Find the value of $k$ if the straight line $2 x + 3 y + 4 + k (6 x - y + 12) = 0$ is perpendicular to the line $7 x + 5 y - 4 = 0 .$

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Solution

Step 1: Find the slope of the lines
Given, the two lines are:
$\begin{array}{lc} l_{1} \to x (2 + 6 k) + y (3 - k) + 4 + 12 k = 0 & \dots (i) \\ l_{2} \to 7 x + 5 y - 4 = 0 & \dots (ii) \end{array}$
To find slope of line $l_{1}$ and $l_{2}$ , convert them into $y = m x + c$ form. Thus,
$\begin{matrix} x (2 + 6 k) + y (3 - k) + 4 + 12 k = 0 \\ \Rightarrow & y = - \frac{2 + 6 k}{3 - k} x - 4 - 12 k \end{matrix} \begin{matrix} \begin{matrix} 7 x + 5 y - 4 = 0 \\ \Rightarrow & y = - \frac{7}{5} x + \frac{4}{5} \end{matrix} \end{matrix}$
Comparing to the standard equation,
$m_{1} = - \frac{2 + 6 k}{3 - k} m_{2} = - \frac{7}{5}$
where $m_{1}$ and $m_{2}$ are the slopes of the lines $l_{1}$ and $l_{2}$ respectively.
Step 2: Use the relationship between slopes of perpendicular lines
We know that, if lines $l_{1}$ and $l_{2}$ are perpendicular.
Then,
$\begin{matrix} m_{1} \cdot m_{2} = - 1 \\ \Rightarrow & (- \frac{2 + 6 k}{3 - k}) . (\frac{- 7}{5}) = - 1 \\ \Rightarrow & 14 + 42 k = - 15 + 5 k \\ \Rightarrow & k = - \frac{29}{37} \end{matrix}$
Hence the required value of $k$ is $\frac{- 29}{37}$ .

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Similar questions

Find the value of k such that the line (k - 2) x + (k + 3) y - 5 = 0 is :

(i) perpendicular to the line 2x - y + 7 = 0

(ii) parallel to it.

Find the value of p if the lines, whose equations are 2x - y + 5 = 0 and px + 3y = 4 are perpendicular to each other.

2 x + 3 y + 4 + k (6 x - y + 12) = 0

7 x + 5 y - 4 = 0

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