The value of k ( 0),such that the angle betw | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The slopes of the lines $4x - y + 7 = 0$ and $kx - 5y - 9 = 0$ are $m_1 = 4$ and $m_2 = \frac{k}{5}$,respectively.Given the angle $	heta = 45^{\c

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The slopes of the lines $4x - y + 7 = 0$ and $kx - 5y - 9 = 0$ are $m_1 = 4$ and $m_2 = \frac{k}{5}$,respectively.Given the angle $	heta = 45^{\circ}$ between the lines,we use the formula:$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$$	an 45^{\circ} = \left| \frac{4 - \frac{k}{5}}{1 + 4 \cdot \frac{k}{5}} ight|$$1 = \left| \frac{20 - k}{5 + 4k} ight|$This implies $\frac{20 - k}{5 + 4k} = 1$ or $\frac{20 - k}{5 + 4k} = -1$.Case $1$: $20 - k = 5 + 4k$ $\Rightarrow 5k = 15$ $\Rightarrow k = 3$.Case $2$: $20 - k = -5 - 4k$ $\Rightarrow 3k = -25$ $\Rightarrow k = -\frac{25}{3}$.Since $k > 0$,the only valid solution is $k = 3$.

The value of $k (> 0)$,such that the angle between the lines $4x - y + 7 = 0$ and $kx - 5y - 9 = 0$ is $45^{\circ}$,is

Similar Questions

The equations of the lines passing through $(1,1)$ and making an angle of $45^{\circ}$ with the line $x+y=0$ are

The angle between the straight lines $x - y\sqrt{3} = 5$ and $\sqrt{3}x + y = 7$ is ............. $^\circ$.

If the lines $y=3x+1$ and $2y=x+3$ are equally inclined to the line $y=mx+4$,then the value of $m$ is equal to

If the line segment joining the points $(5, 2)$ and $(2, a)$ subtends an angle $\frac{\pi}{4}$ at the origin,then the absolute value of the product of all possible values of $a$ is:

The lines $p(p^2 + 1)x - y + q = 0$ and $(p^2 + 1)^2x + (p^2 + 1)y + 2q = 0$ are perpendicular to a common line for which value of $p$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module