Question:

If lines represented by $x + 3y - 6 = 0, 2x+y-4=0$ and $kx - 3y + 1 = 0$ are concurrent, then the value of $k$ is

Updated On: Apr 8, 2024
  • $\frac {6}{19}$
  • $\frac {19}{6}$
  • $\frac {-19}{6}$
  • $\frac {-6}{19}$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Given lines are,
$x+3 y-6=0$
$2 x+y-4= 0$
$k x-3 y+1=0$
these given lines will concurrent
when $\begin{vmatrix}1&3&-6\\ 2&1&-4\\ k&-3&1\end{vmatrix}=0$
Expand with respect to $R_{1}$ :
$1(1-12)-3(2+4 k)-6(-6-k)=0$
$\Rightarrow -11-6-12 k+36+6 k=0$
$\Rightarrow -6 k-17+36=0$
$\Rightarrow 6 k=19$
$\Rightarrow k=\frac{19}{6}$
Was this answer helpful?
1
0

Concepts Used:

Straight lines

A straight line is a line having the shortest distance between two points. 

A straight line can be represented as an equation in various forms,  as show in the image below:

 

The following are the many forms of the equation of the line that are presented in straight line-

1. Slope – Point Form

Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.

y – y0 = m (x – x0)

2. Two – Point Form

Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2)  are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes

The slope of P2P = The slope of P1P2 , i.e.

\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)

Hence, the equation becomes:

y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)

3. Slope-Intercept Form

Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by

y – c =m( x - 0 )

As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if

y = m x +c