SSC CHSL Premium Mock 03 – Q061

Question:

Find the equation of a line that is parallel to \(y = -2x + 3\) and passes through the point \((2, -1)\).

उस रेखा का समीकरण ज्ञात कीजिए जो \(y = -2x + 3\) के समानांतर है और बिंदु \((2, -1)\) से होकर गुजरती है।

Options:

  • \(y = -2x + 3\)
  • \(y = -2x + 1\)
  • \(y = -2x – 5\)
  • \(y = -2x + 5\)
  • \(y = -2x + 3\)
  • \(y = -2x + 1\)
  • \(y = -2x – 5\)
  • \(y = -2x + 5\)

Correct Answer: a)

Explanation:

Parallel line has the same slope \(m = -2\).
Using point-slope form: \(y – y_1 = m(x – x_1)\)
\(y – (-1) = -2(x – 2) \implies y + 1 = -2x + 4 \implies y = -2x + 3\).

समानांतर रेखा की प्रवणता समान होगी: \(m = -2\)।
बिंदु-प्रवणता रूप: \(y – (-1) = -2(x – 2) \implies y + 1 = -2x + 4 \implies y = -2x + 3\)।

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