Need help with this linear algebra problem
1 answer:
You should have
on its own points to the point (2, -3) and lies on the line through this point and the origin. We capture all points on this line by scaling
by
. Then we add to
another vector that points to P(1, 1), which shifts the endpoint of all vectors on the line
to the line through P that runs parallel to
.
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Answer:
See below
Step-by-step explanation:
<h3><u>General outline</u></h3>- Construct the perpendicular bisectors of DE, EF, and DF.
- Identify the circumcenter.
- Draw the circumcircle.
To construct a circle, one must know the center point of the circle, and the circle's radius.
The circle that circumscribes a triangle (called a circumcircle), has a circumcenter (center point) that is the point of concurrency (a point where more than two lines intersect at the same point) of all three perpendicular bisectors (perpendicular lines that happen to cut a given line segment exactly in half) of the sides of the triangle.
Construct the circle with a center at the circumcenter, and a radius out to any one of the vertices of the triangle. The circumcenter is equidistant from all three vertices, so the circle draw will contain all three vertices.
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<h3><u>Constructing perpendicular bisectors</u></h3>To construct perpendicular a bisector, recall that a perpendicular bisector is a line containing all points that are equidistant the end points of a line segment (each point on the line is the same distance from both segment endpoints simultaneously).
So, if we can find two points that are the equidistant from the two segment endpoints, we'll have two points on the line, and to draw any line, one only needs two distinct points, and the straightedge.
<u>Finding a point that is equidistant from two given points</u>
Set the compass to any radius that is larger than half the distance between the two endpoints. If uncertain of a "good" choice, one can simply choose the length of the line segment itself as a radius.
Setting the center of the compass circle on one endpoint, draw an arc such that the arc will pass through the perpendicular line we're trying to construct twice. If uncertain, draw the entire circle. Keeping the radius saved.
With the same radius, setting the center of the compass circle to the second endpoint, draw and arc such that the arc intersects the first circle two times. If it doesn't intersect two times, the radius chosen at the beginning was too small. Start the entire process over with a larger radius.
These two points of intersection are equidistant from the two endpoints of the line segment, and thus, they are on the perpendicular bisector of the line segment.
Using those two points, and a straightedge, draw the line containing those two points, and extend it generously in both directions.
<h3><u>Process</u></h3><u><em>Steps 1-5 in pink; steps 6-10 in green; steps 11-15 in blue (see diagram):</em></u>
- Set compass radius to length DE
- Draw circle, centered at D, through E. Keep radius set.
- Draw circle, centered at E, through D.
- Identify points of intersection of the circles from steps 2 & 3, M & N
- Draw line MN
- Set compass radius to length EF
- Draw circle, centered at E, through F. Keep radius set.
- Draw circle, centered at F, through E.
- Identify points of intersection of the circles from steps 7 & 8, Q & R
- Draw line QR
- Set compass radius to length DF
- Draw circle, centered at D, through F. Keep radius set.
- Draw circle, centered at F, through D.
- Identify points of intersection of the circles from steps 12 & 13, S & T
- Draw line ST
- Identify point of concurrency of line MN, line QR, and line ST; label it point P
- Set compass radius to length DP (or EP, or FP)
- Draw circle, centered at P, through points D, E, and F.
<em>Note: Since all three perpendicular bisectors intersect at the same point, it is only necessary to find two of three, not all three. So, one could skip constructing one of those three perpendicular bisectors (either steps 1-5, 6-10, or 11-15), and still be able to construct the circumcircle correctly.</em>
