# A line segment is bisected by a line with the equation  3 y - 7 x = 3 . If one end of the line segment is at (7 ,8 ), where is the other end?

Jul 2, 2016

The other point is at $\left(\frac{7}{29} , 10 \frac{26}{29}\right)$

#### Explanation:

Write the equation of the given line in standard form:

$3 y = 7 x + 3 \Rightarrow y = \frac{7 x}{3} + 1$

The gradient is $\frac{7}{3}$

We will have to assume that this line is the perpendicular bisector rather than just a bisector, else the question cannot be answered.

The line segment has a gradient of $- \frac{3}{7}$
and passes through the point $\left(7 , 8\right)$

Find its equation using: $y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - 8 = - \frac{3}{7} \left(x - 7\right)$

$y = \frac{- 3 x}{7} + 3 + 8 \Rightarrow y = \frac{- 3 x}{7} + 11$

The midpoint of the line segment is where the lines intersect. Solve them simultaneously by equating the y's.

$y = \frac{7 x}{3} + 1 \mathmr{and} y = \frac{- 3 x}{7} + 11$
$y = y \Rightarrow \frac{7 x}{3} + 1 = \frac{- 3 x}{7} + 11$

$\frac{7 x}{3} + \frac{3 x}{7} = 10 \text{ } \times 21$

$49 x + 9 x = 210$

$58 x = 210 \Rightarrow x = \frac{105}{29}$ This gives y as $\frac{274}{29}$

This point is the midpoint of the line segment.

$\frac{7 + {x}_{2}}{2} = \frac{105}{29} \text{ and } \frac{8 + {y}_{2}}{2} = \frac{274}{29}$

${x}_{2} = \frac{210}{29} - 7 \text{ and } {y}_{2} = \frac{548}{29} - 8$

The other point is at $\left(\frac{7}{29} , 10 \frac{26}{29}\right)$