Write the equation of a line that is perpendicular to the given line and that passes through the given point.
y-3=8/3(x+2);(-2,3)
To find a line that is perpendicular to the given line, we need to find the negative reciprocal of the slope of the given line.
The given line has a slope of 8/3.
The negative reciprocal of 8/3 is -3/8.
Now, we have the slope (-3/8) and the point (-2,3). We can use the point-slope form of the equation of a line to find the equation of the line that passes through the given point and has the perpendicular slope.
y - y₁ = m(x - x₁)
where (x₁, y₁) is the given point and m is the slope.
Substituting the values into the equation:
y - 3 = (-3/8)(x - (-2))
Simplifying,
y - 3 = (-3/8)(x + 2)
Now, we can distribute -3/8 into (x + 2)
y - 3 = (-3/8)x - 3/4
Moving the constant term -3 to the right side by adding 3 on both sides:
y = (-3/8)x - 3/4 + 3
y = (-3/8)x - 3/4 + 12/4
y = (-3/8)x + 9/4
Therefore, the equation of the line that is perpendicular to the given line and passes through the point (-2,3) is y = (-3/8)x + 9/4.