a Let's plot the points on graph paper and connect them.
Examining the diagram, we see that JI ∥ GH. If we also can prove that GJ ∥ HI, we have a quadrilateral with two pairs of parallel sides which means it's a parallelogram. To do that, we should find the segments' slopes.
The slope of GJ and HI are both 43 which means these sides are also parallel. With this information, we know that GHIJ is a parallelogram.
A parallelogram can also be classified as a rhombus if all sides are congruent. Therefore, let's calculate the side's lengths using the Distance Formula. Note that JI and GH are horizontal sides. These lengths can be found by measuring the number of steps between their endpoints. From the diagram, we see that JI=5 and GH=5.
Segment
Points
sqrt((x_2-x_1)^2+(y_2-y_1)^2)
d
GJ
( - 2,2), ( 1,6)
sqrt(( - 2- 1)^2+( 2- 6)^2)
5
HI
( 3,2), ( 6,6)
sqrt(( 2- 6)^2+( 3- 6)^2)
5
Since all sides are 5 units long, GHIJ is in fact a rhombus.
b Let's first draw the diagonals GI and JH.
Both of these lines are straight lines which means we can write them in slope-intercept form.
y=mx+b
In this equation, m is the lines slope and b is the y-intercept. We can find the slope by using the Slope Formula.
segment
points
y_2-y_1/x_2-x_1
m
GI
G(- 2,2), I(6,6)
6- 2/6-( - 2)
1/2
JH
J(1,6), H(3,2)
2- 6/3- 1
- 2
With this information, we have half of what we need to write the equations.
GI:& y= 1/2x+b
JH:& y= - 2x+b
Finally, we must find the y-intercept by substituting any of the points through which the lines passes, into the equations and solving for b. For example, we can substitute I(6,6) in the equation for GI and H(3,2) in the equation for JH.
segment
y=mx+b
substitute point
solve for b
GI
y= 1/2x+b
6= 1/2( 6)+b
b=3
JH
y= - 2x+b
2= - 2( 3)+b
b=8
Now we can finalize the equations.
GI:& y= 1/2x+3
JH:& y= - 2x+8
c If the equations are perpendicular, the product of their slopes equals - 1.
m_1m_2=- 1
By substituting the slopes in this equation, we can determine if they are perpendicular.
d To rotate a point by 90^(∘) clockwise, we first draw a segment form J to the origin. Using a protractor, we can measure a 90^(∘) angle clockwise from this segment. By drawing a second segment that has the same length as the first, we can identify the position of J'.
From the diagram, we see that the rotated point is located at (6,- 1).
e To find the area of a parallelogram, we have to identify its base and height. Observing the diagram, we can identify these dimensions.
Now we can calculate the area by multiplying the rhombus height h= 4 and base b= 5.
A=( 4)( 5) ⇔ A=20 square units
