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Find the value of \(\alpha\) if the line \(2y - \alpha x + 4 = 0\) is perpendicu...


Question

Find the value of \(\alpha\) if the line \(2y - \alpha x + 4 = 0\) is perpendicular to the line \(y + \frac{1}{4}x - 7 = 0\)

Options

A) -4

B) 4

C) 8

D) -8

The correct answer is C.

Explanation:

\(2y - \alpha x + 4 = 0\) is perpendicular to \(y + \frac{1}{4}x - 7 = 0\)
Writing both equations in the gradient-intercept form, gives:
\(y = \cfrac{-\alpha}{2}x - 2\)........(i)
\(y = \cfrac{1}{4}x + 7\).........(ii)
Gradient of first line = \(\frac{-\alpha}{2}\)
Gradient of the second line = \(\frac{1}{4}\)
For two lines to be perpendicular, the product of their gradient is \(-1\)
\(\left( \cfrac{-\alpha}{2} \right)\left( \cfrac{1}{4} \right) = -1\)
\(\cfrac{-\alpha}{8} = -1\)
\(-\alpha = -8\)
\(\alpha = 8\)

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Dicussion (1)

  • \(2y - \alpha x + 4 = 0\) is perpendicular to \(y + \frac{1}{4}x - 7 = 0\)
    Writing both equations in the gradient-intercept form, gives:
    \(y = \cfrac{-\alpha}{2}x - 2\)........(i)
    \(y = \cfrac{1}{4}x + 7\).........(ii)
    Gradient of first line = \(\frac{-\alpha}{2}\)
    Gradient of the second line = \(\frac{1}{4}\)
    For two lines to be perpendicular, the product of their gradient is \(-1\)
    \(\left( \cfrac{-\alpha}{2} \right)\left( \cfrac{1}{4} \right) = -1\)
    \(\cfrac{-\alpha}{8} = -1\)
    \(-\alpha = -8\)
    \(\alpha = 8\)

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