Question

I need help with questions 4 and 5 on the math homework

Answer 1

Question 4

Recall the following facts:


1. If two lines are parallel, their slopes are the same

2. If two lines are perpendicular, the product of their slopes equals 1.

The line given to us is:

$y = \frac{3}{4} x + 12$
By comparing to ;

$y = mx + c$

The slope is
$m = \frac{3}{4}$
Now let us compare this to:

$y = \frac{4}{3} x - 2$
which has slope
$= \frac{4}{3}$
Now let us check to see if the two are parallel,

$\frac{3}{ 4} \neq \frac{4}{3}$
Since the two slopes are not the same, they are not parallel.


Now let us check to see if they are perpendicular

$\frac{3}{4} \times \frac{4}{3} \neq - 1$
Since their product is not -1, the two lines are not perpendicular.

Hence ,
$y = \frac{3}{4} x + 12$
is neither parallel or perpendicular to
$y = \frac{4}{3} x - 2$

The next equation is

$y = - \frac{4}{3} x + 5$
The slope of this equation is
$= - \frac{4}{3}$
Since,

$\frac{3}{4} \times - \frac{4}{3} = - 1$

The equation
$y = \frac{3}{4} x + 12$
is perpendicular to

$y = - \frac{4}{3} x + 5$
The next equation is

$y = \frac{3}{4} x$
Since the slope if the two are equal, that is

$\frac{3}{4} = \frac{3}{4}$
the two equations are parallel.





The next equation is

$y = - \frac{4}{3} x - 6$
Since

$\frac{ 3}{4} \times \frac{ - 4}{3} = -1$
the two equations are perpendicular.


Question 5.

The given line in the graph passes through,

(10,7), (-8,-5) and (1,1).


Using any two points we can determine the slope,

$= \frac{7 - - 5}{10 - - 8}$
$= \frac{7 +5 }{10 + 8}$
$= \frac{12}{18} = \frac{2}{3}$


The line parallel to this line which passes through (5,-1), also has slope

$= \frac{2}{3}$
The equation of this line is

$y - - 1 = \frac{2}{3} (x - 5)$
This implies that,

$y + 1 = \frac{2}{3} x - \frac{10}{3}$
This simplifies to

$y = \frac{2}{3} x - \frac{13}{3}$

To find any point on this line, choose any value for x and solve the corresponding y value. So when
$x = 2$
$y = \frac{2}{3} \times 2 - \frac{13}{3} = - 3$
Hence

$(2 \: \: - 3)$
is a point on this line.