# A line segment is bisected by a line with the equation  5 y + 2 x = 1 . If one end of the line segment is at (6 ,4 ), where is the other end?

Jul 11, 2017

2nd endpoint: $\left(\frac{50}{29} , - \frac{194}{29}\right)$

#### Explanation:

Given: $5 y + 2 x = 1 \text{ and line segment endpoint } \left(6 , 4\right)$.

Line segment is bisected means cut in half. At the intersection point (the midpoint of the segment), the two lines will be perpendicular. This means their slopes will be negative reciprocals of each other.

1. Put the line equation of the bisecting line in slope-intercept form $y = m x + b$, so we can find it's slope:

5y = -2x + 1; " "$y = - \frac{2}{5} x + \frac{1}{5}$

1. Find the slope of the line segment ${m}_{s e g m e n t} = - \frac{1}{m} = \frac{5}{2}$

2. Find the equation of the line segment that goes through the point $\left(6 , 4\right)$:

$y = \frac{5}{2} x + b$

$4 = \frac{5}{2} \cdot \frac{6}{1} + b$

$4 = 15 + b$

$b = 4 - 15 = - 11$

${y}_{s e g m e n t} = \frac{5}{2} x - 11$

3. Find the midpoint of the line segment - the intersection of the two lines, by setting the two equations equal:

$- \frac{2}{5} x + \frac{1}{5} = \frac{5}{2} x - 11 \text{ Move like terms to the same side}$

$\frac{1}{5} + 11 = \frac{2}{5} x + \frac{5}{2} x \text{ Find common denominators}$

$\frac{1}{5} + \frac{11}{1} \cdot \frac{5}{5} = \frac{2}{5} \cdot \frac{2}{2} x + \frac{5}{2} \cdot \frac{5}{5} x$

$\frac{56}{5} = \frac{29}{10} x \text{ Multiply by the reciprocal}$

${x}_{m} = {x}_{\text{midpoint}} = \frac{56}{5} \cdot \frac{10}{29} = \frac{56}{1} \cdot \frac{2}{29} = \frac{112}{29}$

${y}_{m} = {y}_{\text{midpoint}} = \frac{5}{2} \cdot \frac{112}{29} - \frac{11}{1} = \frac{5}{1} \cdot \frac{56}{29} - \frac{11}{1} \cdot \frac{29}{29}$

${y}_{m} = - \frac{39}{29}$

4. Use the midpoint formula to find the 2nd endpoint:

$\left({x}_{m} , {y}_{m}\right) = \left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$

$\left(\frac{112}{29} , - \frac{39}{29}\right) = \left(\frac{6 + {x}_{2}}{2} , \frac{4 + {y}_{2}}{2}\right)$

112/29 = (6+ x_2)/2; " " -39/29 = (4 + y_2)/2

$$112/29 *2/1 = 6 + x_2; " " -39/29 * 2/1 = 4 + y_2

224/29 = 6 + x_2; " " -78/29 = 4 + y_2

x_2 = 224/29 - 6/1 *29/29; " " y_2 = -78/29 - 4/1 * 29/29

x_2 = 50/29; " " y_2 = -194/29


2nd endpoint: $\left(\frac{50}{29} , - \frac{194}{29}\right)$

CHECK:

1. The 2nd endpoint should lie on the line segment:
$- \frac{194}{29} = \frac{5}{2} \cdot \frac{50}{29} - 11$

$- \frac{194}{29} = \frac{5}{1} \cdot \frac{25}{29} - \frac{11}{1} \cdot \frac{29}{29}$

$- \frac{194}{29} = - \frac{194}{29}$

2. Both segments to the midpoint should be equal in length:

$d = \sqrt{{\left(6 - \frac{112}{29}\right)}^{2} + {\left(4 - - \frac{39}{29}\right)}^{2}} \approx 5.756555483$

$d = \sqrt{{\left(\frac{50}{29} - \frac{112}{29}\right)}^{2} + {\left(- \frac{194}{29} - - \frac{39}{29}\right)}^{2}} \approx 5.756555483$