Question

Indicate the equation of the given line in standard form. The line that is the perpendicular bisector of the segment whose endpoints are R(-1, 6) and S(5, 5)

Answer 1

Equation of the line joining the points R(-1,6) and S(5,5)

Answer 2

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

$x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}$

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

$x=\frac{-1+5}{2},y=\frac{6+5}{2}$

$x=2,y=\frac{11}{2}$

The coordinates of mid point P$(2,\frac{11}{2})$ of the line joining points  R and S

Slope of line RS,$m_1=\frac{y_2-y_1}{x_2-x_1}$

$y_1=6,y_2=5,x_1=-1,x_2=5$

Slope of line RS,$m_1=\frac{5-6}{5+1}=-\frac{1}{6}$

Slope of perpendicular line is opposite reciprocal of the line RS

Hence, the slope of perpendicular line $m_2=-\frac{1}{m_1}$

Slope of perpendicular line,$m_2=6$

The perpendicular line is passing through the mid point P$(2,\frac{11}{2})$ because it bisect the line RS at mid point P.

The equation of perpendicular line passing through the point P with slope 6

$y-\frac{11}{2}=6(x-2)$

The equation of a line which is perpendicular to RS

$\frac{2y-11}{2}=6x-12$

The equation of  line which is perpendicular to the line RS is given by

$2y-11=12x-24$

The  equation of a line which is  perpendicular to the  line RS

$12x-2y=24-11$

Hence, the required equation of a line which is perpendicular to the line RS

$12x-2y=13$