Anwers

INCLINATION AND SLOPE OF A LINE

1. Given the points, show that ABCD is a parallelogram.

a. A(-1,-1), B(5,0), C(4,3) and D(-2,2).

Solution:-

Part a :-

First of all to show that a quadrilateral is a parallelogram, it is sufficient to show that the

diagonals of the quadrilateral bisect each other.

Plotting the points on the X and Y axis as per the given co-ordinates.

Let P is the middle point of AC line.

P co-ordinates will be (−1+4

2,−1+3

2) = (

3

2, 2)

So P Co-ordinates = (3

2, 2)

Let Q is the middle point of DB line.

Q co-ordinates will be (−2+5

2,−2+0

2) = (

3

2, 2)

So Q Co-ordinates = (3

2, 2)

As we can see the both co-ordinate for P and Q are same that means the midpoint of

both line are bisect to each other. Hence ABCD is a parallelogram

Part b :-

b. A(1,5), B(3,4), C(4,8), F(6,7).

First of all to show that a quadrilateral is a parallelogram, it is sufficient to show that the

diagonals of the quadrilateral bisect each other.

Plotting the points on the X and Y axis as per the given co-ordinates.

Let P is the middle point of AD line.

P co-ordinates will be (1+6

2,5+7

2) = (

7

2, 6)

So P Co-ordinates = (7

2, 6)

Let Q is the middle point of BC line.

Q co-ordinates will be (3+4

2,4+8

2) = (

7

2, 6)

So Q Co-ordinates = (7

2, 6)

As we can see the both co-ordinate for P and Q are same that means the midpoint of

both line are bisect to each other. Hence ABCD is a parallelogram

2. Find the tangents of the angles of the triangle whose vertices are A(3,-2), B(-5,8), and

C(4,5). Then use the inverse tangent to find each angle and express it to the nearest

degree.

Let the give points be A(3,-2) and B(-5,8)

Here slope MAB (tanӨ1) =(8)−(−2)

(−5)−(3)) =

(8)+(2)

8=

5

4

Ө1=tan-1(5/4)= 51º

Let the give points be B(-5,8) and C(4,5)

Here slope MBC (tanӨ2) =(5)−(8)

(4)−(−5)) =

(−3)

9=

−1

3

Ө2= tan-1(-1/3)= -18.41 º

Let the give points be C(4,5) and A(3,-2)

Here slope MCA (tanӨ3) =(−2)−(5)

(3)−(4)) =

(−7)

−1= 7

Ө3= tan-1(7)= 81.86 º

Solution 4.

Make a sketch and divide it into two right triangles, with the camera on a common side.

One triangle has legs of 20 and 70 yards. The angle at the camera has tangent 70/20. So

that angle is arctan (70/20) which is approximately 70.05 degrees. Actually, 70.05 is

rounded down a little bit so let's say 70.06 to make sure you have complete coverage of

the end zone.

The other triangle has legs of 20 and 50. The angle at the camera has tangent 50/20. So

that angle is arctan (50/20) which is approximately 68.12 degrees. That angle is already

rounded up a little bit, so it is sufficient to cover the end zone.

The camera must be able to pan approximately 68.12 + 70.06 degrees, or 88.18 degrees.

That is a little high but you didn't specify how many decimal places were needed.

Solution 5. Find the slope of the line passing through the two points. Find also the

inclination to the nearest degree.

a. Slope of the line passing through the (2,3) , (3,7) points

Here slope (tanӨ) =(3)−(7)

(2)−(3)) =

(−4)

(−1)= 4

Ө=tan-1(4)= 75.9637

b. Slope of the line passing through the (−2,8), (4, −3)points

Here slope (tanӨ) =(8)−(−3)

(−2)−(4)) =

(11)

(6)=

Ө=tan-1(11/6)= 61.389

c. Slope of the line passing through the (11.7142,4.0015) and (−3.8014 , −2.8117)

points

Here slope (tanӨ) =(4.0015)−(−2.8117)

(11.7142)−(−2.8117)) =

(6.8132)

(14.5259)=

Ө=tan-1 (6.8132)

(14.5259)= 25.126577

Solution 6 Show that each of the four points are vertices of the parallelogram ABCD.

a. A(3,0) , B(7,0) , C(5,3) , D(1,3)

A parallelogram is a quadrilateral with both pairs of opposite sides parallel.

A parallelogram has these properties

A- 2 sets of parallel sides

B- 2 sets of congruent sides

C- opposite angles congruent

D- consecutive angles supplementary

E- diagonals bisect each other

F- diagonals form 2 congruent triangles

You need to remember that parallel lines have the same slope and perpendicular

lines have slopes that are negative reciprocal (the product of the slopes of two

perpendicular lines equals -1. So find the slopes is the segments AB and CD. Then find the

slopes of segments BC and DA.

slope when given two points= (𝑌2−𝑌1)

(𝑋2−𝑋1)

Slope AB is given by mAB=(0−0)

(7−3)=0

Slope CD is given by mCD=(3−3)

(5−1)=0

Both are 0, so they are parallel

Slope BC is given by mBC=(3−0)

(5−7)=(3)

(2)

Slope AD is given by mAD=(3−0)

(1−3)=(3)

(2)

Both are (3)

(2), so they are parallel

Hence A figure with two sets of parallel lines is a a prallelogram

b. A(-1,-2) , B(3,-6) , C(11,-1) , D(7,3)

slope when given two points= (𝑌2−𝑌1)

(𝑋2−𝑋1)

Slope AB is given by mAB=(−6)−(−2)

(3)−(−1)=-1

Slope CD is given by mCD=(3)−(−1)

(7)−(11)=-1

Both are -1, so they are parallel

Slope BC is given by mBC=(−1)−(−6)

(11)−(3)=(5)

(8)

Slope AD is given by mAD=(3)−(−2)

(7)−(−1)=(5)

(8)

Both are (5)

(8), so they are parallel

Hence A figure with two sets of parallel lines is a a parallelogram

Solution 7

a. (4,-4) , (4,4) , (0,0)

you basically need to find the equations for the lines that form the triangle and then

determine if any of them are perpendicular to any of the others.To determine whether

they are perpendicular, you need to determine if the slope of any one of the lines is

perpendicular to the slope of any of the other lines. you probably don't even need the

equation of the lines. all you really need is the slope of the lines, since that is what will

allow you to determine whether the lines are perpendicular to each other or not. the

pairs of points that you want to examine for their slope are:

(4,-4) and (4,4)

(4,4) and (0,0

(0,0) and (4,-4)

b. (-1,-1) , (16,-1), (0,3)

since that is what will allow you to determine whether the lines are perpendicular to

each other or not. the pairs of points that you want to examine for their slope are:

(-1,-1) and (16,-1)

(16,-1) and (0,3)

(0,3) and (-1,-1),