Chapter 4, Problem 16MCQ

To determine

To verify the given pair of equations are paralle1, perpendicular or neither.

Answer to Problem 16MCQ

Given lines are parallel.

Explanation of Solution

Given:

  $\begin{array}{l}y=-6x+8\\ 3x+\frac{1}{2}y=-3\end{array}$

Concept used:

Parallel lines have same slope.

Two lines are perpendicular if and only if product of their slopes is $-1$ .

Calculation:

Let m₁ and m₂ be the slopes of $y=-6x+8\&3x+\frac{1}{2}y=-3$ respectively.

Now comparing $y=-6x+8$ with $y=mx+c$ , we get

  $\text{Slope}=m_1=-6$

Now consider,

  $\begin{array}{l}3x+\frac{1}{2}y=-3\\ \Rightarrow \frac{1}{2}y=-3x-3\\ \Rightarrow y=-6x-6\end{array}$

Now comparing $y=-6x-6$ with $y=mx+c$ , we get

  $\text{Slope}=m_2=-6$

Clearly m₁ and m₂ are equal, therefore given lines are parallel.