Question

find the equation of a straight line passing through the point of intersection lines x-2y-4=0 and 4x-y-4=0 and parallel is the line 16x-11y+3=0?​

Answer 1

Answer:

The equation of the line passing the point of intersection given two lines is

16x-11y-28=0

Step-by-step explanation:

Given lines are

x-2y-4=0 and 4x-y-4=0

The equation of the line passing through the point of intersection of the given two lines can be written as

$x-2y-4+\lambda(4x-y-4)=0$

$(1+4\lambda)x+(-2-\lambda)y+(-4-4{\lambda})=0$.........(1)

As per given data, line (1) is parallel to 16x-11y+3=0

Then, their slopes are equal

$\frac{-(1+4\lambda)}{-2-\lambda}=\frac{-16}{-11}$

$\frac{-1-4\lambda}{-2-\lambda}=\frac{16}{11}$

$-11-44\lambda=-32-16\lambda$

$-28\lambda=-21$

$\bf\lambda=\frac{3}{4}$

The equation of the required line is

$x-2y-4+\frac{3}{4}(4x-y-4)=0$

$4(x-2y-4)+3(4x-y-4)=0$

$\implies\boxed{\bf\;16x-11y-28=0}$