What is the equation of the line having the specified slope and contained the indicated point. M=-3/5;(12,-6)
Just use the good old point-slope form:
(y+6)/(x-12) = -3/5
rearrange things as you will; one form of the equation is as good as another.
Given: m = -3/5, P(12,-6).
Y = mx + b,
Y = (-3/5)12 + b = -6,
-36/5 + b = -6,
b = -6 + 36/5,
b = -30/5 + 36/5 = 6/5.
Eq: Y = (-3/5)x + 6/5.
To find the equation of a line given its slope and a point, you can use the point-slope form of a linear equation.
The point-slope form of a linear equation is given by:
y - y1 = m(x - x1)
Where:
m is the slope of the line,
(x1, y1) is the given point on the line.
In this case, the given slope (m) is -3/5, and the given point is (12, -6).
Substituting the values into the point-slope form, we get:
y - (-6) = (-3/5)(x - 12)
Simplifying this equation, we have:
y + 6 = (-3/5)(x - 12)
To get the final equation in slope-intercept form (y = mx + b), we can simplify further:
y + 6 = (-3/5)(x - 12)
Multiply through by 5 to eliminate the fraction:
5(y + 6) = -3(x - 12)
Distribute 5 on the left side and -3 on the right side:
5y + 30 = -3x + 36
Rearrange the equation to isolate y:
5y = -3x + 36 - 30
5y = -3x + 6
Divide through by 5 to solve for y:
y = (-3/5)x + 6
Thus, the equation of the line with a slope of -3/5 and containing the point (12, -6) is y = (-3/5)x + 6.