Question

If the pair of straight lines 6x2 +14xy+ky2 =0 are at right angles, find k.

Answer 1

Step-by-step explanation:

[

a

h

h

b

g

f

]=0

m

1

+m

2

=

b

−2n

and m

1

m

2

=a/b=

k

4

m

1

+m

2

=

k

−6

and m

1

m

2

=

k

4

As one line bisects angle between co-ordinates

So, m

1

=1 or −1

If m

1

=1

m

2

=

k

4

⇒1+

k

4

=

k

−6

⇒1=

k

−10

⇒k=−10

Now, if m

1

=−1

m

2

=

k

−4

⇒−1−

k

4

=

k

−6

⇒−1=

k

−2

⇒k=2

∴k∈{2,−10}

Answer 2

Answer:

$6x^{2} + 14xy +ky^{2} = 0$

we know, the  standard equation of the pair of straight lines is

$ax^{2} + 2hxy +by^{2} = 0$

a = 6, h = 6 and b = k

We also know that the relation for the angle between the pair of straight line (θ) is

tanθ = $\frac{2\sqrt{h^{2} - ab}}{a+b}$ = $\frac{2\sqrt{6^{2} - 6k }}{6+k}$

Well, you can get the answer from here right?