analytic geometry (part 2)
analytic geometry (part 2)
Equation of a Tangent
MPM2D: Principles of Mathematics
Recap
Determine the equation of the tangent to x 2 + y 2 = 74 at
point P(5, −7).
Find the slope of the radius from the origin to P, then the
slope of the tangent.
Right Bisectors
mOP = − 57
J. Garvin
mT =
5
7
Substitute mT and the coordinates of P into y = mx + b.
− 7 = 75 (5) + b
− 49 = 25 + 7b
b = − 74
7
The equation of the tangent is y = 57 x −
J. Garvin — Right Bisectors
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analytic geometry (part 2)
74
7 .
analytic geometry (part 2)
Right Bisectors
Right Bisectors
A right bisector (sometimes called a perpendicular bisector)
is a line that intersects a line segment at its midpoint at 90◦ .
To determine an equation for a right bisector of some line
segment, PQ, we use a similar strategy to that used for
finding an equation for a tangent to a circle.
Since the right bisector is perpendicular to PQ, its slope,
mRB , must be the negative reciprocal of the slope of the line
segment, mPQ .
Since PQ and the right bisector intersect at PQ’s midpoint,
MPQ , we can substitute that point into the equation of a line
along with mRB to determine the y -intercept.
J. Garvin — Right Bisectors
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J. Garvin — Right Bisectors
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analytic geometry (part 2)
Right Bisectors
analytic geometry (part 2)
Right Bisectors
Example
Show that the line y = 23 x + 31 is the right bisector of PQ,
given P(−6, 5) and Q(2, −7).
The line has a slope, mRB , of 32 .
PQ has a slope, mPQ , of
−7−5
2−(−6)
= − 32 .
Therefore, the line and the line segment are perpendicular.
−6+2 5+(−7)
= (−2, −1).
2 ,
2
The midpoint of PQ, MPQ , is
Test if MPQ satisfies the line.
y = 23 (−2) +
1
3
= −1
Since MPQ is on the line, the line is the right bisector of PQ.
J. Garvin — Right Bisectors
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J. Garvin — Right Bisectors
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analytic geometry (part 2)
Right Bisectors
analytic geometry (part 2)
Right Bisectors
Example
Determine the equation of the line that bisects AB, given
A(−5, −3) and B(7, 1).
Find the midpoint and slope of AB.
−5+7 −3+1
2 , 2
−3−1
1
−5−7 = 3 .
The midpoint, MAB , is
The slope, mAB , is
= (1, −1).
The slope of the right bisector, mRB , is −3.
Use MAB and mRB in y = mx + b.
−1 = −3(1) + b
b=2
The equation of the right bisector of AB is y = −3x + 2.
J. Garvin — Right Bisectors
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J. Garvin — Right Bisectors
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analytic geometry (part 2)
analytic geometry (part 2)
Right Bisectors
Right Bisectors
Example
The line segment forms the hypotenuse of a right triangle,
shown in green.
The line y = − 34 x + 5 is the right bisector of PQ,
intersecting at M(4, 2). If |PQ| = 20, find P and Q.
Since mRB = − 34 , mPQ = 34 .
The arms of this triangle have lengths 3k and 4k, where k is
some real scaling factor.
20
2
Since M is the midpoint of PQ, |PM| =
= 10 units.
Use the Pythagorean Theorem to solve for k.
(3k)2 + (4k)2 = 102
9k 2 + 16k 2 = 100
25k 2 = 100
k2 = 4
k=2
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J. Garvin — Right Bisectors
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analytic geometry (part 2)
Right Bisectors
analytic geometry (part 2)
Right Bisectors
Thus, the arms have lengths 3k = 3 × 2 = 6 and
4k = 4 × 2 = 8 units.
From M(4, 2), we can draw these triangles to determine the
two endpoints of PQ.
In this case, P is at (−2, −6) and Q is at (10, 10).
J. Garvin — Right Bisectors
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J. Garvin — Right Bisectors
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analytic geometry (part 2)
Questions?
J. Garvin — Right Bisectors
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