# A line segment is bisected by a line with the equation  2 y -4 x = 1 . If one end of the line segment is at (3 ,8 ), where is the other end?

Oct 21, 2016

The other end of the line segment is (4.2, 7.4)

#### Explanation:

Let the other end of the line segment be $\left({x}_{1} , {y}_{1}\right)$
So the mid-point of $\left(3 , 8\right)$ and $\left({x}_{1} , {y}_{1}\right)$ is $\left(\frac{{x}_{1} + 3}{2} , \frac{{y}_{1} + 8}{2}\right)$
The mid point belongs to the line $2 y - 4 x = 1$
The slope of this line is$= 2$

Sustituting the values of the mid point in this equation
$\frac{2 \left({y}_{1} + 8\right)}{2} - 4 \frac{\left({x}_{1} + 3\right)}{2} = 1$
Simplifying the equation ${y}_{1} + 8 - 2 \left({x}_{1} + 3\right) = 1$
${y}_{1} + 8 - 2 {x}_{1} - 6 = 1$
${y}_{1} - 2 {x}_{1} = 1 - 8 + 6 = - 1$

We need the equation of the line segment
the slope $m = - \frac{1}{2}$ since the two lines are perpendicular and the product of the slopes is ${m}_{1} \cdot {m}_{2} = - 1$

so the equation is $\frac{{y}_{1} - 8}{{x}_{1} - 3} = - \frac{1}{2}$
$2 {y}_{1} + {x}_{1} = 19$
${y}_{1} - 2 {x}_{1} = - 1$
Solving we get $\left({y}_{1} = 2 {x}_{1} - 1\right)$
Substituting in other equation
$2 \left(2 {x}_{1} - 1\right) + {x}_{1} = 19$
$4 {x}_{1} + {x}_{1} = 21$ $\implies$${x}_{1} = \frac{21}{5} = 4.2$
and ${y}_{1} = - 1 + 2 \cdot 4.2 = 7.4$

So the point is (4.2,7.4)