Which system of equations is graphed below?
1 answer:
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9514 1404 393
Answer:
(a) 4/3
(b) y -3 = 4/3(x -1)
(c) y -3 = -3/4(x -1)
(d) r = 5
Step-by-step explanation:
a) The slope is given by the slope formula:
m = (y2 -y1)/(x2 -x1)
m = (7 -3)/(4 -1) = 4/3
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b) The radius is normal to the circle. The point-slope form of the equation for a line can be useful here:
y -k = m(x -h) . . . . . line with slope m through point (h, k)
For slope 4/3, the line through point (1, 3) will have the equation ...
y -3 = 4/3(x -1) . . . . point-slope equation of the normal
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c) The tangent is perpendicular to the radius. It will have a slope that is the opposite reciprocal of the slope of the radius: -1/(4/3) = -3/4.
y -3 = -3/4(x -1) . . . . point-slope equation of the tangent
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d) The radius can be found from the distance formula.
d = √((x2 -x1)² +(y2 -y1)²)
d = √((4 -1)² +(7 -3)²) = √(3² +4²) = √25 = 5
The radius of the circle is 5.
