Solution:
Given the graph:
(a) The coordinates of P(x,y) and Q(x,y) are;
[tex]P(0,3)\text{ and }Q(-2,0)[/tex]
(b) The gradient, m of the line segment is;
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ \\ \text{ Where }x_1=0,y_1=3,x_2=-2,y_2=0 \end{gathered}[/tex]
Thus;
[tex]\begin{gathered} m=\frac{0-3}{-2-0} \\ \\ m=\frac{3}{2} \end{gathered}[/tex]
(c) The equation of the line is;
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ \\ \text{ Where }x_1=0,y_1=3,m=\frac{3}{2} \end{gathered}[/tex]
Thus;
[tex]\begin{gathered} y-3=\frac{3}{2}(x-0) \\ \\ y=\frac{3}{2}x+3 \end{gathered}[/tex]
(d) The length of the line segment PQ is;
[tex]\begin{gathered} PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ \text{ Where }x_1=0,y_1=3,x_2=-2,y_2=0 \end{gathered}[/tex]
Thus;
[tex]\begin{gathered} PQ=\sqrt{(-2-0)^2+(0-3)^2} \\ \\ PQ=\sqrt{13} \end{gathered}[/tex]
(e) The midpoint, MP of the line segment is;
[tex]\begin{gathered} MP=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ \\ \text{ Where }x_1=0,y_1=3,x_2=-2,y_2=0 \end{gathered}[/tex]
Thus;
[tex]\begin{gathered} MP=(\frac{0+(-2)}{2},\frac{3+0}{2}) \\ \\ MP=(-1,\frac{3}{2}) \end{gathered}[/tex]
