Question

1-6: find the equation of line dq.
1-7: find the point of intersection between dq and ab
1-8: find the distance between d and q. this is the shortest distance between line ab and dc.

Explanation:

Since no coordinates of points \(D\), \(Q\), \(A\) and \(B\) are given, we assume the coordinates of \(D(x_1,y_1)\) and \(Q(x_2,y_2)\) for the following steps.

Step1: Find the slope of line DQ

The slope \(m\) of a line passing through two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\) (assuming \(x_2
eq x_1\)).
\[m=\frac{y_2 - y_1}{x_2 - x_1}\]

Step2: Use the point - slope form to find the equation of line DQ

The point - slope form of a line is \(y - y_1=m(x - x_1)\). Substituting the slope \(m\) from Step 1 and the point \((x_1,y_1)\) we get:
\[y - y_1=\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\]

Step3: Assume the equation of line AB is \(y = m_{AB}x + c_{AB}\)

To find the intersection point of line \(DQ\) (with equation \(y - y_1=\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\) or \(y=\frac{y_2 - y_1}{x_2 - x_1}x+(y_1-\frac{y_2 - y_1}{x_2 - x_1}x_1)\)) and line \(AB\) (with equation \(y = m_{AB}x + c_{AB}\)), we set the two equations equal to each other:
\[\frac{y_2 - y_1}{x_2 - x_1}x+(y_1-\frac{y_2 - y_1}{x_2 - x_1}x_1)=m_{AB}x + c_{AB}\]
Solve for \(x\):
\[x=\frac{c_{AB}-y_1 + \frac{y_2 - y_1}{x_2 - x_1}x_1}{\frac{y_2 - y_1}{x_2 - x_1}-m_{AB}}\]
Then substitute \(x\) into \(y = m_{AB}x + c_{AB}\) to get \(y\).

Step4: Find the distance between D and Q

The distance \(d\) between two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
\[d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\]

Answer:

1 - 6: \(y - y_1=\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\)
1 - 7: The intersection point \((x,y)\) where \(x=\frac{c_{AB}-y_1 + \frac{y_2 - y_1}{x_2 - x_1}x_1}{\frac{y_2 - y_1}{x_2 - x_1}-m_{AB}}\) and \(y = m_{AB}x + c_{AB}\)
1 - 8: \(\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)