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Standard X

Mathematics

Slope of a Line

A value of ...

Question

A value of $k$ such that the straight lines $y - 3 k x + 4 = 0$ and $(2 k - 1) x - (8 k - 1) y - 6 = 0$ are perpendicular is

\frac{1}{6}

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- \frac{1}{6}

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Solution

The correct option is A $\frac{1}{6}$
Let $m_{1}$ and $m_{2}$ be slopes of the two lines respectively.
Then, $m_{1} = 3 k$ and $m_{2} = \frac{2 k - 1}{8 k - 1}$
For the two lines to perpendicular,
$m_{1} m_{2} = - 1$
$⟹ (3 k) (\frac{2 k - 1}{8 k - 1}) = - 1$
$⟹ 6 k^{2} + 5 k - 1 = 0$
$⟹ (6 k - 1) (k + 1) = 0$
$k = \frac{1}{6} o r k = - 1$
Hence, the answer is (A)

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A value of k such that the straight lines y−3kx+4=0 and (2k−1)x−(8k−1)y−6=0 are perpendicular is

The correct option is A 16Let m1 and m2 be slopes of the two lines respectively.Then, m1=3k and m2=2k−18k−1For the two lines to perpendicular, m1m2=−1⟹(3k)(2k−18k−1)=−1⟹6k2+5k−1=0⟹(6k−1)(k+1)=0k=16ork=−1Hence, the answer is (A)

A value of $k$ such that the straight lines $y - 3 k x + 4 = 0$ and $(2 k - 1) x - (8 k - 1) y - 6 = 0$ are perpendicular is