If m=1 is the slope of a line L, then the product of the slopes of non-parallel lines which are inclined at an angle of 60° with L is

Question

If m=1 is the slope of a line L, then the product of the slopes of non-parallel lines which are inclined at an angle of 60° with L is (A) 1 (B) -1 (C) √3 (D) -(1/2)

Tardigrade Admin · Accepted Answer

Let slope of line, which inclined with angle 60° with line L is 'n', so tan 60°=|(n-1/1+n)|=√3 ⇒ (n-1)2=3(n+1)2 ⇒ 2 n2+8 n+2=0, which roots are slope of required lines so product of slopes =1

Q. If $m = 1$ is the slope of a line $L$ , then the product of the slopes of non-parallel lines which are inclined at an angle of $6 0^{\circ}$ with $L$ is

1

-1

$3$

$- \frac{1}{2}$

Let slope of line, which inclined with angle $6 0^{\circ}$ with line $L$ is $^{'} n^{'}$ ,
so $tan 6 0^{\circ} = ∣ ∣ \frac{n - 1}{1 + n} ∣ ∣ = 3$
$\Rightarrow (n - 1)^{2} = 3 (n + 1)^{2}$
$\Rightarrow 2 n^{2} + 8 n + 2 = 0$ ,
which roots are slope of required lines so product of slopes $= 1$

Q. If m=1 is the slope of a line L, then the product of the slopes of non-parallel lines which are inclined at an angle of 60∘ with L is

1

-1

3​

−21​

Let slope of line, which inclined with angle 60∘ with line L is ′n′, so tan60∘=∣∣​1+nn−1​∣∣​=3​ ⇒(n−1)2=3(n+1)2 ⇒2n2+8n+2=0, which roots are slope of required lines so product of slopes =1

Q. If $m = 1$ is the slope of a line $L$ , then the product of the slopes of non-parallel lines which are inclined at an angle of $6 0^{\circ}$ with $L$ is

$3$

$- \frac{1}{2}$