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

Coordinates of points:
[tex]\[
A = (14, -1), \quad B = (2, 1)
\][/tex]

1. The [tex]\( y \)[/tex]-intercept of [tex]\(\overleftrightarrow{AB} \)[/tex] is [tex]\(\square\)[/tex].

2. The equation of [tex]\(\overleftrightarrow{BC}\)[/tex] is [tex]\( y = \square x + \square \)[/tex].

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



Answer :

GinnyAnswer
To solve this problem, we need to find several values based on the coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex].

1. Determine the slope of line [tex]\(\overleftrightarrow{A B}\)[/tex]:

The coordinates of [tex]\( A \)[/tex] are [tex]\((14, -1)\)[/tex] and the coordinates of [tex]\( B \)[/tex] are [tex]\((2, 1)\)[/tex].

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

Applying the coordinates of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ \text{slope}_{AB} = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \][/tex]

So, the slope of [tex]\(\overleftrightarrow{A B}\)[/tex] is [tex]\(-\frac{1}{6}\)[/tex].

2. Find the y-intercept of [tex]\(\overleftrightarrow{A B}\)[/tex]:

The equation of a line in slope-intercept form is [tex]\( y = mx + b \)[/tex].

Using point [tex]\( B \)[/tex] ([tex]\(2, 1\)[/tex]) to find [tex]\( b \)[/tex]:
[tex]\[ 1 = -\frac{1}{6} \cdot 2 + b \][/tex]
[tex]\[ 1 = -\frac{1}{3} + b \][/tex]
[tex]\[ b = 1 + \frac{1}{3} = \frac{4}{3} \][/tex]

So, the y-intercept of [tex]\(\overleftrightarrow{A B}\)[/tex] is [tex]\( \frac{4}{3} \)[/tex] or approximately [tex]\( 1.3333333333333333 \)[/tex].

3. Determine the slope of line [tex]\(\overleftrightarrow{B C}\)[/tex]:

Since [tex]\(\overleftrightarrow{B C}\)[/tex] is perpendicular to [tex]\(\overleftrightarrow{A B}\)[/tex], the slope of [tex]\(\overleftrightarrow{B C}\)[/tex] is the negative reciprocal of the slope of [tex]\(\overleftrightarrow{A B}\)[/tex].

The slope of [tex]\(\overleftrightarrow{A B}\)[/tex] is [tex]\(-\frac{1}{6}\)[/tex], so:
[tex]\[ \text{slope}_{BC} = -1 \div -\frac{1}{6} = 6 \][/tex]

Therefore, the slope of [tex]\(\overleftrightarrow{B C}\)[/tex] is 6.

4. Find the y-intercept of [tex]\(\overleftrightarrow{B C}\)[/tex]:

Using point [tex]\( B \)[/tex] ([tex]\(2, 1\)[/tex]) to find [tex]\( b \)[/tex]:
[tex]\[ 1 = 6 \cdot 2 + b \][/tex]
[tex]\[ 1 = 12 + b \][/tex]
[tex]\[ b = 1 - 12 = -11 \][/tex]

So, the y-intercept of [tex]\(\overleftrightarrow{B C}\)[/tex] is [tex]\( -11 \)[/tex].

5. Find the x-coordinate of point [tex]\( C \)[/tex] if its y-coordinate is 13:

Using the equation of [tex]\(\overleftrightarrow{B C}\)[/tex]:
[tex]\[ y = 6x - 11 \][/tex]

Substitute [tex]\( y = 13 \)[/tex]:
[tex]\[ 13 = 6x - 11 \][/tex]
[tex]\[ 24 = 6x \][/tex]
[tex]\[ x = \frac{24}{6} = 4 \][/tex]

So, the solutions are:
- The y-intercept of [tex]\(\overleftrightarrow{A B}\)[/tex] is [tex]\( \frac{4}{3} \)[/tex] or [tex]\( 1.3333333333333333 \)[/tex].
- The equation of [tex]\(\overleftrightarrow{B C}\)[/tex] is [tex]\( y = 6x - 11 \)[/tex].
- The x-coordinate of point [tex]\( C \)[/tex] is [tex]\( 4 \)[/tex].

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