# A line segment is bisected by a line with the equation  4 y - 3 x = 2 . If one end of the line segment is at ( 7 , 5 ), where is the other end?

Jul 7, 2016

The other point is on the line $\textcolor{g r e e n}{4 y - 3 x = 5}$
(any point on this line will satisfy the given requirement)

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}}$The line $\textcolor{red}{4 y - 3 x = 2}$
and a point color(purple)(""(7,5))

Consider the vertical line through color(purple)(""(7,5))
This vertical line will intersect $\textcolor{red}{4 y - 3 x = 2}$ at color(red)(""(x=color(purple)(7),y))
where $\textcolor{red}{y}$ can be determined by solving
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{4 y - 3 \times \textcolor{p u r p \le}{7} = 2}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow \textcolor{red}{y = \frac{23}{4}}$

The distance from color(purple)(""(7,5)) to color(red)(""(7,23/4))
is $\textcolor{red}{\frac{23}{4}} - \textcolor{p u r p \le}{5} = \frac{3}{4}$

So a point color(green)(""(7,5+2xx3/4) = (7,13/2)) will be twice as far away from color(purple)(""(7,5)) as the point color(red)(""(7,23/4)) on the same vertical line

Note that any point on a line parallel to color(red)(4y-3x=2 through color(green)(""(7,13/2)) will also be twice as far away from color(purple)(""(7,5)) as the point from color(purple)(""(7,5)) on $\textcolor{red}{4 y - 3 x = 2}$ to that point.

If $\textcolor{red}{\text{L1}}$ parallel to $\textcolor{g r e e n}{\text{L2}}$
then $\triangle A B C \cong \triangle A D E$
$\rightarrow \left\mid A B \right\mid : \left\mid A D \right\mid = \left\mid A C \right\mid : \left\mid A E \right\mid$

Since $\textcolor{red}{4 y - 3 x = 2}$ has a slope of $\textcolor{red}{\frac{3}{4}}$

Our required line will also have a slope of $\textcolor{g r e e n}{\frac{3}{4}}$
and since it passes through (""(7,13/2))
using the slope-point form, we have
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{y - \frac{13}{2} = \frac{3}{4} \left(x - 7\right)}$
or
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{4} y - 3 x = 5$

Jul 7, 2016

The other end-pt. lies on the line given by the eqn. $4 y - 3 x = 5.$

#### Explanation:

Suppose that the other end-pt. is $P \left(X , Y\right)$.

Let the given end-pt. be $Q \left(7 , 5\right) ,$ and the given line be $L : 4 y - 3 x = 2.$

If $M$ is the mid-pt. of the segment $P Q$, then co-ords. of $M$ as obtained by using Section Formula for Mid-pt. are $= M \left(\frac{7 + X}{2} , \frac{5 + Y}{2}\right)$

Now, $P Q$ is bisected by $L$ at $M$, so, $M \in L .$

Therefore, co-ords. of $M$ must satisfy the eqn. of $L .$

Hence, $4 \left\{\frac{5 + Y}{2}\right\} - 3 \left\{\frac{7 + X}{2}\right\} = 2 \Rightarrow 20 + 4 Y - 21 - 3 X = 4 ,$ i.e., $4 Y - 3 X = 5 ,$ as Sir Alan P. has readily derived!

This shows that :
(i) the co-ords. of other end-pt. can not be uniquely derived under the given conds.
(ii) What we can say about it (the other end-pt.) is that it lies on $: 4 y - 3 x = 5.$ This eqn. represents a line $| |$ to $L$.