Please solve 2y²-18xy+18x²
1 answer:
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equation for the perpendicular Bisector of the line segment whose endpoints are (-9,-8) and (7,-4)
Perpendicular bisector lies at the midpoint of a line
Lets find mid point of (-9,-8) and (7,-4)
midpoint formula is
midpoint is (-1, -6)
Now find the slope of the given line
(-9,-8) and (7,-4)
Slope of perpendicular line is negative reciprocal of slope of given line
So slope of perpendicular line is -4
slope = -4 and midpoint is (-1,-6)
y - y1 = m(x-x1)
y - (-6) = -4(x-(-1))
y + 6 = -4(x+1)
y + 6 = -4x -4
Subtract 6 on both sides
y = -4x -4-6
y= -4x -10
equation for the perpendicular Bisector y = -4x - 10
If you are asking what is the graph of y = 3x^2 -2x+1.
Then, the attached file would be the answer.
To check, b^2 - 4(a)(c), for each equation and use these facts:
If b^2 - 4(a)(c) = 0, there is only one real root meaning, the graph touches the x-axis only in one point.
If b^2 - 4ac > 0, there are two real roots meaning, the graph touches the x-axis in two different points.
If b2 - 4ac < 0, there are no real roots then the graph does not touch the x-axis. This would be the case for y = 3x^2 - 2x + 1.
Solution:
(-2)^2 -4(3)(1) = 4 - 12 = -8 < 0 will result in not real roots.
Then, the attached file would be the answer.
To check, b^2 - 4(a)(c), for each equation and use these facts:
If b^2 - 4(a)(c) = 0, there is only one real root meaning, the graph touches the x-axis only in one point.
If b^2 - 4ac > 0, there are two real roots meaning, the graph touches the x-axis in two different points.
If b2 - 4ac < 0, there are no real roots then the graph does not touch the x-axis. This would be the case for y = 3x^2 - 2x + 1.
Solution:
(-2)^2 -4(3)(1) = 4 - 12 = -8 < 0 will result in not real roots.