Chapter 4.1, Problem 44PPS

(a)

To determine

To Graph :

  $y=\frac{3}{4}x+1$

Explanation of Solution

Given Information:

Equation of line is

  $y=\frac{3}{4}x+1$

Concept Used:

The slope-intercept form of a line is :

  $\begin{array}{l}y=mx+c\\ where\\ m=slope\\ c=y-\mathrm{intercept}\end{array}$

Graph:

  $y=\frac{3}{4}x+1$

This equation is in slope-intercept form.

Here , $\begin{array}{l}m=\frac{3}{4}\\ c=1\end{array}$

The y-intercept is the point (0,1).

Now

Steps to draw the line :

1.Start at the origin and move 1 unit up. Draw a point at (0,1) = (0,c)

2.The slope is $\frac{3}{4}=\frac{a}{b}=\frac{rise}{run}$ . This means change in y is 3 = a and change in x is 4 = b.From the point (0,1) , go 3 units up and 4 units to the right.

You are now at (4,4) = (b,c+a). Draw a point here.

3. Continue this process. Now use a ruler and connect the points and extend the line in both directions.

  

(b)

To determine

To Draw:

Use protrector to draw a line perpendicular to the line drawn in (a)

Explanation of Solution

Concept Used:

Construction of perpendicular lines with the help of protractor:

a. Let l be the given line and A the given point on it.

  

b. Place the protractor on the line l such that its base line coincides with l, and its centre falls on A.

c. Mark a point B against the 90° mark on the protractor.

d. Remove the protractor and draw a line m passing through A and B.

Then line m is perpendicular to line l at A.

Graph:

Draw a line with the help of Protractor :

Keeping (0,1) as point A in the protractor and $y=\frac{3}{4}x+1$ as the line l.

  

(c)

To determine

To Find :

Equation of line drawn in part (b)

Explanation of Solution

Concept Used:

The slope-intercept form of a line is :

  $\begin{array}{l}y=mx+c\\ where\\ m=slope\\ c=y-\mathrm{intercept}\end{array}$

Calculation:

1.Find the slope

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\mathrm{.........................}[slope\_formula]\\ m=\frac{-3-1}{3-0}\mathrm{..........................}[(x_1,y_1)=(0,1)and(x_2,y_2)=(3,-3)]\\ m=-\frac{4}{3}\end{array}$

Choose (0,1) as the point through which line passes and find the y-intercept

  $\begin{array}{l}y=mx+c\mathrm{.......................}[(x,y)=(0,1);m=-\frac{4}{3}]\\ \Rightarrow 1=-\frac{4}{3}(0)+c\\ \Rightarrow c=1\end{array}$

Write the equation using Slope Intercept form

  $\begin{array}{l}y=mx+c\\ m=-\frac{4}{3}\\ c=1\\ \Rightarrow y=-\frac{4}{3}x+1\end{array}$

Hence, required equation of line is $y=-\frac{4}{3}x+1$

(d)

To determine

To Compare :

The slopes of lines in part (a) and (c)

Explanation of Solution

Given Information :

Equations of lines from part (a) and (c) :

  $\begin{array}{l}y=\frac{3}{4}x+\mathrm{1........................}(a)\\ y=-\frac{4}{3}x+\mathrm{1......................}(c)\end{array}$

Interpretation:

From part (a) and (c)

We have slopes of lines

  $\begin{array}{l}y=\frac{3}{4}x+1\\ m_1=\frac{3}{4}\\ y=-\frac{4}{3}x+1\\ m_2=-\frac{4}{3}\\ m_1\cdot m_2=\frac{3}{4}\cdot (-\frac{4}{3})\\ m_1\cdot m_2=-1\end{array}$

So, we have the relation between the slopes of perpendicular lines.

The product of slopes of two perpendicular lines is -1.