Answer with Step-by-step explanation:
We are given that a line passing through the point (p-4,2) and (-2,9) and other line passing through the point (p,-1) and (-1,1).
Slope-formula:[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
By using the formula
Slope of line which passing through the point (p-4,2) and (-2,9)
[tex]m_1=\frac{9-2}{-2-p+4}=\frac{7}{2-p}[/tex]
Slope of other line which passing through the point (p,-1) and (-1,1)
[tex]m_2=\frac{1+1}{-1-p}=\frac{2}{-1-p}[/tex]
When two lines are parallel then their slopes are equal
a.[tex]m_1=m_2[/tex]
[tex]\frac{7}{2-p}=\frac{2}{-1-p}[/tex]
[tex]-7-7p=4-2p[/tex]
[tex]-7-4=-2p+7p[/tex]
[tex]5p=-11[/tex]
[tex]p=\frac{-11}{5}[/tex]
b.If the two lines are perpendicular then their slopes is opposite reciprocal to each other.
[tex]m_1=-\frac{1}{m_2}[/tex]
[tex]\frac{7}{2-p}=-\frac{-1-p}{2}[/tex]
[tex]\frac{7}{2-p}=\frac{1+p}{2}[/tex]
[tex]14=(1+p)(2-p)[/tex]
[tex]14=2-p+2p-p^2[/tex]
[tex]p^2+14=p+2[/tex]
[tex]p^2-p+14-2\implies p^2-p+12=0[/tex]
It is quadratic equation in variable p
[tex]D=b^2-4ac[/tex]
[tex]D=(-1)^2-4(1)(12)=1-48=-47<0[/tex]
The root of equation are imaginary which is not possible .
If the lines are perpendicular then the value of p does not exist.
