Name: Condition for Concurrency of Three Lines
Uploaded: 2019-07-19
Duration: 10 min 26 s
Description: Well, this video beautifully explains the conditions for concurrency of three lines with some interesting examples. Come, let's learn and watch the video...!!

Question

A line is such that its segment between the straight lines $5 x - y - 4 = 0$ and $3 x + 4 y - 4 = 0$ is bisected at the point $(1, 5) .$ Obtain its equation.

Accepted Answer

Let $P_{1} P_{2}$ be the intercept between the lines $5 x - y - 4 = 0$ and $3 x + 4 y - 4 = 0 .$
Let $P_{1} P_{2}$ make an angle $θ$ with the positive $x -$ axis.

Here, $(x_{1}, y_{1}) = A (1, 5)$

So, the equation of the line passing through $A (1, 5)$ is

$\frac{x - x_{1}}{\cos θ} = \frac{y - y_{1}}{\sin θ}$

$\Rightarrow \frac{x - 1}{\cos θ} = \frac{y - 5}{\sin θ}$

$\Rightarrow \frac{y - 5}{x - 1} = \tan θ$

Let $A P_{1} = A P_{2} = r$

Then, the coordinates of $P_{1}$ and $P_{2}$ are given by

$\frac{x - 1}{\cos θ} = \frac{y - 5}{\sin θ} = r and \frac{x - 1}{\cos θ} = \frac{y - 5}{\sin θ} = - r$

So, the coordinates of $P_{1}$ and $P_{2}$ are

$(1 + r \cos θ, 5 + r \sin θ) and (1 - r \cos θ, 5 - r \sin θ)$ , respectively.

Clearly, $P_{1}$ and $P_{2}$ lie on $5 x - y - 4 = 0$ and $3 x + 4 y - 4 = 0$ , respectively.

$∴ 5 (1 + r \cos θ) - 5 - r \sin θ - 4 = 0 and 3 (1 - r \cos θ)$ $+ 4 (5 - r \sin θ) - 4 = 0$

$\Rightarrow r = \frac{4}{5 \cos θ - \sin θ} and r = \frac{19}{3 \cos θ + 4 \sin θ}$

$\Rightarrow \frac{4}{5 \cos θ - \sin θ} = \frac{19}{3 \cos θ + 4 \sin θ}$

$\Rightarrow 95 \cos θ - 19 \sin θ = 12 \cos θ + 16 \sin θ$

$\Rightarrow 83 \cos θ = 35 \sin θ$

$\Rightarrow \tan θ = \frac{83}{35}$

Thus, the equation of the required line is
$\frac{y - 5}{x - 1} = \tan θ$

$\Rightarrow \frac{y - 5}{x - 1} = \frac{83}{35}$ [Since slope $(m)$ of the line Joining any two points $(x_{1}, y_{1}),$ and $(x_{2}, y_{2})$ is $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ ]

$\Rightarrow 83 x - 35 y + 92 = 0$

A line is such that its segment between the straight lines $5 x - y - 4 = 0$ and $3 x + 4 y - 4 = 0$ is bisected at the point $(1, 5) .$ Obtain its equation.

Important Questions on The Straight Lines

Learn and Prepare for any exam you want !

A line is such that its segment between the straight lines 5x − y − 4 = 0 and 3x + 4y − 4 = 0 is bisected at the point (1, 5). Obtain its equation.

Important Questions on The Straight Lines

Learn and Prepare for any exam you want !

A line is such that its segment between the straight lines $5 x - y - 4 = 0$ and $3 x + 4 y - 4 = 0$ is bisected at the point $(1, 5) .$ Obtain its equation.