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Pretest: Coordinate Geometry

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[tex]$\overleftrightarrow{AB}$[/tex] and [tex]$\overleftrightarrow{BC}$[/tex] form a right angle at their point of intersection, [tex]$B$[/tex].

If the coordinates of [tex]$A$[/tex] and [tex]$B$[/tex] are [tex]$(14, -1)$[/tex] and [tex]$(2, 1)$[/tex], respectively, the [tex]$y$[/tex]-intercept of [tex]$\overleftrightarrow{AB}$[/tex] is [tex]$\square$[/tex] and the equation of [tex]$\overleftrightarrow{AB}$[/tex] is [tex]$y=$[/tex] [tex]$\square x + \square$[/tex].

If the [tex]$y$[/tex]-coordinate of point [tex]$C$[/tex] is 13, its [tex]$x$[/tex]-coordinate is [tex]$\square$[/tex].


Sagot :

Let's solve the problem step-by-step.

First, we are given the coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex] as [tex]\( A(14, -1) \)[/tex] and [tex]\( B(2, 1) \)[/tex] respectively.

1. Calculate the slope of [tex]\(\overleftrightarrow{AB}\)[/tex]:

The formula for the slope [tex]\( m \)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]

Substituting in the given coordinates:
[tex]\[ m_{AB} = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \][/tex]

2. Determine the y-intercept of [tex]\(\overleftrightarrow{AB}\)[/tex]:

The equation of the line can be written as [tex]\( y = mx + c \)[/tex], where [tex]\( c \)[/tex] is the y-intercept. Using the slope [tex]\( m_{AB} = -\frac{1}{6} \)[/tex] and point [tex]\( B(2, 1) \)[/tex]:
[tex]\[ 1 = \left( -\frac{1}{6} \right) \cdot 2 + c \][/tex]
[tex]\[ 1 = -\frac{2}{6} + c \][/tex]
[tex]\[ 1 = -\frac{1}{3} + c \][/tex]
Adding [tex]\(\frac{1}{3}\)[/tex] to both sides:
[tex]\[ c = 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} \][/tex]

So, the y-intercept [tex]\( c \)[/tex] of [tex]\(\overleftrightarrow{AB}\)[/tex] is:
[tex]\[ c = \frac{4}{3} \approx 1.333 \][/tex]

3. The equation of [tex]\(\overleftrightarrow{AB}\)[/tex]:

Now that we have the slope [tex]\( m_{AB} \)[/tex] and the y-intercept [tex]\( c \)[/tex]:
[tex]\[ y = -\frac{1}{6}x + \frac{4}{3} \][/tex]

4. Calculate the slope of [tex]\(\overleftrightarrow{BC}\)[/tex]:

Since [tex]\(\overleftrightarrow{AB}\)[/tex] and [tex]\(\overleftrightarrow{BC}\)[/tex] form a right angle, the slope [tex]\( m_{BC} \)[/tex] is the negative reciprocal of [tex]\( m_{AB} \)[/tex]:
[tex]\[ m_{BC} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{1}{6}} = 6 \][/tex]

5. Determine the x-coordinate of point [tex]\( C \)[/tex]:

Given the y-coordinate of point [tex]\( C \)[/tex] is [tex]\( 13 \)[/tex], use the point-slope form of the line [tex]\(\overleftrightarrow{BC}\)[/tex] passing through point [tex]\( B(2, 1)\)[/tex]:
[tex]\[ y - 1 = m_{BC}(x - 2) \][/tex]
Substituting [tex]\( y = 13 \)[/tex] and [tex]\( m_{BC} = 6 \)[/tex]:
[tex]\[ 13 - 1 = 6(x - 2) \][/tex]
[tex]\[ 12 = 6(x - 2) \][/tex]
[tex]\[ 12 = 6x - 12 \][/tex]
Adding 12 to both sides:
[tex]\[ 24 = 6x \][/tex]
Dividing both sides by 6:
[tex]\[ x = 4 \][/tex]

So, the answers are:
- The [tex]\( y \)[/tex]-intercept of [tex]\(\overleftrightarrow{AB}\)[/tex] is approximately [tex]\( 1.333 \)[/tex].
- The equation of [tex]\(\overleftrightarrow{AB}\)[/tex] is [tex]\( y = -\frac{1}{6}x + \frac{4}{3} \)[/tex].
- The [tex]\( x \)[/tex]-coordinate of point [tex]\( C \)[/tex] is [tex]\( 4 \)[/tex].
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