# Area of Triangle

• Aug 8th 2010, 07:48 PM
ilovemymath
Area of Triangle
The tangent and normal to the curve $\displaystyle y=4(sqroot(x+2))$ at the point A(7,12) cut the x-axis at points B and C respectively. Calculate the area of the triangle ABC.
• Aug 8th 2010, 08:01 PM
eumyang
$\displaystyle y = 4\sqrt{x + 2}$
$\displaystyle y' = \dfrac{2}{\sqrt{x +2}}$

At point A(7, 12), the slope of the tangent line is $\displaystyle m = \dfrac{2}{\sqrt{7 +2}} = \dfrac{2}{3}$. The slope of the normal line would be $\displaystyle \dfrac{-1}{m} = -\dfrac{3}{2}$.

The equation of the tangent line would be
$\displaystyle y = m(x - x_1) + y_1$

Plug in m, x1, y1, and also y = 0 to solve for x to get your point B:
$\displaystyle 0 = \frac{2}{3}(x - 7) + 12$
...

The equation of the normal line would be
$\displaystyle y = \frac{-1}{m}(x - x_1) + y_1$

Plug in -1/m, x1, y1, and also y = 0 to solve for x to get your point C:
$\displaystyle 0 = -\frac{3}{2}(x - 7) + 12$
...

ABC would be a right triangle so let AB be your base and AC be your height (it doesn't matter which is which). Find the distances AB and AC using the distance formula, then use the area of the triangle formula $\displaystyle A = \frac{1}{2}bh$ to find the area.