MHT CET · Maths · Three Dimensional Geometry
The equation of line passing through the points \((3,4,-7)\) and \((6,-1,1)\) is
- A \(\frac{x-3}{3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
- B \(\frac{x-3}{3}=\frac{y-4}{5}=\frac{z+7}{8}\)
- C \(\frac{x-3}{-3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
- D \(\frac{x-3}{3}=\frac{y-4}{-5}=\frac{z-7}{8}\)
Answer & Solution
Correct Answer
(A) \(\frac{x-3}{3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
Step-by-step Solution
Detailed explanation
Required eq. of line is
\(\frac{x-3}{6-3}=\frac{y-4}{-1-4}=\frac{z+7}{1+7} \text { i.e. } \frac{x-3}{3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
\(\frac{x-3}{6-3}=\frac{y-4}{-1-4}=\frac{z+7}{1+7} \text { i.e. } \frac{x-3}{3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
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