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Without using the Pythagorean Theorem nor the Distance Formula, can it be determined whether the quadrilateral shown below is a rectangle?
In mathematics, there are usually several ways of proving or disproving a statement. The following example will explore two ways of determining whether a point is on a circle.
Ramsha and Dylan are taking Geometry together. They were asked to determine whether the point P(1,3) is on the circle that passes through Q(0,2) and whose center is at the origin. They decided to do it using different methods.The standard equation of the circle with center at (h,k) and radius r is (x−h)2+(y−k)2=r2. The distance d between two points with coordinates (x1,y1) and (x2,y2) is d=(x2−x1)2+(y2−y1)2.
Ramsha has decided to write the standard equation of the circle, and then check whether the coordinates of P satisfy this equation. Conversely, Dylan has decided to use the Distance Formula. These two options will be explored one at a time.
As already said, Ramsha has decided first to write the standard equation of the circle and use it to see whether the point is on that circle. To do so, she will follow two steps.
These steps will be done one at a time.
As previously stated, Dylan has decided to use the Distance Formula. To do so, he will follow two steps.
These steps will be done one at a time.
Real life problems can also be solved by setting a coordinate plane and using geometric properties.
A small country has three main roads connecting the beach, the mountains, and a national park. To avoid traffic jams and car accidents during summer holidays, the government of the country is planning to build a fourth road to connect the midpoints of Road 1 and Road 2.
After placing the diagram on a coordinate plane, use the Midpoint Formula to find the coordinates of the midpoints of Road 1 and Road 2.
Midpoint Formula: (2x1+x2,2y1+y2) | |||
---|---|---|---|
Road | Points | Substitute | Simplify |
Road 1 | B(4,0) and M(-4,4) | M1(24+(-4),20+4) | M1(0,2) |
Road 2 | M(-4,4) and P(-2,-2) | M2(2-4+(-2),24+(-2)) | M2(-3,1) |
Distance Formula: d=(x2−x1)2+(y2−y1)2 | |||
---|---|---|---|
Points | Substitute | Simplify | |
M1(0,2) and M2(-3,1) | dM1M2=(-3−0)2+(1−2)2 | dM1M2=10 | |
B(4,0) and P(-2,-2) | dBP=(-2−4)2+(-2−0)2 | dBP=210 |
Slope Formula: m=x2−x1y2−y1 | |||
---|---|---|---|
Points | Substitute | Simplify | |
M1(0,2) and M2(-3,1) | mM1M2=-3−01−2 | mM1M2=31 | |
B(4,0) and P(-2,-2) | mBP=-2−4-2−0 | mBP=31 |
The result obtained in the previous example can be generalized to any triangle.
Not only can coordinates be used to prove properties of triangles, but also they can be used to prove properties of quadrilaterals.
Davontay lives in a neighborhood in Boston where the stadium, the amusement park, the airport, and the train station are the vertices of a quadrilateral that seems to be a parallelogram.See solution.
Place the diagram on a coordinate plane.
d=(x2−x1)2+(y2−y1)2 | |||
---|---|---|---|
Side | Endpoints | Substitute | Simplify |
SP | S(-2,-3) and P(-5,4) | SP= (-5−(-2))2+(4−(-3))2 | SP=58 |
PA | P(-5,4) and A(3,5) | PA= (3−(-5))2+(5−4)2 | PA=69 |
AT | A(3,5) and T(6,-2) | AT= (6−3)2+(-2−5)2 | AT=58 |
TS | T(6,-2) and S(-2,-3) | TS= (-2−6)2+(-3−(-2))2 | TS=69 |
Substitute (-5,4) & (6,-2)
M(2x1+x2,2y1+y2) | |||
---|---|---|---|
Diagonal | Endpoints | Substitute | Simplify |
PT | P(-5,4) and T(6,-2) | M1(2-5+6,24+(-2)) | M1(21,1) |
SA | S(-2,-3) and A(3,5) | M2(2-2+3,2-3+5) | M2(21,1) |
The challenge presented at the beginning of this lesson can be solved by using coordinates.
Using neither the Pythagorean Theorem nor the Distance Formula, it is desired to determine whether the quadrilateral below is a rectangle.
By the Slopes of Perpendicular Lines Theorem, if the slopes of two lines or segments are opposite reciprocals, then lines are perpendicular.
For simplicity, the vertices will be labeled A, B, C, and D.
Next, the slopes of the sides will be found by using the Slope Formula. The slope of AB will be calculated first.Substitute values
m=x2−x1y2−y1 | |||
---|---|---|---|
Side | Endpoints | Substitute | Simplify |
AB | A(-6,2) and B(-3,6) | mAB=-3−(-6)6−2 | mAB=34 |
BC | B(-3,6) and C(9,-3) | mBC=9−(-3)-3−6 | mBC=-43 |
DC | D(6,-7) and C(9,-3) | mDC=9−6-3−(-7) | mDC=34 |
AD | A(-6,2) and D(6,-7) | mAD=6−(-6)-7−2 | mAD=-43 |
By the Slopes of Perpendicular Lines Theorem, if the slopes of two line segments are opposite reciprocals, then the segments are perpendicular. The slopes of consecutive sides will be multiplied to see if they are opposite reciprocals. Recall that, in a polygon, two sides are consecutive if they share a common vertex.
Sides | Product of Slopes |
---|---|
AB and BC | 34(-43)=-1 ✓ |
BC and DC | -43(34)=-1 ✓ |
DC and AD | 34(-43)=-1 ✓ |
AD and AB | -43(34)=-1 ✓ |
Since the product of their slopes is -1, the quadrilateral's consecutive sides are perpendicular. Therefore, they form right angles. Consequently, by its definition, the quadrilateral is a rectangle.