The line segment RT has coordinates at R(-4, -7) and T(6, -1). Find the slope of this line using the formula;
m= change in y-coordinate values / change in x-coordinate values
m= -1 --7 / 6--4
m= -1+7 /10
m=6/10 --------you can simplify to 3/5
The slope of segment RT is m1= 3/5
The line perpendicular segment RT will have a slope m2 where ;
m1*m2 = -1
Find m2 , which is the gradient of the line perpendicular to segment RT
3/5 * m2 = -1
m2= -5/3
This means the gradient of the perpendicula line to the line segment RT is ;
m2 = -5/3
Now using the slope m2 = -5/3 , the midpoint S, (1, -4) , and imaginary point on the line (x,y) then the equation will be;
m= change in y-values / change in x-values
-5/3 = y--4/x-1
-5/3 = y+4 /x-1
-5{x-1} = 3{y+4}
-5x+5 =3y +12 ------collect like terms and write in form of y= mx + c
-5x+5-12=3y
-5x-7 =3y
-5/3 x - 7/3 = y
y= -5/3 x - 7/3
y-intercept is - 7/3
Answer
-7/3