Question

Math help Please!!!!

Answer 1

Part A. What is the slope of a line that is perpendicular to a line whose equation is −2y=3x+7?

Rewrite the equation  −2y=3x+7 in the form $y=-\dfrac{3}{2}x-\dfrac{7}{2}.$ Here the slope of the given line is  $m_1=-\dfrac{3}{2}.$ If $m_2$ is the slope of perpendicular line, then

$m_1\cdot m_2=-1,\\ \\m_2=-\dfrac{1}{m_1}=\dfrac{2}{3}.$

Answer 1: $\dfrac{2}{3}$

Part B. The slope of the line y=−2x+3 is -2. Since $-\dfrac{3}{2}\neq -2\quad \text{and}\quad \dfrac{2}{3}\neq -2,$ then lines from part A are not parallel to line a.

Since $-2\cdot \left(-\dfrac{3}{2}\right)=3\neq -1\quad \text{and}\quad -2\cdot \dfrac{2}{3}=-\dfrac{4}{3}\neq -1,$ both lines are not perpendicular to line a.

Answer 2: Neither parallel nor perpendicular to line a

Part C. The line parallel to the line 2x+5y=10 has the equation 2x+5y=b. This line passes through the point (5,-4), then

2·5+5·(-4)=b,

10-20=b,

b=-10.

Answer 3: 2x+5y=-10.

Part D. The slope of the line $y=\dfrac{x}{4}+5$ is $\dfrac{1}{4}.$ Then the slope of perpendicular line is -4 and the equation of the perpendicular line is y=-4x+b. This line passes through the point (2,7), then

7=-4·2+b,

b=7+8,

b=15.

Answer 4: y=-4x+15.

Part E. Consider vectors $\vec{p}_1=(-c-0,0-(-d))=(-c,d)\quad \text{and}\quad \vec{p}_2=(0-b,a-0)=(-b,a).$ These vectors are collinear, then

$\dfrac{-c}{-b}=\dfrac{d}{a},\quad \text{or}\quad -\dfrac{a}{b}=-\dfrac{d}{c}.$

Answer 5: $-\dfrac{a}{b}=-\dfrac{d}{c}.$

Answer 2

Answer:

(a) 2/3

(b) neither parallel nor perpendicular ; parallel; perpendicular

(c) y = (-2/5)*x - 2

(d) y = -4*x + 15

(e) -a*c = -d*b

Step-by-step explanation:

(a)

What is the slope of a line that is perpendicular to a line whose equation is −2y=3x+7 ?

First you need to rearrange the equation so that y term is isolated, in this case you should pass the -2 (which multiplies in one side) dividing to the other side.

y = (3*x + 7)/(-2)  

y = (-3/2)*x + 7

The slope of any line is the term next to x, in this case -3/2

Two lines are perpendicular if the multiplication of each line slopes is equal to -1. For example, given lines y = m1*x+b1 and y = m2*x+b2 where m1 is the slope of the first line and m2 of the second one, they are perpendicular if m1*m2 = -1

In this case we should solve:

(-3/2)*m2 = -1

m2 = 2/3

(b)

Line a is represented by the equation y=−2x+3 . How do these equations compare to line a? Drag and drop the equations into the boxes to complete the table. Parallel to line a Perpendicular to line a Neither parallel nor perpendicular to line a y=2x−1, y=−2x+5, y=1/2x+7

Two lines are parallel if they have the same slope.  

a's slope is -2  

Any line perpendicular to a must have an slope of 1/2,  so that -2*1/2 = -1

in y=2x−1 slope is 2 -> they are neither parallel nor perpendicular

in y=−2x+5 slope is -2 -> they are parallel  

in y=1/2x+7 slope is 1/2 -> they are perpendicular

(c)

What is the equation of a line that passes through the point (5, −4) and is parallel to the line whose equation is 2x + 5y = 10?

Rearranging the equation:

5*y = -2*x + 10  

y = (-2*x + 10)/5

y = (-2/5)*x + 2

Our equation has the form y = m*x + b where m = -2/5. In order to find b we replace the known point (5, -4) in the equation:

y = (-2/5)*x + b  

-4 = (-2/5)*5 + b  

-4 = -2 + b

-2 = b

Finally, y = (-2/5)*x - 2

(d)

What is the equation of a line that passes through the point (2, 7) and is perpendicular to the line whose equation is y=x/4+5?

Calling m1 = 1/4 the slope of the known equation, we have to find a slope m2 which fulfill m1*m2 = -1. Replacing:

1/4*m2 = -1

m2 = -4

In the same way as before, we replace the point (2, 7) in the general equation form:

y = -4*x + b

7 = -4*2 + b

7 = -8 +b

15 = b

Finally, y = -4*x + 15

(e)

In this diagram, which equation could you prove to be true in order to conclude that the lines are parallel?

The slope of a line can be  computed by m = (y1 - y2)/(x1 -x2) where (x1, y1) and (x2, y2) are points in the line

For the above line  

(x1, y1) = (b, 0)

(x2, y2) = (0, a)

Replacing

m1 = (0 - a)/(b -0) = -a/b

For the below line  

(x1, y1) = (-c, 0)

(x2, y2) = (0, -d)

Replacing

m2 = (0 - (-d) )/( (-c) - 0) = -d/c

m1 = m2,  so that both lines are parallel

-a/b = -d/c

-a*c = -d*b