1. Find the x‐intercept and y‐intercept of the graph of 2x – 5y = ‐10
2. x‐int ___ Write the equation of the tangent line to the circle (x ‐ 5)² + (y + 2)² = 41 at the point (1, 3).
Center = (5,­2)
y–int ____
T
The line is to the radius at (1,3).
The slope between (5,­2) and (1,3) is ­5/4 so the slope
of the tangent is 4/5
use this for your point
y­3=4/5(x­1)
....y=4/5x+11/5
3. If C is a centroid of ΔDOG, find the length of DY.
D
Y is midpoint. If you find the distance
from CY, you can triple it to get DY.
C (­1,­2)
O (­3,­5)
Y
(1,­5)
CY = (­1­1)2 + (­2­­5)2
G (5,­5)
=
13
Therefore, DY = 3 13
May 18­10:55 AM
4. Given triangle ABC with A(6, ‐5), B(‐2, 10), and C(4, 5), find the coordinates of the centroid.
5. Write an equation for each line in the specified form.
a) Slope ‐3/2 containing (0, ‐2) in slope‐intercept form.
is the y-int
(6+­2+4) , (­5+10+5) = 8 , 10
____ ____ _ _
3 3
3 3
y = ­3/2x ­ 2
or is pt­slope form
y ­ ­2 = ­3/2(x ­ 0), when solved for y becomes
y = ­3/2x ­ 2
May 18­12:42 PM
b) Y‐intercept = ‐8, x‐intercept = 4, in point‐slope form.
(0, ­8)
c) Parallel to 4x – 6y = ‐1
containing (2, 3) in slope‐
intercept form.
(4, 0)
slope between them is 8/4 = 2
so using pt­slope gives you
is y = 2/3x + 1/6 when solved for y
so, it has a slope 2/3. With (2,3) in
pt­slope form becomes: y ­ 3 = 2/3(x ­ 2). when solved for y,
y = 2/3x + 5/3
y ­ ­8 = 2(x ­ 0) or
y ­ 0 = 2(x ­ 4)
May 18­12:42 PM
d) Containing (5, ‐3) and (4, ‐1) in point slope form. e) slope between them is ­2/1 = ­2
so using pt­slope gives you
Containing ( ‐6, 2) and (‐6, 7) in any form.
The slope is undefined (5/0), so the
line must be vertical. Vertical lines
are in the form x = a, and for this case,
x = ­6.
y ­ ­3 = ­2(x ­ 5) or
y ­­ 1 = ­2(x ­ 4)
May 18­12:51 PM
g) With intercepts three time those of ‐7x ‐ 2y = 14 in slope‐
intercept form.
has intercepts (0, ­7) and (­2, 0). Threetimes those are (0, ­21) and (­6,0).
The slope between those two = ­7/2.
Perpendicular to y = ‐4/3x + 5 containing (1, ‐2) in point‐
slope form.
T
f) slope ­4/3 is to 3/4.
y + 2 = 3/4(x ­ 1)
Since you already know they y­int is (0, ­21), it's eqn is y = ­7/2x ­ 21
May 21­12:29 PM
6. Given triangle ABC with A(‐2, ‐2), B(2, 6), and C(12, ‐2), find the circumcenter.
May 18­12:48 PM
Given triangle ABC with A(‐2, ‐2), B(2, 6), and C(12, ‐2), find the orthocenter.
6. May 18­12:48 PM
7.
A(‐2,2), B(4,10), C(1,5) length of altitude to BC.
May 18­12:48 PM
May 21­10:59 AM