ab=bc for a(-3,2) and c(2,7). which of the following could be the coordinates of b? select all that apply. a. (-2,2) b. (-3,7) c. (-4,5) d. (1,3) e. (-1,-1) f. (8,0)

Question

Accepted Answer

# Explanation:
To determine if a point $ B(x,y) $ lies on the line segment $ AC $ (or the line $ AC $), we first find the equation of the line passing through $ A(-3,2) $ and $ C(2,7) $.

## Step 1: Find the slope of $ AC $
The slope $ m $ between two points $ (x_1,y_1) $ and $ (x_2,y_2) $ is given by $ m=\frac{y_2 - y_1}{x_2 - x_1} $.
For $ A(-3,2) $ and $ C(2,7) $, we have:
$ m=\frac{7 - 2}{2 - (-3)}=\frac{5}{5} = 1 $

## Step 2: Find the equation of the line $ AC $
Using the point - slope form $ y - y_1=m(x - x_1) $, with $ (x_1,y_1)=(-3,2) $ and $ m = 1 $:
$ y-2=1\times(x + 3) $
Simplify to get the slope - intercept form $ y=x + 5 $

## Step 3: Check each option
- **Option A: $(-2,2)$**
Substitute $ x=-2 $ into $ y=x + 5 $, we get $ y=-2 + 5=3
eq2 $. So, $ (-2,2) $ is not on the line $ AC $.
- **Option B: $(-3,7)$**
Substitute $ x = - 3 $ into $ y=x + 5 $, we get $ y=-3 + 5 = 2
eq7 $. So, $ (-3,7) $ is not on the line $ AC $.
- **Option C: $(-4,5)$**
Substitute $ x=-4 $ into $ y=x + 5 $, we get $ y=-4 + 5 = 1
eq5 $. Wait, let's re - calculate. Wait, $ y=x + 5 $, when $ x=-4 $, $ y=-4 + 5=1$. Wait, maybe I made a mistake. Wait, no, let's check the slope again. Wait, the slope between $ A(-3,2) $ and $ C(2,7) $ is $ m = 1 $. The equation is $ y-2=(x + 3)$, so $ y=x + 5 $. For $ x=-4 $, $ y=-4 + 5 = 1
eq5 $. Wait, maybe the problem is about $ B $ being a point such that $ AB\perp BC $? Wait, the original problem says $ AB\perp BC $? Wait, the user's problem statement: "AB⊥BC for A(-3,2) and C(2,7). Which of the following could be the coordinates of B?"

Oh! I misread the problem. If $ AB\perp BC $, then the product of the slopes of $ AB $ and $ BC $ is $ - 1 $.

Let $ B=(x,y) $. The slope of $ AB$, $ m_{AB}=\frac{y - 2}{x + 3} $, and the slope of $ BC$, $ m_{BC}=\frac{y - 7}{x - 2} $. Since $ AB\perp BC $, $ m_{AB}\times m_{BC}=-1 $, so $ \frac{y - 2}{x + 3}\times\frac{y - 7}{x - 2}=-1 $, which simplifies to $ (y - 2)(y - 7)=-(x + 3)(x - 2) $, or $ y^{2}-9y + 14=-x^{2}-x + 6 $, or $ x^{2}+y^{2}+x-9y + 8 = 0 $

Now check each option:

- **Option A: $(-2,2)$**
Substitute $ x=-2,y = 2 $ into $ x^{2}+y^{2}+x-9y + 8 $:
$ (-2)^{2}+2^{2}+(-2)-9\times2 + 8=4 + 4-2-18 + 8=-4
eq0 $
- **Option B: $(-3,7)$**
Substitute $ x=-3,y = 7 $ into $ x^{2}+y^{2}+x-9y + 8 $:
$ (-3)^{2}+7^{2}+(-3)-9\times7 + 8=9 + 49-3-63 + 8=-0$ (since $ 9 + 49=58,58-3 = 55,55-63=-8,-8 + 8 = 0 $)
- **Option C: $(-4,5)$**
Substitute $ x=-4,y = 5 $ into $ x^{2}+y^{2}+x-9y + 8 $:
$ (-4)^{2}+5^{2}+(-4)-9\times5 + 8=16 + 25-4-45 + 8=0$ ( $ 16+25 = 41,41-4 = 37,37-45=-8,-8 + 8 = 0 $)
- **Option D: $(1,3)$**
Substitute $ x = 1,y = 3 $ into $ x^{2}+y^{2}+x-9y + 8 $:
$ 1^{2}+3^{2}+1-9\times3 + 8=1 + 9+1-27 + 8=-8
eq0 $
- **Option E: $(-1,-1)$**
Substitute $ x=-1,y=-1 $ into $ x^{2}+y^{2}+x-9y + 8 $:
$ (-1)^{2}+(-1)^{2}+(-1)-9\times(-1)+8=1 + 1-1 + 9+8=18
eq0 $
- **Option F: $(8,0)$**
Substitute $ x = 8,y = 0 $ into $ x^{2}+y^{2}+x-9y + 8 $:
$ 8^{2}+0^{2}+8-9\times0 + 8=64 + 0+8 + 8=80
eq0 $

Wait, maybe the problem is about $ B $ being a point such that $ B $ is on the perpendicular bisector of $ AC $? The mid - point of $ AC $ is $ M(\frac{-3 + 2}{2},\frac{2+7}{2})=M(-\frac{1}{2},\frac{9}{2}) $. The slope of $ AC $ is $ 1 $, so the slope of the perpendicular bisector is $ - 1 $. The equation of the perpendicular bisector is $ y-\frac{9}{2}=-1(x+\frac{1}{2}) $, or $ y=-x + 4 $

Check each option:

- **Option A: $(-2,2)$**
Substitute $ x=-2 $ into $ y=-x + 4 $, $ y=-(-2)+4=6
eq2 $
- **Option B: $(-3,7)$**
Substitute $ x=-3 $ into $ y=-x + 4 $, $ y=-(-3)+4=7 $. So, $ (-3,7) $ is on the perpendicular bisector.
- **Option C: $(-4,5)$**
Substitute $ x=-4 $ into $ y=-x + 4 $, $ y=-(-4)+4=8
eq5 $
- **Option D: $(1,3)$**
Substitute $ x = 1 $ into $ y=-x + 4 $, $ y=-1 + 4=3 $. So, $ (1,3) $ is on the perpendicular bisector.
- **Option E: $(-1,-1)$**
Substitute $ x=-1 $ into $ y=-x + 4 $, $ y=-(-1)+4=5
eq-1 $
- **Option F: $(8,0)$**
Substitute $ x = 8 $ into $ y=-x + 4 $, $ y=-8 + 4=-4
eq0 $

Since the problem says "AB⊥BC", if we consider the geometric meaning that $ \angle ABC = 90^{\circ} $, then $ B $ lies on the circle with $ AC $ as diameter (by Thales' theorem). The center of the circle is the mid - point of $ AC $: $ M(-\frac{1}{2},\frac{9}{2}) $, and the radius $ r=\frac{1}{2}\sqrt{(2 + 3)^{2}+(7 - 2)^{2}}=\frac{1}{2}\sqrt{25 + 25}=\frac{5\sqrt{2}}{2} $

The equation of the circle is $ (x+\frac{1}{2})^{2}+(y-\frac{9}{2})^{2}=\frac{25\times2}{4}=\frac{25}{2} $

Check option B: $ (-3,7) $
$ (-3+\frac{1}{2})^{2}+(7-\frac{9}{2})^{2}=(-\frac{5}{2})^{2}+(\frac{5}{2})^{2}=\frac{25}{4}+\frac{25}{4}=\frac{50}{4}=\frac{25}{2} $. So, $ (-3,7) $ is on the circle.

Check option C: $ (-4,5) $
$ (-4+\frac{1}{2})^{2}+(5-\frac{9}{2})^{2}=(-\frac{7}{2})^{2}+(\frac{1}{2})^{2}=\frac{49}{4}+\frac{1}{4}=\frac{50}{4}=\frac{25}{2} $. So, $ (-4,5) $ is on the circle.

Check option D: $ (1,3) $
$ (1+\frac{1}{2})^{2}+(3-\frac{9}{2})^{2}=(\frac{3}{2})^{2}+(-\frac{3}{2})^{2}=\frac{9}{4}+\frac{9}{4}=\frac{18}{4}=\frac{9}{2}
eq\frac{25}{2} $

Check option A: $ (-2,2) $
$ (-2+\frac{1}{2})^{2}+(2-\frac{9}{2})^{2}=(-\frac{3}{2})^{2}+(-\frac{5}{2})^{2}=\frac{9}{4}+\frac{25}{4}=\frac{34}{4}=\frac{17}{2}
eq\frac{25}{2} $

Check option E: $ (-1,-1) $
$ (-1+\frac{1}{2})^{2}+(-1-\frac{9}{2})^{2}=(-\frac{1}{2})^{2}+(-\frac{11}{2})^{2}=\frac{1}{4}+\frac{121}{4}=\frac{122}{4}=\frac{61}{2}
eq\frac{25}{2} $

Check option F: $ (8,0) $
$ (8+\frac{1}{2})^{2}+(0-\frac{9}{2})^{2}=(\frac{17}{2})^{2}+(-\frac{9}{2})^{2}=\frac{289}{4}+\frac{81}{4}=\frac{370}{4}=\frac{185}{2}
eq\frac{25}{2} $

So, the points $ B $ that satisfy $ AB\perp BC $ are $ B(-3,7) $ and $ B(-4,5) $ and $ B(1,3) $? Wait, no, when we checked the circle equation, $ (1,3) $ did not satisfy. Wait, my mistake in the circle equation calculation for $ (1,3) $:

$ (x+\frac{1}{2})^{2}+(y - \frac{9}{2})^{2} $

For $ (1,3) $:

$ (1+\frac{1}{2})^{2}+(3-\frac{9}{2})^{2}=(\frac{3}{2})^{2}+(-\frac{3}{2})^{2}=\frac{9 + 9}{4}=\frac{18}{4}=\frac{9}{2}$, and $ \frac{25}{2}=12.5 $, $ \frac{9}{2}=4.5 $, so it's not on the circle.

But when we checked the perpendicular bisector, $ (1,3) $ was on the perpendicular bisector. There is a confusion here. The correct condition for $ \angle ABC = 90^{\circ} $ is that $ B $ lies on the circle with $ AC $ as diameter (Thales' theorem). So, let's recalculate the radius:

The distance between $ A(-3,2) $ and $ C(2,7) $ is $ d=\sqrt{(2 + 3)^{2}+(7 - 2)^{2}}=\sqrt{25 + 25}=\sqrt{50}=5\sqrt{2} $, so the radius $ r=\frac{d}{2}=\frac{5\sqrt{2}}{2} $, and $ r^{2}=\frac{50}{4}=\frac{25}{2}=12.5 $

For point $ B(-3,7) $:

$ (x + \frac{1}{2})^{2}+(y-\frac{9}{2})^{2}=(-3+\frac{1}{2})^{2}+(7-\frac{9}{2})^{2}=(-\frac{5}{2})^{2}+(\frac{5}{2})^{2}=\frac{25 + 25}{4}=\frac{50}{4}=\frac{25}{2} $. Correct.

For point $ B(-4,5) $:

$ (-4+\frac{1}{2})^{2}+(5-\frac{9}{2})^{2}=(-\frac{7}{2})^{2}+(\frac{1}{2})^{2}=\frac{49+1}{4}=\frac{50}{4}=\frac{25}{2} $. Correct.

For point $ B(1,3) $:

$ (1+\frac{1}{2})^{2}+(3-\frac{9}{2})^{2}=(\frac{3}{2})^{2}+(-\frac{3}{2})^{2}=\frac{9 + 9}{4}=\frac{18}{4}=\frac{9}{2}
eq\frac{25}{2} $. So, it's not on the circle.

For point $ B(-2,2) $:

$ (-2+\frac{1}{2})^{2}+(2-\frac{9}{2})^{2}=(-\frac{3}{2})^{2}+(-\frac{5}{2})^{2}=\frac{9 + 25}{4}=\frac{34}{4}=\frac{17}{2}
eq\frac{25}{2} $

For point $ B(-1,-1) $:

$ (-1+\frac{1}{2})^{2}+(-1-\frac{9}{2})^{2}=(-\frac{1}{2})^{2}+(-\frac{11}{2})^{2}=\frac{1+121}{4}=\frac{122}{4}=\frac{61}{2}
eq\frac{25}{2} $

For point $ B(8,0) $:

$ (8+\frac{1}{2})^{2}+(0-\frac{9}{2})^{2}=(\frac{17}{2})^{2}+(-\frac{9}{2})^{2}=\frac{289 + 81}{4}=\frac{370}{4}=\frac{185}{2}
eq\frac{25}{2} $

So the correct options are B. $(-3,7)$ and C. $(-4,5)$ and let's check the slope condition again.

Slope of $ AB $ for $ B(-3,7) $: $ m_{AB}=\frac{7 - 2}{-3+3} $, which is undefined (vertical line). Slope of $ BC $: $ m_{BC}=\frac{7 - 7}{2+3}=0 $ (horizontal line). So, $ AB\perp BC $ (vertical and horizontal lines are perpendicular).

For $ B(-4,5) $:

Slope of $ AB $: $ m_{AB}=\frac{5 - 2}{-4 + 3}=\frac{3}{-1}=-3 $

Slope of $ BC $: $ m_{BC}=\frac{5 - 7}{-4 - 2}=\frac{-2}{-6}=\frac{1}{3} $

Product of slopes: $ (-3)\times\frac{1}{3}=-1 $, so $ AB\perp BC $

For $ B(1,3) $:

Slope of $ AB $: $ m_{AB}=\frac{3 - 2}{1+3}=\frac{1}{4} $

Slope of $ BC $: $ m_{BC}=\frac{3 - 7}{1 - 2}=\frac{-4}{-1}=4 $

Product: $ \frac{1}{4}\times4 = 1
eq-1 $,

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QUESTION IMAGE

Question

Explanation:

Step 1: Find the slope of \( AC \)

Step 2: Find the equation of the line \( AC \)

Step 3: Check each option

Answer:

Step 1: Find the slope of \( AC \)

Step 2: Find the equation of the line \( AC \)

Step 3: Check each option