Answer:
Part A:
[tex]m_1m_2=-1[/tex]
[tex]\frac{b}{a}(\frac{-a}{b})=-1\\-1=-1[/tex]
Hence proved that Vector= ai + bj is perpendicular to the line ax + by = c.
Part B:
Slope of vector = [tex]\frac{b}{a}[/tex]
Step-by-step explanation:
Condition for perpendicular is:
[tex]m_1m_2=-1[/tex]
Part A:
Consider the vector v = ai + bj
x component of vector=a
y component of vector=b
Slope of vector=[tex]m_1=\frac{y}{x}=\frac{b}{a}[/tex]
Consider the line ax + by = c:
Rearranging the equation:
ax+by=c
by=c-ax
y=[tex]\frac{-ax}{b}+\frac{c}{b}[/tex]
According to general equation of line: [tex]y=mx+c[/tex]
Where m is the slope
In our case the slope of above line is:
[tex]m_2=\frac{-a}{b}[/tex]
According to the condition of perpendicular:
[tex]m_1m_2=-1[/tex]
[tex]\frac{b}{a}(\frac{-a}{b})=-1\\-1=-1[/tex]
Hence proved that Vector= ai + bj is perpendicular to the line ax + by = c.
Part B:
Slope of vector is also calculated above.
Since the slope of vector is negative reciprocal of the slope of the given line:
According to equation of line ax + by = c
y=[tex]\frac{-ax}{b}+\frac{c}{b}[/tex]
According to general equation of line: [tex]y=mx+c[/tex]
Where m is the slope
Slope of given line=m=[tex]\frac{-a}{b}[/tex]
negative reciprocal of the slope of the given line = [tex]\frac{b}{a}[/tex]
Slope of vector = [tex]\frac{b}{a}[/tex]
