Solve the problem.
-Find the centroid of the region cut from the first quadrant by the line $\frac { x } { 7 } + \frac { y } { 6 } = 1$.
A) $\bar { x } = \frac { 7 } { 3 } , \bar { y } = 2$
B) $\bar { x } = \frac { 3 } { 2 } , \bar { y } = \frac { 7 } { 4 }$
C) $\bar { x } = 2 , \bar { y } = \frac { 7 } { 3 }$
D) $\bar { x } = \frac { 7 } { 4 } , \bar { y } = \frac { 3 } { 2 }$

Question

Quizplus · Accepted Answer

The centroid $(\bar{x}, \bar{y})$ of a region bounded by $y = f(x)$, $x = a$, $x = b$, and the x-axis is given by $\bar{x} = \frac{1}{A}\int_{a}^{b} x \cdot f(x) dx$ and $\bar{y} = \frac{1}{2A}\int_{a}^{b} [f(x)]^2 dx$, where $A$ is the area of the region. For the given line $\frac{x}{7} + \frac{y}{6} = 1$, rearranging gives $y = 6 - \frac{6x}{7}$. The line intersects the x-axis at $x = 7$ and the y-axis at $y = 6$, forming a right triangle in the first quadrant. The area of this triangle is $A = \frac{1}{2} \cdot 7 \cdot 6 = 21$. The centroid of a right triangle is located at $\frac{1}{3}$ the distance from each axis, giving $\bar{x} = \frac{1}{3} \cdot 7 = \frac{7}{3}$ and $\bar{y} = \frac{1}{3} \cdot 6 = 2$, which matches option A.

Solve the Problem x7+y6=1\frac { x } { 7 } + \frac { y } { 6 } = 17x​+6y​=1

Solve the Problem $\frac { x } { 7 } + \frac { y } { 6 } = 1$