Answer:
Step-by-step explanation:
We are given that a line that is perpendicular bisector of the line whose end points are R(-1,6) and S(5,5).
To find the equation of perpendicular line then we have find the slope of line and a point through which perpendicular line is passing .
To find the point through which perpendicular line is passing then we have find the mid point of the line joining points R(-1,6) and S(5,5)
Mid point formula: The coordinates of mid point
![x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D%2Cy%3D%5Cfrac%7By_1%2By_2%7D%7B2%7D)
Let P is the mid point of the line joining the points R (-1,6) and S(5,5)
Therefore, the coordinates of mid point P by using mid point formula
![x=\frac{-1+5}{2},y=\frac{6+5}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-1%2B5%7D%7B2%7D%2Cy%3D%5Cfrac%7B6%2B5%7D%7B2%7D)
![x=2,y=\frac{11}{2}](https://tex.z-dn.net/?f=x%3D2%2Cy%3D%5Cfrac%7B11%7D%7B2%7D%20)
The coordinates of mid point P
of the line joining points R and S
Slope of line RS,![m_1=\frac{y_2-y_1}{x_2-x_1}](https://tex.z-dn.net/?f=m_1%3D%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
![y_1=6,y_2=5,x_1=-1,x_2=5](https://tex.z-dn.net/?f=y_1%3D6%2Cy_2%3D5%2Cx_1%3D-1%2Cx_2%3D5)
Slope of line RS,![m_1=\frac{5-6}{5+1}=-\frac{1}{6}](https://tex.z-dn.net/?f=m_1%3D%5Cfrac%7B5-6%7D%7B5%2B1%7D%3D-%5Cfrac%7B1%7D%7B6%7D)
Slope of perpendicular line is opposite reciprocal of the line RS
Hence, the slope of perpendicular line ![m_2=-\frac{1}{m_1}](https://tex.z-dn.net/?f=m_2%3D-%5Cfrac%7B1%7D%7Bm_1%7D)
Slope of perpendicular line,![m_2=6](https://tex.z-dn.net/?f=m_2%3D6)
The perpendicular line is passing through the mid point P
because it bisect the line RS at mid point P.
The equation of perpendicular line passing through the point P with slope 6
![y-\frac{11}{2}=6(x-2)](https://tex.z-dn.net/?f=%20y-%5Cfrac%7B11%7D%7B2%7D%3D6%28x-2%29)
The equation of a line which is perpendicular to RS
![\frac{2y-11}{2}=6x-12](https://tex.z-dn.net/?f=%5Cfrac%7B2y-11%7D%7B2%7D%3D6x-12)
The equation of line which is perpendicular to the line RS is given by
![2y-11=12x-24](https://tex.z-dn.net/?f=2y-11%3D12x-24)
The equation of a line which is perpendicular to the line RS
![12x-2y=24-11](https://tex.z-dn.net/?f=12x-2y%3D24-11)
Hence, the required equation of a line which is perpendicular to the line RS
![12x-2y=13](https://tex.z-dn.net/?f=12x-2y%3D13)