Answer:
a. (7, 5)
b. See answer below line
Step-by-step explanation:
Focusing on A, we want to figure out how far away B is from P. To do this, let's divide the distance from A to P by 2 (since the ratio splits the segment so that AP = 2/5 of the entire segment). We'll then multiply that number by 3 (accounting for the other 3/5) and add the coordinate number from P to get the coordinate for B.
Let's first do this with the x-coordinates. P lies on x = 1 and A lies on x = -3. 1 and -3 are 4 units away from each other, so one half of 4 is 2. This means that B's x-coordinate is 3 * 2 units, or 6 units away from P horizontally. 1 + 6 = 7, so the x-coordinate is 7.
Let's do the same thing for the y-coordinates. P lies on y = 2 and A lies on y = 0. 0 and 2 are 2 units away from each other and 2/2 = 1, so B's y-coordinate is 3 * 1, or 3 units away from P vertically. 2 + 3 = 5, so the y-coordinate is 5. This means that B exists at (7, 5).
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Now onto part B. I'll try to make this as short as possible. Let's find the slope of AP for starters:
[tex]\frac{2-0}{1+3} = \frac{2}{4} = \frac{1}{2}[/tex]
Next, let's make a slope-intercept form equation with point P:
[tex]y=mx+b[/tex]
[tex]2 = \frac{1}{2} (1) + b[/tex]
[tex]2 = \frac{1}{2} + b[/tex]
[tex]2 - \frac{1}{2} = \frac{1}{2} - \frac{1}{2} + b[/tex]
[tex]\frac{3}{2} = b[/tex]
[tex]y=\frac{1}{2}x + \frac{3}{2}[/tex]
Finally, let's put the coordinates for B into our equation:
[tex]5 = \frac{1}{2}(7) + \frac{3}{2}[/tex]
[tex]5 = \frac{7}{2} + \frac{3}{2} = \frac{10}{2} = 5[/tex]
When the coordinates for B are placed into the expression, the expression is true. This means that the slope at the point B is the same slope as on the line AP. Part A also showed how we used AP to find the position of B, meaning that B is intersected by the line AP. Combined, these show that B lies on the line AP.
