what is the equation in point-slope form for the line parallel to y=-2x+10 that contains j(6,8)?
y-8 = -2(x-6)
To find the equation of a line parallel to the line y = -2x + 10 and passing through the point j(6,8), we can use the point-slope form of the equation of a line.
The point-slope form of a linear equation is y - y₁ = m(x - x₁), where (x₁, y₁) represents a point on the line, and m represents the slope of the line.
First, let's determine the slope (m) of the given line y = -2x + 10. In this form, the slope is the coefficient of x, which is -2.
Since we want a line parallel to this line, the slope of the new line will also be -2.
Now, we can substitute the values of the point j(6,8) and the slope (-2) into the point-slope form equation:
y - y₁ = m(x - x₁)
y - 8 = -2(x - 6)
Simplifying further:
y - 8 = -2x + 12
To convert this equation to slope-intercept form (y = mx + b), let's isolate y:
y = -2x + 12 + 8
y = -2x + 20
Therefore, the equation in point-slope form for the line parallel to y = -2x + 10 that contains j(6,8) is y - 8 = -2(x - 6) and in slope-intercept form, it is y = -2x + 20.