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Sagot :
Let's solve this problem step by step.
### Step 1: Finding the Equation of Line AB
To find the equation of the line passing through points [tex]\(A(14, -1)\)[/tex] and [tex]\(B(2, 1)\)[/tex], we first need to determine its slope.
The slope of line [tex]\(AB\)[/tex] is given by the formula:
[tex]\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substitute the coordinates of points [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ m_{AB} = \frac{1 - (-1)}{2 - 14} = \frac{1 + 1}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \][/tex]
Next, we use the point-slope form of a line equation, [tex]\(y - y_1 = m_{AB}(x - x_1)\)[/tex], to find the equation of line [tex]\(AB\)[/tex]. Using point [tex]\(B(2, 1)\)[/tex] and the slope [tex]\(-\frac{1}{6}\)[/tex], we get:
[tex]\[ y - 1 = -\frac{1}{6}(x - 2) \][/tex]
Simplify this to the slope-intercept form [tex]\(y = mx + b\)[/tex]:
[tex]\[ y - 1 = -\frac{1}{6}x + \frac{1}{3} \][/tex]
[tex]\[ y = -\frac{1}{6}x + \frac{1}{3} + 1 \][/tex]
[tex]\[ y = -\frac{1}{6}x + \frac{4}{3} \][/tex]
Thus, the y-intercept ([tex]\(b\)[/tex]) of line [tex]\(AB\)[/tex] is [tex]\(\frac{4}{3}\)[/tex].
### Step 2: Finding the Equation of Line BC
Since [tex]\(AB\)[/tex] and [tex]\(BC\)[/tex] are perpendicular, the slope of line [tex]\(BC\)[/tex] ([tex]\(m_{BC}\)[/tex]) is the negative reciprocal of the slope of line [tex]\(AB\)[/tex]:
[tex]\[ m_{BC} = -\frac{1}{-\frac{1}{6}} = 6 \][/tex]
Using the point-slope form of a line equation [tex]\(y - y_1 = m_{BC}(x - x_1)\)[/tex], we use point [tex]\(B(2, 1)\)[/tex] and the slope [tex]\(6\)[/tex]:
[tex]\[ y - 1 = 6(x - 2) \][/tex]
Simplify this to the slope-intercept form [tex]\(y = mx + b\)[/tex]:
[tex]\[ y - 1 = 6x - 12 \][/tex]
[tex]\[ y = 6x - 12 + 1 \][/tex]
[tex]\[ y = 6x - 11 \][/tex]
Thus, the equation of line [tex]\(BC\)[/tex] is [tex]\(y = 6x - 11\)[/tex].
### Step 3: Finding the x-coordinate of Point C
Given that the y-coordinate of point [tex]\(C\)[/tex] is 13, we substitute [tex]\(y = 13\)[/tex] into the equation of line [tex]\(BC\)[/tex]:
[tex]\[ 13 = 6x - 11 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ 13 + 11 = 6x \][/tex]
[tex]\[ 24 = 6x \][/tex]
[tex]\[ x = \frac{24}{6} = 4 \][/tex]
Thus, the x-coordinate of point [tex]\(C\)[/tex] is 4.
### Final Answers
1. The [tex]\(y\)[/tex]-intercept of [tex]\(\overleftrightarrow{A B}\)[/tex] is [tex]\(\frac{4}{3}\)[/tex].
2. The equation of [tex]\(\overleftrightarrow{B C}\)[/tex] is [tex]\(y = 6x - 11\)[/tex].
3. The [tex]\(x\)[/tex]-coordinate of point [tex]\(C\)[/tex] is 4.
So, the final boxes should be filled with:
1. [tex]\( \frac{4}{3} \)[/tex]
2. [tex]\( 6 \)[/tex]
3. [tex]\( -11 \)[/tex]
4. [tex]\( 4 \)[/tex]
### Step 1: Finding the Equation of Line AB
To find the equation of the line passing through points [tex]\(A(14, -1)\)[/tex] and [tex]\(B(2, 1)\)[/tex], we first need to determine its slope.
The slope of line [tex]\(AB\)[/tex] is given by the formula:
[tex]\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substitute the coordinates of points [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ m_{AB} = \frac{1 - (-1)}{2 - 14} = \frac{1 + 1}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \][/tex]
Next, we use the point-slope form of a line equation, [tex]\(y - y_1 = m_{AB}(x - x_1)\)[/tex], to find the equation of line [tex]\(AB\)[/tex]. Using point [tex]\(B(2, 1)\)[/tex] and the slope [tex]\(-\frac{1}{6}\)[/tex], we get:
[tex]\[ y - 1 = -\frac{1}{6}(x - 2) \][/tex]
Simplify this to the slope-intercept form [tex]\(y = mx + b\)[/tex]:
[tex]\[ y - 1 = -\frac{1}{6}x + \frac{1}{3} \][/tex]
[tex]\[ y = -\frac{1}{6}x + \frac{1}{3} + 1 \][/tex]
[tex]\[ y = -\frac{1}{6}x + \frac{4}{3} \][/tex]
Thus, the y-intercept ([tex]\(b\)[/tex]) of line [tex]\(AB\)[/tex] is [tex]\(\frac{4}{3}\)[/tex].
### Step 2: Finding the Equation of Line BC
Since [tex]\(AB\)[/tex] and [tex]\(BC\)[/tex] are perpendicular, the slope of line [tex]\(BC\)[/tex] ([tex]\(m_{BC}\)[/tex]) is the negative reciprocal of the slope of line [tex]\(AB\)[/tex]:
[tex]\[ m_{BC} = -\frac{1}{-\frac{1}{6}} = 6 \][/tex]
Using the point-slope form of a line equation [tex]\(y - y_1 = m_{BC}(x - x_1)\)[/tex], we use point [tex]\(B(2, 1)\)[/tex] and the slope [tex]\(6\)[/tex]:
[tex]\[ y - 1 = 6(x - 2) \][/tex]
Simplify this to the slope-intercept form [tex]\(y = mx + b\)[/tex]:
[tex]\[ y - 1 = 6x - 12 \][/tex]
[tex]\[ y = 6x - 12 + 1 \][/tex]
[tex]\[ y = 6x - 11 \][/tex]
Thus, the equation of line [tex]\(BC\)[/tex] is [tex]\(y = 6x - 11\)[/tex].
### Step 3: Finding the x-coordinate of Point C
Given that the y-coordinate of point [tex]\(C\)[/tex] is 13, we substitute [tex]\(y = 13\)[/tex] into the equation of line [tex]\(BC\)[/tex]:
[tex]\[ 13 = 6x - 11 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ 13 + 11 = 6x \][/tex]
[tex]\[ 24 = 6x \][/tex]
[tex]\[ x = \frac{24}{6} = 4 \][/tex]
Thus, the x-coordinate of point [tex]\(C\)[/tex] is 4.
### Final Answers
1. The [tex]\(y\)[/tex]-intercept of [tex]\(\overleftrightarrow{A B}\)[/tex] is [tex]\(\frac{4}{3}\)[/tex].
2. The equation of [tex]\(\overleftrightarrow{B C}\)[/tex] is [tex]\(y = 6x - 11\)[/tex].
3. The [tex]\(x\)[/tex]-coordinate of point [tex]\(C\)[/tex] is 4.
So, the final boxes should be filled with:
1. [tex]\( \frac{4}{3} \)[/tex]
2. [tex]\( 6 \)[/tex]
3. [tex]\( -11 \)[/tex]
4. [tex]\( 4 \)[/tex]
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