Question

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

• A. $83x-35y+92=0$
• B. $83x-85y+92=0$
• C. $83x+35y+92=0$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

$5x_1-y_1+4=0$

$3x_2+4y_2-4=0$

Also $\frac{x_1+x_2}{2}=1$ & $\frac{y_1+y_2}{2}=5$

$x_1=2-x_2$ $y_1=10-y_2$

$-5x_2+4+y_2=0]\times 3$

$3x_2-4+4y_2=0]\times 5$

$y_2=8/23x_2=\frac{20}{23}$

$y_1=\frac{222}{23}{x}_1=\frac{26}{23}$

for equation of line

$(y-\frac{222}{23})=\frac{\frac{8-222}{23}}{\frac{20-26}{23}}(x-\frac{26}{23})\left[use\right(y-y_1)=m(x-x_1\left)\right]$

$107y-3y-92=0$