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3 years ago

Find all the points, if any, where the graph of 12x-5y=0 intersects (x+12)^2+(y-5)^2=169.

2 answers:

6 0

I: 12x-5y=0
II:(x+12)^2+(y-5)^2=169

with I:
12x=5y
x=(5/12)y
-> substitute x in II:
((5/12)y+12)^2+(y-5)^2=169
(25/144)y^2+10y+144+y^2-10y+25=169
(25/144)y^2+y^2+10y-10y+144+25=169
(25/144)y^2+y^2+144+25=169
(25/144)y^2+y^2+169=169
(25/144)y^2+y^2=0
y^2=0
y=0

insert into I:
12x=0
x=0

-> only intersection is at (0,0) = option B

Serga [27]3 years ago

5 0

Answer:

(0,0)

Step-by-step explanation:

If you graph $12x-5y=0$ and $(x+12)^2+(y-5)^2=169$ along with the points (0,0) and (4.5, 10,9) you with see that the equations intersect at (0,0)

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