Question

Plzz can u guys help me with this....ASAP​

Answer 1

First question: Since this shape is a square, the midpoint of the two diagonals shall coincide with each other.

Second question: assume that by "midpoint" the question refers to the centroid of the triangle. The centroid of a triangle is on its median 2/3 the way from the corresponding vertice.

Explanation:

First question

Midpoint of the diagonal between (x1, y1) and (x3, y3):

`\displaystyle \left((x_1 + x_3)/(2), (y_1 + y_3)/(2)\right)`.

Similarly, midpoint of the diagonal between (x2, y2) and (x4, y4):

`\displaystyle \left((x_2 + x_4)/(2), (y_2 + y_4)/(2)\right)`.

The two midpoints shall coincide. Therefore,

`\displaystyle (x_1 + x_3)/(2) = (x_2 + x_4)/(2) \implies x_1 + x_3 = x_2 + x_4`.

Similarly,

`\displaystyle (y_1 + y_3)/(2) = (y_2 + y_4)/(2) \implies y_1 + y_3 = x_2 + x_4`.

Second question

The centroid of a triangle divides all three of its medians at a 2:1 ratio. If the length of a median of the triangle is
`L`, the centroid of that triangle is at a distance of
`2/3\;L` from the vertex on that median.

Start with the median that goes through the vertex
`(x_1, y_1)`. That median also goes through the midpoint between
`(x_2, y_2)` and
`(x_3, y_3)`.

• Vertex:
  `(x_1, y_1)`.
• Midpoint of the opposite side:
  `\displaystyle \left ((x_2 + x_3)/(2), (y_2 + y_3)/(2)\right)`

The centroid will be located at

`\displaystyle \left (\phantom{\frac{\phantom{x_(1)}}{2}}\right.x_1 + (2)/(3)\underbrace{\left((x_2 + x_3)/(2) - x_1\right)}_{\begin{gathered}\text{Separation}\\[-0.5em]\text{in }x\text{-}\\[-0.5em]\text{direction}\end{gathered}}, \quad y_1 + (2)/(3)\underbrace{\left((y_2 + y_3)/(2) - y_1\right)}_{{\begin{gathered}\text{Separation}\\[-0.5em]\text{in }y\text{-}\\[-0.5em]\text{direction}\end{gathered}}}\left.\phantom{\frac{\phantom{x_(1)}}{2}}\right)\right)`.

Simplify this expression to obtain:

`\displaystyle \left ((x_1 + x_2 + x_3)/(3), (y_1 +y_2 + y_3)/(3)\right)\right)`.