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 $(1, 5)$ .If the equation is $83 x + 3 y - c = 0$ . Then, what is the value of $c$ ?

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Solution

Given
$5 x - y - 4 = 0$ ..................... $e q 1$
and
$3 x + 4 y - 4 = 0$ ....................... $e q 2$
Let $A (c, d); B (a, b) a n d P (1, 5)$

$A P = P B$

$\frac{c + a}{2} = 1 = (c + a) = 2$

$\frac{d + b}{2} = 5 ⟹ (d + b) = 10$

$c = 2 - a$ and $d = 10 - b$

Putting (c,d) in $e q 1$
$5 x - y - 4 = 0 e q : (3)$

$= 5 (2 - a) - (10 - b) - 4 = 0$

$⟹ 5 a + b = 6$

Putting (a,b) in $e q 2$
$3 x + 4 y - 4 = 0$

$= 3 a + 4 b = 4 e q : 4$

Multiplying $e q 3$ with $4$ and subtracting from $e q 2$ we get
$a = \frac{20}{17}$

$4 b = 4 - 3 \times \frac{20}{17}$

$b = \frac{2}{17}$

$B (\frac{20}{17}, \frac{2}{17})$ and $P (1, 5)$

Equation of $A B$ is
$y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$

$y - \frac{2}{17} = \frac{5 - \frac{2}{17}}{1 - \frac{20}{17}} (x - \frac{20}{17})$

$y - \frac{2}{17} = - \frac{83}{3} x + \frac{1660}{51}$

$83 x + 3 y - 98 = 0$
Comparing with $83 x - 35 y - c = 0$ we get $c = 99$

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Similar questions

5 x - y + 4 = 0

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5 x - y - 4 = 0

3 x + 4 y - 4 = 0

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

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