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Sagot :
The equation of a line that is perpendicular to 2y 3x 10 and passes through (- 6 1) is -3y + 2x -1 with slope.
1. Start by finding the slope of the given line by taking the coefficient of x and dividing it by the coefficient of y.
2. To find the slope of the line perpendicular to the given line, we need to find the negative reciprocal of the slope of the given line.
Slope of the perpendicular line: m = -2/3
3. We know that the equation of a line in slope intercept form is: y = mx + b, where m is the slope and b is the y-intercept.
4. So, the equation of the line perpendicular to 2y 3x 10 and passes through (-6 1) is:
y = -2/3x + b
5. To find the y-intercept, substitute the given point (-6 1) in the equation and solve for b.
-1 = -2/3(-6) + b
b = 1
6. Therefore, the equation of a line that is perpendicular to 2y 3x 10 and passes through (- 6 1) is -3y + 2x -1
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