Step 1: Calculate the point P(any variable) on the curve  

             NOTE: Substitute the given coordinate in the curve equation to get the  

             point

             EXAMPLE: y = 9 - 4x - \frac{8}{x}, x-coordinate = 2

                               So,   y = 9 - 4*2 - \frac{8}{2}

                                                 y = 9 - 8 - 4 = -3

Step 1: Calculate the slope of the curve

            Skill 1: Note down the given curve equation and apply the differentiation on

            both sides

                  NOTE: Multiple both sides of the function by \frac{dy}{dx}  

            Skill 2:  Differentiate both sides of the function with respect

            to x (any  variable).

            EXAMPLE: y = x^3 - 2x^2 - x + 9

                              [math] \frac{d}{dx} y = \frac{d}{dx}[x^3] - \frac{d}{dx} [2x^2] -\frac{d}{dx} x + \frac{d}{dx} 9][/math]

                               \frac{dy}{dx} = 3x^2 - 4x - 1

            Skill 3: Plug in the (x, y) values to find ( \frac{dy}{dx})

            for that point. 

           Skill 4: To know the slope (gradient), apply the BODMAS rules to simplify

            the  equation.

Step 2: Find the equation of tangent line and the point

          Skill 1: Use the slope of curve and a point on curve to find its y-intercept.

                   NOTE:

         Skill 2: Use the slope of line and the y-intercept of line to find the equation

         of the line.

              EXAMPLE: Plug the slope m = -2 and the y-intercept b = -8 into the

                slope- intercept formula.

                         y = mx + b

                         y = -2x + -8

Step 3: Find the equation of the normal line to the tangent

NOTE: i) Repeat the step 2 but change the slope.

ii) Perpendicular lines have slopes that are opposite  

  reciprocals, like \frac{a}{b} \text{ and } \frac{-b}{a}. The slopes also have a product of -1

Step 4:  Skill 3: Substitute the corresponding intercept of above lines, to get  required point.

NOTE: i) If the line intersects with the x-axis put y = 0 then simplify for the x-coordinator  

ii)If the line intersects with the y-axis put x = 0 then simplify for the y-coordinator  

Step 5: Calculate the area of the triangle

Method 1:

Step 1: Locate the coordinates of the endpoints.

            EXAMPLE: The given points M (2, 3) and N (-1, 0), (2, -4.)

            Therefore, (x_1, y_1) = (2, 3),(x_2, y_2) = (-1, 0) and (x_3, y_3) = (2, -4).

Step 2: Plug the corresponding coordinates into the Area of triangle formula

            FORMULA:

            Area of the triangle

             \frac{1}{2} {x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)}

Step 3: Simplify further

            NOTE: Apply the BODMAS rules

Method 2:

Step 1: Calculate the base and height lengths of the triangle  

          NOTE: Use the distance formula to calculate lengths.

Step 2: Substitute the either values (base and height lengths) in the formula

              Area of the right angle triangle = \frac{1}{2} base * height

Step 3: Simplify further

            NOTE: Apply the BODMAS rules