Find the equation of a line through (0,2) and perpendicular to (i) y = 2x – 3 (ii) y = / + 2
Answers
EXPLANATION.
Equation of line passes through point ( 0,2).
perpendicular to the line = y = 2x - 3.
equation of line = 2x - y - 3 = 0
slope of the perpendicular line = b/a = -1/2.
Equation of line = ( y - y₁) = m ( x - x₁)
⇒ ( y - 2 ) = -1/2 ( x - 0 ).
⇒ 2 ( y - 2 ) = -1 ( x - 0 ).
⇒ 2y - 4 = -x.
⇒ 2y + x - 4 = 0. = equation of line.
MORE INFORMATION.
FIRST DERIVATIVE TEST.
A point x₀ is a point of local maxima ( local minima) if,
(1) = f'(x₀) = 0.
(2) = f'(x) changes sign from positive ( negative) to negative ( positive) while passing through x₀.
SECOND DERIVATIVE TEST.
A point x₀ is a point of local maxima ( local minima) if,
(1) = f'(x₀) = 0.
(2) = f''(x₀) < 0 ( > 0 ).
(3) = if f''(x₀) = 0, then second derivative test fails and find f'''(x₀).
if f'''(x₀) ≠ 0 then it is a point of inflection. if f'''(x₀) = 0, find f⁴(x₀). if f⁴(x₀) is
< 0 then x₀ is the point of local maxima. if f⁴(x₀) > 0 then x₀ is the point of local minima. if f⁴(x₀) = 0 then we find f⁵(x₀). if f⁵(x₀) ≠ 0 then x₀ given the point of inflection. if f⁵(x₀) = 0 then we find f⁶(x₀) is so on in similar way.