ANALYTIC GEOMETRY TEST SOLUTIONS
Solve these questions. Show your work.
A/ Points and Lines:
1) Triangle ABC has vertices at A (2, 7), B (0, 0), and C (5, 1):
(a) AC = 
(b) mid-pt. of AB = 
(c) Equation of BC is 
‹ perpendicular distance from (2, 7) to BC = 
(d) equation of AC. 
which becomes 2x + y – 11 = 0
(e) the x-intercept is 11/2, the y-intercept is 11.
(f) slope of AB = 7/2, slope of BC = 1/5 and slope AC = – 2
since no two slopes are negative reciprocals, the triangle is not right-angled.
.
2) (a) through ( – 7, 1) and (2, -6): 
which is 7x + 9y + 40 = 0
(b) x-int. (5, 0) slope = 3/4 so equation is 
3x – 4y – 15 = 0
(c) y-int. = (0, 3) and slope = 1, so we want slope – 1 or y = – x + 3
(d) through ( – 2, 1), parallel to the y-axis: x = – 2
(e) through the y-intercept of y = 2x – 7, parallel to the x-axis: y = – 7
.
B/ Circles:
1) Write the equation of the circle:
(a) center (0, 0), radius = 6: x2 + y2 = 36
(b) radius = distance from (1, – 1) to ( – 2, 3) = 
equation is (x + 2)2 + (y – 3)2 = 25
(c) center = mid-pt = ( – 1, 4) radius = distance from ( – 1, 4) to (3, 6)
so the equation is (x + 1)2 + ( y – 4)2 = 20
.
2) Find the center and radius of the circle:
| (a) center (0,0) r = 7 | (b) center ( – 1, – 3) r = 6 | (c) center (0, 5) r = 10 |
3) (a) center is (0,0) so slope of radius to ( – 4, 3) = – 3/4
slope of tangent = 4/3, equation of tangent is 
4x – 3y + 25 = 0
.
(b) center is ( – 2, 1), slope of radius to (1, 5) is 4/3, so slope of tangent = – 3/4
equation of tangent is 
3x + 4y – 23 = 0
.
c) x2 + y2 + 4x – 2y – 20 = 0 from the external point (3, – 6) so 2 tangents.
center is (– 2, 1), r = 5, and y + 6 = m(x – 3) is the equation of any line through (3, – 6)
rearranged, this is m x – y – (6 + 3m) = 0 and we know that the perpendicular distance
from the center to this line = radius so 
| – 5m – 7 | = 
[– (5m + 7)]2 = 25m2 + 25
‹ 25m2 + 70m + 49 = 25m2 + 25 u m = – 12/35
so the tangent equation is 12x + 35y + 174 = 0
but what about the other tangent? There are always 2 tangents from an external point.
Let's make a diagram and figure it out.
the 2nd tangent is the vertical line with equation x = 3 which touches the circle at (3, 1).
C/ Parabolas:
1) y = 3x2 - 12x + 16
(a) y = 3(x2 – 4x) + 16 u y = 3(x2 – 4x + 4 – 4) + 16 u y = 3(x – 2)2 + 4
(b) since a = 3 > 0, it opens upwards.
(c) coordinates of the vertex: (2, 4).
(d) a minimum y value.
(e) minimum value: y = 4
(f) the equation of the axis of symmetry is x = 2
(g) Since y = 4 is the minimum, there are no x-intercepts. The y-intercept is (0, 16).
2) Find the x-intercepts for:
| (a) y = x2 – 6x – 7 0 = (x – 7)(x + 1) so x = 7 or x = – 1 |
(b) y = 2x2 + x – 15 0 = (2x – 5)(x + 3) so x = 5/2 or x = – 3 |
(c) y = – x2 – x + 3 doesn't factor so we use the quadratic formula.
D/ Graphing:
1) (a) y = – (x2 – x) + 6 u y = – (x2 – x + ¼ – ¼) + 6 u y = – (x – ½)2 + 
(b) y = – x2 + x + 6 u y = 6 + x – x2 u y = (3 – x)(2 + x)
‹ the x-intercepts are 3 and – 2 the y-intercept is 6.
2) (a) The lines intersect at (2, 7).
2 (b) The points of intersection are:(3, 0) and (1, 2)
(c) y = x2 – 3x – 4 u y = (x – 4)(x + 1) or y = (x – 3/2)2 – 25/4
points of intersection are (– 1, 0) and (3, – 4).
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