# A line segment is bisected by a line with the equation  -7 y + 5 x = 1 . If one end of the line segment is at (1 ,4 ), where is the other end?

May 11, 2018

I get $\left(\frac{157}{37} , - \frac{20}{37}\right)$

#### Explanation:

$- 7 y + 5 x = 1$

I find it less confusing this way:

$5 x - 7 y = 1$

The perpendicular family is gotten by swapping the coefficients on $x$ and $y$, negating one. The constant is gotten by plugging in the point $\left(1 , 4\right)$ on the perpendicular:

$7 x + 5 y = 7 \left(1\right) + 5 \left(4\right) = 27$

We find the meet by multiplying the first by 5 and the second by 7:

$25 x - 35 y = 5$

$49 x + 35 y = 7 \cdot 27 = 189$

$74 x = 194$

$x = \frac{194}{74} = \frac{97}{37}$

$y = \frac{1}{5} \left(27 - 7 \left(\frac{97}{37}\right)\right) = \frac{64}{37}$

If we call our endpoint E and our meet M we get an informal equation for the other endpoint F that's

$F = M - \left(E - M\right) = M + \left(M - E\right) = 2 M - E$

So our other endpoint is

$\left(2 \left(\frac{97}{37}\right) - 1 , 2 \left(\frac{64}{37}\right) - 4\right) = \left(\frac{157}{37} , - \frac{20}{37}\right)$

Check:

Let's see if we can get the grapher to graph it:

$- 7 y + 5 x = 1$

$\left(y - 4\right) \left(\frac{157}{37} - 1\right) = \left(x - 1\right) \left(- \frac{20}{37} - 4\right)$

$\left(- 7 y + 5 x - 1\right) \left(\left(y - 4\right) \left(\frac{157}{37} - 1\right) - \left(x - 1\right) \left(- \frac{20}{37} - 4\right)\right) = 0$

graph{ ( -7 y + 5x - 1) ( (y-4)(157/37 -1) - (x-1)( -20/37 -4 ) ) = 0 [-7.83, 12.17, -2.44, 7.56]}

Looks pretty good.