Question

triangle abc is located on three lines such that the vertices occur at the points of intersection of pairs of the lines, as shown on the graph. line ab, line bc, line ac. type the answer in the space provided. explain your reasoning. then, graph the three lines that contain △abc on the coordinate - plane.

Explanation:

Step1: Recall line - slope form

The equation of a line in slope - intercept form is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept. To graph the lines with given slopes (\(m_{A'C'}=5\), \(m_{B'C'}=\frac{3}{2}\), \(m_{A'B'}=-\frac{2}{3}\)), we can assume a y - intercept (for simplicity, we can assume \(b = 0\) first and then adjust). For a line \(y = mx + b\), if we know the slope \(m\) and a point \((x_1,y_1)\) on the line, we can use the point - slope form \(y - y_1=m(x - x_1)\) to find other points on the line.

Step2: Find points on the lines

For the line \(A'C'\) with \(m = 5\), if we start from the origin \((0,0)\) (assuming \(b = 0\)), when \(x = 1\), \(y=5\times1 = 5\), so \((1,5)\) is on the line. We can plot these two points and draw the line. For the line \(B'C'\) with \(m=\frac{3}{2}\), if \(x = 2\), then \(y=\frac{3}{2}\times2=3\), starting from the origin \((0,0)\) we can plot \((0,0)\) and \((2,3)\) and draw the line. For the line \(A'B'\) with \(m =-\frac{2}{3}\), if \(x = 3\), then \(y=-\frac{2}{3}\times3=- 2\), starting from the origin \((0,0)\) we can plot \((0,0)\) and \((3,-2)\) and draw the line.

Step3: Consider the triangle formation

The three lines will intersect at three points which will form \(\triangle A'B'C'\). The reason for this is that non - parallel lines (lines with different slopes) in a two - dimensional plane will intersect at a single point. Since the slopes of \(A'C'\), \(B'C'\) and \(A'B'\) are all different, they will intersect pairwise to form the vertices of the triangle.

To graph:

1. For line \(A'C'\) with slope \(m = 5\), start at the origin \((0,0)\), move 1 unit to the right (increase \(x\) by 1) and 5 units up (increase \(y\) by 5) to get the point \((1,5)\), then draw a straight line through these two points.
2. For line \(B'C'\) with slope \(m=\frac{3}{2}\), start at the origin \((0,0)\), move 2 units to the right (increase \(x\) by 2) and 3 units up (increase \(y\) by 3) to get the point \((2,3)\), then draw a straight line through these two points.
3. For line \(A'B'\) with slope \(m =-\frac{2}{3}\), start at the origin \((0,0)\), move 3 units to the right (increase \(x\) by 3) and 2 units down (decrease \(y\) by 2) to get the point \((3,-2)\), then draw a straight line through these two points.

The intersection points of these three lines will form \(\triangle A'B'C'\).

Answer:

Graph the lines as described above. The lines with slopes \(5\), \(\frac{3}{2}\) and \(-\frac{2}{3}\) will intersect pairwise to form \(\triangle A'B'C'\) because non - parallel lines in a plane intersect at a point and these lines have different slopes.