If k_1 k_2 are the two values of k such tha | vedclass

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

Question

(D) The slope of the line $y - 3kx + 4 = 0$ is $m_1 = 3k$. The slope of the line $(2k - 1)x - (8k - 1)y - 6 = 0$ is $m_2 = \frac{2k - 1}{8k - 1}$. Sin

Vedclass · Accepted Answer

$6x + y = 0$

Vedclass · Answer

(D) The slope of the line $y - 3kx + 4 = 0$ is $m_1 = 3k$. The slope of the line $(2k - 1)x - (8k - 1)y - 6 = 0$ is $m_2 = \frac{2k - 1}{8k - 1}$. Since the lines are perpendicular,$m_1 	imes m_2 = -1$. $3k 	imes \left(\frac{2k - 1}{8k - 1}ight) = -1$. $6k^2 - 3k = -8k + 1$. $6k^2 + 5k - 1 = 0$. $(6k - 1)(k + 1) = 0$. Thus,$k = \frac{1}{6}$ or $k = -1$. Given $k_1 > k_2$,we have $k_1 = \frac{1}{6}$ and $k_2 = -1$. The slope of the required line is $m = \frac{k_2}{k_1} = \frac{-1}{1/6} = -6$. The equation of the line passing through $(k_1, k_2) = (\frac{1}{6}, -1)$ with slope $m = -6$ is: $y - (-1) = -6(x - \frac{1}{6})$. $y + 1 = -6x + 1$. $6x + y = 0$.

If $k_1 > k_2$ are the two values of $k$ such that the lines $y - 3kx + 4 = 0$ and $(2k - 1)x - (8k - 1)y - 6 = 0$ are perpendicular,then the equation of the line passing through $(k_1, k_2)$ and having the slope $\left(\frac{k_2}{k_1}\right)$ is

Similar Questions

The number of integral values of $m$,for which the $x$-coordinate of the point of intersection of the lines $3x + 4y = 9$ and $y = mx + 1$ is an integer,is:

If the equation of the straight line passing through the point of intersection of $x+2y-19=0$ and $x-2y-3=0$ and which is at a perpendicular distance of $5$ units from the point $(-2,4)$ is $5x+by+c=0$,then a possible value of $5+b+c$ is

If a straight line is passing through the point of intersection of the lines $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ and making equal non-zero intercepts on the coordinate axes,then the area (in sq. units) of the triangle formed by this line with the coordinate axes is

The equation of the line joining the point $(3, 5)$ to the point of intersection of the lines $4x + y - 1 = 0$ and $7x - 3y - 35 = 0$ is equidistant from the points $(0, 0)$ and $(8, 34)$.

Consider the fourteen lines in the plane given by $y=x+r$ and $y=-x+r$,where $r \in \{0, 1, 2, 3, 4, 5, 6\}$. The number of squares formed by these lines,whose sides are of length $\sqrt{2}$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module