Respuesta :
We will solve as follows:
*
We first determine the slope of AB, that is:
[tex]m_{AB}=\frac{0-2}{-3-1}\Rightarrow m_{AB}=\frac{1}{2}[/tex]And using the properties of perpendicular lines, we have the following:
[tex]m_{CD}=-\frac{1}{m_{AB}}[/tex]Where mCD is the slope of the points C & D, therefore:
[tex]m_{CD}=-\frac{1}{\frac{1}{2}}\Rightarrow m_{CD}=-2[/tex]So, now we find the slope of CD using its coordinates, that is:
[tex]m_{CD}=\frac{k-3}{3-5}\Rightarrow m_{CD}=-\frac{k-3}{2}[/tex]Now, we equal both values of mCD:
[tex]-\frac{k-3}{2}=-2[/tex]Now, we solve for k:
[tex]\Rightarrow k-3=4\Rightarrow k=7[/tex]So, the points such that AB is perpendicular to CD are A(1,2), B(-3, 0), C(5, 3) & D(3, 5).
k = 7.
**
We determine the slope of BC:
[tex]m_{BC}=\frac{3-0}{5-(-3)}\Rightarrow m_{BC}=\frac{3}{8}[/tex]From this, we have that the slope of AD is:
[tex]m_{AD}=-\frac{1}{\frac{3}{8}}\Rightarrow m_{AD}=-\frac{8}{3}[/tex]And now, we calculate the slope of AD using its coordinates, that is:
[tex]m_{AD}=\frac{k-2}{3-1}\Rightarrow m_{AD}=\frac{k-2}{2}[/tex]Now, we equal both values of the slope of AD, that is:
[tex]\frac{k-2}{2}=-\frac{8}{3}[/tex]Now, we solve for k:
[tex]k-2=-\frac{16}{3}\Rightarrow k=-\frac{10}{3}[/tex]So, we have that the points such that AD is perpendicular to BC are A(1, 2), B(-3, 0), C(5, 3) & D(3, -10/3).
k = -10/3.
