Question

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

Answer 1

Given
$5x-y-4=0$.....................$eq1$
and
$3x+4y-4=0$.......................$eq2$

Let $A(c,d);B(a,b)andP(1,5)$

$AP=PB$

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

$\frac{d+b}{2}=5⟹(d+b)=10$

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

Putting (c,d) in $eq1$

$5x-y-4=0eq:\left(3\right)$

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

$⟹5a+b=6$

Putting (a,b) in $eq2$

$3x+4y-4=0$

$=3a+4b=4eq:4$

Multiplying $eq3$ with $4$ and subtracting from $eq2$ we get
$a=\frac{20}{17}$

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

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

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

Equation of $AB$ is

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

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

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

$83x+3y-98=0$

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