You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- \frac{1}{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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)

Suggest Corrections

Similar questions

Q. A value of

k

such that the straight lines

y - 3 x + 4 = 0

and

(2 k - 1) x - (8 k - 1) y - 6 = 0

are perpendicular is

Find k, if the straight lines y-3kx+4=0,(2k-1)x-(8k-1) y-6=0

Q. Find the values of k for which the roots are real and equal in each of the following equations:

(i)

k x^{2} + 4 x + 1 = 0

(ii)

k x^{2} - 2 \sqrt{5} x + 4 = 0

(iii)

3 x^{2} - 5 x + 2 k = 0

(iv)

4 x^{2} + k x + 9 = 0

(v)

2 k x^{2} - 40 x + 25 = 0

(vi)

9 x^{2} - 24 x + k = 0

(vii)

4 x^{2} - 3 k x + 1 = 0

(viii)

x^{2} - 2 (5 + 2 k) x + 3 (7 + 10 k) = 0

(ix)

(3 k + 1) x^{2} + 2 (k + 1) x + k = 0

(x)

k x^{2} + k x + 1 = - 4 x^{2} - x

(xi)

(k + 1) x^{2} + 2 (k + 3) x + (k + 8) = 0

(xii)

x^{2} - 2 k x + 7 k - 12 = 0

(xiii)

(k + 1) x^{2} - 2 (3 k + 1) x + 8 k + 1 = 0

(xiv)

(2 k + 1) x^{2} + 2 (k + 3) x + (k + 5) = 0

(xvii)

4 x^{2} - 2 (k + 1) x + (k + 4) = 0

(xviii)

x^{2} - 2 (k + 1) x + k^{2} = 0

(xix)

k^{2} x^{2} - 2 (2 k - 1) x + 4 = 0

(xx)

(k + 1) x^{2} - 2 (k - 1) x + 1 = 0

(xxi)

2 x^{2} + k x + 3 = 0

(xxii)

k x (x - 2) + 6 = 0

(xxiii)

x^{2} - 4 k x + k = 0

(xxiv)

k x (x - 2 \sqrt{5}) + 10 = 0

(xxv)

p x (x - 3) + 9 = 0

(xxvi)

4 x^{2} + p x + 3 = 0

Q. The lines

x + y = 1,

(k - 1) x + (k^{2} - 7) y - 5 = 0

and

(k - 2) x + (2 k - 5) y = 0

are

Q. If x < 0, y < 0 such that xy = 1, then write the value of tan⁻¹ x + tan⁻¹ y.

Join BYJU'S Learning Program

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