Question

A line segment is bisected by line with the equation `6 y + 7 x = 4`. If one end of the line segment is at `(2 ,4 )`, where is the other end?

Answer 1

End point coordinates `((6/7),-6)`

Explanation:

It is assumed that the intersecting line is a perpendicular bisector.
`6y+7x=4`
`y=-(7/6)x+(4/6)=-(7/6)x+(2/3)`
Slope `= -(7/6)`
Slope of the line segment `=-1/-(7/6)=6/7`
Equation of line segment is
`(y-4)=(6/7)(x-2)`
`7y-28=6x-12`
`7x-6y=16`
Solving the equations we will get the coordinates of the mid point (intersection point).
`14x=20`
`x=(10/7)`
`10-6y=16`
`y=-1`
Mid point coordinates `((10/7),-1)`
But mid point `(4+y1)/2=-1` & `(2+x1)/2=10/7`
`y1=-6` & `x1=(20/7)-2=6/7`
End point coordinates `((6/7),-6)`

Answer 2

`(-(18)/5,-4/5)`

Explanation:

Let `L1` be the perpendicular bisector and `L2` (segment `AB`) be the bisected line, as shown in the figure.
Given that the equation of the bisector `L1` is `6y+7x=4`,
`=> y=-7/6x+2/3`
Let `m_1` be the slope of `L1`, and `m_2` the slope of `L2`,
`=> m_1=-7/6`
Recall that the product of the slopes of two perpendicular lines is `-1`,
`=> m_2xxm_1=-1, => m_2=6/7`
Given `A=(2,4)`,
`=>` equation of `L2` is : `y-4=6/7(x-2)`
`=> y=6/7x+(16)/7`
Set the equations of `L1 and L2` equal to each other to find the intersection point `M(x_m, y_m)`, which is also the midpoint of `L2`.
`=> -7/6x+2/3=6/7x+16/7`
`=> x=-4/5`
`=> y=-7/6x+2/3=-7/6xx(-4/5)+2/3=8/5`
`=> M(x_m,y_m)=(-4/5,8/5)`
Let the other end point of `L2` be `B(x,y)`,
Since `M` is the midpoint of `L2`,
`=> ((x+2)/2, (y+4)/2) = (-4/5,8/5)`
`=> (x,y)=(-(18)/5,-4/5)`