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).

Then the equation is $107x-3y-92=0$

Enter 0 if True and 1 if False.

Answer 1

$5x-y+4=\mathrm{0...}\left(i\right)$

$3x+4y-4=0...\left(ii\right)$

$\to$ Let the required line intersect line (i) & (ii) at point

$(\alpha ,\beta )$ & $(a_2,{\beta }_2)$ respectively.

$5{\alpha }_1-{\beta }_1+4=0\to {\beta }_1=5{\alpha }_1+\mathrm{4...}\left(iii\right)$

$3{\alpha }_2+4{\beta }_2-4=0\to {\beta }_2=\frac{4-3{\alpha }_2}{4}...\left(iv\right)$

It is given that the mid point of line bet${}^n$

$({\alpha }_1,{\beta }_1)$ & $({\alpha }_2,{\beta }_2)$ is $(1,5)$

$\frac{{\alpha }_1+{\alpha }_2}{2}=1$ & $\frac{{\beta }_1+{\beta }_2}{2}=5$

$\to$ using equation (iii) & (iv) we get ${\alpha }_1$ & ${\alpha }_2$

${\alpha }_1+{\alpha }_2=2$ and $20{\alpha }_1-3{\alpha }_2=20$

$\to$ Solving the above two equation, we get ${\alpha }_1$ & ${\alpha }_2$

${\alpha }_1=\frac{26}{23}$ & ${\alpha }_2=\frac{20}{23},{\beta }_1=\frac{222}{23}$

$\to$ the equation on of the required line passing

through $(1,5)$ & $(\alpha ,\beta )$ is

$y-5=\frac{{\beta }_1-5}{{\alpha }_1-1}(x-1)$

$\therefore 107x-3y-92=0$ Ans