Question

y = -x + 7\
y = \frac{1}{4}x - 3\
plot two lines by clicking the graph.\
click a line to delete it.\
\
y\
10\
9\
8\
7\
6\
5\
4\
3\
2

Explanation:

Step1: Find intersection of lines

Set $-x + 7 = \frac{1}{4}x - 3$. Solve:
$-x - \frac{1}{4}x = -3 -7$ → $-\frac{5}{4}x = -10$ → $x=8$. Substitute $x=8$ into $y=-x+7$: $y=-1$. So intersection is (8,-1).

Step2: Identify points for $y=-x+7$

When $x=0$, $y=7$ (point (0,7)). When $y=0$, $x=7$ (point (7,0)).

Step3: Identify points for $y=\frac{1}{4}x -3$

When $x=0$, $y=-3$ (point (0,-3)). When $y=0$, $\frac{1}{4}x=3$ → $x=12$ (point (12,0)).

Step4: Plot lines

For $y=-x+7$, click (0,7) and (7,0) to draw. For $y=\frac{1}{4}x-3$, click (0,-3) and (12,0) to draw. The lines intersect at (8,-1).

Answer:

Intersection point: (8, -1); Line 1 passes through (0,7) and (7,0); Line 2 passes through (0,-3) and (12,0)