If the two pairs of lines $3 x^{2}$ $- 5 x y$ $+ 2 y^{2} = 0$ and $6 x^{2} - x y + k y^{2} = 0$ have one line in common, then $k^{2} + 7 k - 10$ $=$

0

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

- 20

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

- 10

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

20

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

Open in App

Solution

The correct option is B $- 20$
$3 x^{2} - 5 x y + 2 y^{2} = 0$
$t = \frac{y}{x},$
$2 t^{2} - 5 t + 3 = 0$
$t = \frac{5 \pm \sqrt{25 - 24}}{4}$
$t = \frac{5 \pm 1}{4}$
$t = \frac{3}{2}, 1$
given $3 x^{2} - 5 x y + 2 y^{2} = 0$ and $6 x^{2} - x y + k y^{2} = 0$ have one line in common
So, If $t = \frac{3}{2},$
$k t^{2} - 1 t + 6 = 0$
$k \frac{9}{4} - \frac{3}{2} + 6 = 0$
$k = - 2$ ----(1)
If, $t = 1$ , $k - 1 + 6 = 0$
$k = - 5$ ----(2)
When $k = - 2,$ $k^{2} + 7 k - 10 = 4 - 14 - 10$
$= - 20$
When, $k = - 5$ $k^{2} + 7 k - 10 = 25 - 35 - 10$
$k^{2} + 7 k - 10 = - 20$

Suggest Corrections

Similar questions

Q. If the pair of lines

3 x^{2} - 5 x y + p y^{2} = 0

and

6 x^{2} - x y - 5 y^{2} = 0

have one line in common, then

p =

If the acute angles between the pairs of lines $3 x^{2} - 7 x y + 4 y^{2} = 0$ and $6 x^{2} - 5 x y + y^{2} = 0$ be $θ_{1}$ and $θ_{2}$ respectively, then

Q. Type of quadrilateral formed by the two pairs of lines

6 x^{2} - 5 x y - 6 y^{2} = 0

and

6 x^{2} - 5 x y - 6 y^{2} + x + 5 y - 1 = 0

Q. For the pair of straight lines given by the equation

3 x^{2} - 5 x y + 2 y^{2} - 11 x + 9 y + 10 = 0

which of the following condition holds?

Q. The quadrilateral formed by the two pair of lines

6 x^{2} - 5 x y - 6 y^{2} = 0

and

6 x^{2} - 5 x y - 6 y^{2} + x + 5 y - 1 = 0

is a

Join BYJU'S Learning Program

If the two pairs of lines 3x2 −5xy +2y2=0 and 6x2−xy+ky2=0 have one line in common, then k2+7k−10 =

If the two pairs of lines $3 x^{2}$ $- 5 x y$ $+ 2 y^{2} = 0$ and $6 x^{2} - x y + k y^{2} = 0$ have one line in common, then $k^{2} + 7 k - 10$ $=$