Question

A. Show that the vector v = ai + bj is perpendicular to the line ax + by = c by establishing that the slope of v is the negative reciprocal of the slope of the given line.
b. Determine the slope of the vector : v = ai + bj

Answer 1

Answer:

Part A:

$m_1m_2=-1$

$\frac{b}{a}(\frac{-a}{b})=-1\\-1=-1$

Hence proved that Vector= ai + bj is perpendicular to the line ax + by = c.

Part B:

Slope of vector = $\frac{b}{a}$

Step-by-step explanation:

Condition for perpendicular is:

$m_1m_2=-1$

Part A:

Consider the vector v = ai + bj

x component of vector=a

y component of vector=b

Slope of vector=$m_1=\frac{y}{x}=\frac{b}{a}$

Consider the line ax + by = c:

Rearranging the equation:

ax+by=c

by=c-ax

y=$\frac{-ax}{b}+\frac{c}{b}$

According to general equation of line: $y=mx+c$

Where m is the slope

In our case the slope of above line is:

$m_2=\frac{-a}{b}$

According to the condition of perpendicular:

$m_1m_2=-1$

$\frac{b}{a}(\frac{-a}{b})=-1\\-1=-1$

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=$\frac{-ax}{b}+\frac{c}{b}$

According to  general equation of line: $y=mx+c$

Where m is the slope

Slope of given line=m=$\frac{-a}{b}$

negative reciprocal of the slope of the given line = $\frac{b}{a}$

Slope of vector = $\frac{b}{a}$