Question

find the distance between the point and the line. round your answer to the nearest tenth.
1.
2.

Explanation:

1. Explanation for the first - type of problem (finding distance from a point to a line in a coordinate - plane):

• First, assume the equation of the line is in the general form \(Ax + By+C = 0\) and the point is \((x_0,y_0)\). The distance \(d\) from the point \((x_0,y_0)\) to the line \(Ax + By + C=0\) is given by the formula \(d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^{2}+B^{2}}}\).
• However, if we don't want to use the general - form formula, we can also use the following geometric approach:
• Step 1: Identify two points on the line and the given point
• Let's assume we have two points \((x_1,y_1)\) and \((x_2,y_2)\) on the line and the point \((x_0,y_0)\) whose distance to the line we want to find. First, find the vector \(\vec{v}=(x_2 - x_1,y_2 - y_1)\) along the line and the vector \(\vec{u}=(x_0 - x_1,y_0 - y_1)\) from one of the points on the line to the given point.
• For example, if the line passes through points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) and the point is \(P(x_0,y_0)\).
• Calculate the cross - product (in 2 - D, we can use the formula \(\vec{v}\times\vec{u}=x_1y_2 - x_2y_1\)) of the vectors \(\vec{v}\) and \(\vec{u}\). The magnitude of the cross - product \(\vert\vec{v}\times\vec{u}\vert\) is related to the area of the parallelogram formed by \(\vec{v}\) and \(\vec{u}\).
• The length of the vector \(\vec{v}\) is \(l=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\).
• The distance \(d\) from the point \(P\) to the line is \(d = \frac{\vert\vec{v}\times\vec{u}\vert}{\vert\vec{v}\vert}\).
• Step 2: Calculate the values
• Let's say the line passes through points \(A(0,4)\) and \(B(4,0)\) (assuming we can read these from the graph), and the point \(F(4,7)\).
• \(\vec{v}=(4 - 0,0 - 4)=(4,-4)\), \(\vec{u}=(4 - 0,7 - 4)=(4,3)\).
• The cross - product (in 2 - D) \(\vec{v}\times\vec{u}=4\times3-4\times(- 4)=12 + 16 = 28\).
• The magnitude of \(\vec{v}\) is \(\vert\vec{v}\vert=\sqrt{4^{2}+(-4)^{2}}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2}\).
• The distance \(d=\frac{\vert\vec{v}\times\vec{u}\vert}{\vert\vec{v}\vert}=\frac{28}{4\sqrt{2}}=\frac{7}{\sqrt{2}}\approx4.9\).

1. For the second problem (finding the distance from point \(D\) to the line):

• Step 1: Identify two points on the line and the given point
• Suppose the line passes through \(A(-4,0)\) and \(B(-2,3)\), and the point \(D(-1,-1)\).
• \(\vec{v}=(-2+4,3 - 0)=(2,3)\), \(\vec{u}=(-1 + 4,-1 - 0)=(3,-1)\).
• Step 2: Calculate the cross - product and magnitudes
• The cross - product \(\vec{v}\times\vec{u}=2\times(-1)-3\times3=-2 - 9=-11\).
• The magnitude of \(\vec{v}\) is \(\vert\vec{v}\vert=\sqrt{2^{2}+3^{2}}=\sqrt{4 + 9}=\sqrt{13}\).
• The distance \(d=\frac{\vert\vec{v}\times\vec{u}\vert}{\vert\vec{v}\vert}=\frac{\vert - 11\vert}{\sqrt{13}}=\frac{11}{\sqrt{13}}\approx3.1\).

Since we don't have the exact coordinates from the graph clearly, if we assume we can read the coordinates accurately:

Answer:

1. Approximately \(4.9\)
2. Approximately \(3.1\)