Chapter 0.1, Problem 41E

(a)

To determine

To find: Find the value of $k$ for which two lines $2x+ky=3$ and $x+y=1$ .

Answer to Problem 41E

When given lines are parallel then $k=-2$ .

Explanation of Solution

Given information:

Given equations are $2x+ky=3$ and $x+y=1$ .

Concept used: Condition when two lines are parallel $m_1=m_2$ .

Calculation:

If two line are parallel, then their slopes will be equal.

Two equations are given as $2x+ky=3$ , $x+y=1$ .

The general equation of line is $y=mx+c$ .

Compare the line $2x+ky=3$ with the general equation the slope will be $m_1=\frac{-2}{k}$

Now compare the second equation with the general equation of line the slope will be $m_2=1$ .

If two lines are parallel then their slope will be equal that is $m_1=m_2$ .

Substitute the values of $m_1=\frac{-2}{k}$ and $m_2=1$ it gives:

  $\begin{array}{l}\frac{-2}{k}=1\\ -2=k\\ k=-2\end{array}$

If two lines are parallel then $k=-2$ .

(b)

To determine

To find: Find the value of $k$ for which two lines $2x+ky=3$ and $x+y=1$ are perpendicular.

Answer to Problem 41E

When the given lines are perpendicular then $k=2$ .

Explanation of Solution

Given information:

Given equations are $2x+ky=3$ and $x+y=1$ .

Concept used: Condition when two lines are perpendicular then $m_1\cdot m_2=-1$ .

Calculation:

Two equations are given as $2x+ky=3$ , $x+y=1$ .

The general equation of line is $y=mx+c$ .

Compare the line $2x+ky=3$ with the general equation the slope will be $m_1=\frac{-2}{k}$

Now compare the second equation with the general equation of line the slope will be $m_2=1$ .

If two lines are perpendicular, then the product of their slope is equal to $-1$ .

  $m_1\cdot m_2=-1$

Substitute the values of $m_1=\frac{-2}{k}$ and $m_2=1$ it gives:

  $\begin{array}{l}\frac{-2}{k}\cdot 1=-1\\ -2=-k\\ k=2.\end{array}$

If two lines are perpendicular then $k=2$ .