Question

Point A(7,-3) and B(1,9), find:
a. Slope of AB.
b. Equation of line perpendicular bisector of the line AB.
c. He value of 'p' of (-2,p) lies on it.

Answer 1

$\textsf{Given points are A(7,-3) and B(1,9)}$

$\textsf{Slope of line joining }\mathsf{(x_1,y_1)}\textsf{ and }\mathsf{(x_2,y_2)}\textsf{ is }\mathsf{\frac{y_2-y_1}{x_2-x_1}}$

$\textsf{Slope of line joining }\mathsf{(7,-3)}\textsf{ and }\mathsf{(1,9)}\textsf{ is }\mathsf{\frac{9+3}{1-7}=\frac{12}{-6}=-2}$

$\textsf{The midoint of the line joining }\mathsf{(x_1,y_1)}\textsf{ and }\mathsf{(x_2,y_2)}\textsf{ is }\mathsf{(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})}$

$\textsf{The midoint of the line joining }\mathsf{(7,-3)}\textsf{ and }\mathsf{(1,9)}\textsf{ is }\mathsf{(\frac{7+1}{2},\frac{9-3}{2})=(4,3)}$

$\textsf{Slope of line perpendicular to the}$$\;\textsf{line jpoining (7,-3) and (1,9) is }\;\mathsf{\frac{1}{2}}$

$\textsf{Equation of perpendicular bisector is }$

$\mathsf{y-y_1=m(x-x_1)}$

$\mathsf{y-3=\frac{1}{2}(x-4)}$

$\mathsf{2y-6=x-4}$

$\implies\boxed{\mathsf{x-2y+2=0}}$

$\textsf{Also, the point (-2,p) lies on the line x-2y+2=0}$

$\textsf{-2-2p+2=0}$

$\implies\boxed{\textsf{p=0}}$