Question

point d is located on $overline{mv}$. the coordinates of d are $(0,-\frac{3}{4})$. what ratio relates dv to md?

Explanation:

Step1: Assume coordinates of M and V

Let's assume from the graph that \(M=(- 4,3)\) and \(V=(2,-2)\) (by observing the grid - points).

Step2: Use the section - formula

If a point \(D(x,y)\) divides the line segment joining \(M(x_1,y_1)\) and \(V(x_2,y_2)\) in the ratio \(k:1\), then \(x=\frac{kx_2 + x_1}{k + 1}\) and \(y=\frac{ky_2+y_1}{k + 1}\). Here \(x = 0,y=-\frac{3}{4},x_1=-4,y_1 = 3,x_2=2,y_2=-2\).
Since \(x = 0=\frac{k\times2+( - 4)}{k + 1}\), we cross - multiply: \(0\times(k + 1)=2k-4\), so \(2k=4\), \(k = 2\).
We can also use the \(y\) - coordinate formula \(y=-\frac{3}{4}=\frac{k\times(-2)+3}{k + 1}\). Cross - multiply: \(-\frac{3}{4}(k + 1)=-2k + 3\). Multiply both sides by 4 to get \(-3(k + 1)=-8k + 12\). Expand: \(-3k-3=-8k + 12\). Add \(8k\) to both sides: \(5k-3=12\). Add 3 to both sides: \(5k=15\), \(k = 3\). Let's use the distance formula.
The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
The distance \(MD=\sqrt{(0 + 4)^2+(-\frac{3}{4}-3)^2}=\sqrt{16+(-\frac{15}{4})^2}=\sqrt{16+\frac{225}{16}}=\sqrt{\frac{256 + 225}{16}}=\sqrt{\frac{481}{16}}\).
The distance \(DV=\sqrt{(2 - 0)^2+(-2+\frac{3}{4})^2}=\sqrt{4+(-\frac{5}{4})^2}=\sqrt{4+\frac{25}{16}}=\sqrt{\frac{64 + 25}{16}}=\sqrt{\frac{89}{16}}\).
Another way:
Let the ratio of \(DV:MD\) be \(r\). If we consider the line segment \(MV\) and the point \(D\) on it.
We can use the concept of similar right - triangles formed by the vertical and horizontal displacements.
The vertical displacement from \(M\) to \(D\) is \(3-(-\frac{3}{4})=\frac{12 + 3}{4}=\frac{15}{4}\).
The vertical displacement from \(D\) to \(V\) is \(-\frac{3}{4}-(-2)=\frac{-3 + 8}{4}=\frac{5}{4}\).
The ratio of the vertical displacements (which is the same as the ratio of the line - segment lengths since the line is straight) is \(\frac{DV}{MD}=\frac{\frac{5}{4}}{\frac{15}{4}}=\frac{1}{3}\).

Answer:

The ratio of \(DV\) to \(MD\) is \(1:3\).