# A line segment is bisected by a line with the equation  - 3 y + 5 x = 2 . If one end of the line segment is at ( 7 , 9 ), where is the other end?

Nov 4, 2016

Long explanation !!

#### Explanation:

Slope ${m}_{1}$ (say) of the line $- 3 y + 5 x = 2$ is :
$- 3 y = - 5 x + 2$ ...............(i)
$\therefore y = \frac{5}{3} x - \frac{2}{3}$
$\therefore {m}_{1} = \frac{5}{3.}$

Let slope of the line whose one end is $\left(7 , 9\right)$ be ${m}_{2}$ (say).
$\therefore {m}_{1} \times {m}_{2} = - 1.$ [The two lines are perpendicular to each other].

$\therefore \frac{5}{3} \times {m}_{2} = - 1$
$\therefore {m}_{2} = - \frac{3}{5.}$

$\therefore$ The equation of the line whose one end is $\left(7 , 9\right)$ is :
$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$
$\therefore$ $\therefore \left(y - 9\right) = - \frac{3}{5} \left(x - 7\right)$
$\therefore 3 x + 5 y = 66.$ is the equation. ..............(ii)

Now, solving equations (i) & (ii), we get the value $\left(x , y\right)$ which represents the midpoint of the line whose one end is $\left(7 , 9\right)$.

Now to find out the coordinates say, $\left(a , b\right)$ of the other end, use Distance-Section formula.

:.d=sqrt[(y_2-y_1)^2+(x_2-x_1)^2.

:.sqrt[(x-a)^2+(y-b)^2]=sqrt[(x-7)^2+(y-9)^2.

Here, $\left(x , y\right)$ is midpoint & $\left(a , b\right)$ is the coordinate of the other required end.

Now, I leave it to you. Just put the values and solve.
Best of Luck.