end{array}

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