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[tex]$\overleftrightarrow{CD}$[/tex] is perpendicular to [tex]$\overleftrightarrow{AB}$[/tex] and passes through point [tex]$C(5, 12)$[/tex]. If the coordinates of [tex]$A$[/tex] and [tex]$B$[/tex] are [tex]$(-10, -3)$[/tex] and [tex]$(7, 14)$[/tex], respectively, the x-intercept of [tex]$\overleftrightarrow{CD}$[/tex] is [tex]$\square$[/tex]. The point [tex]$\square$[/tex] lies on [tex]$\overleftrightarrow{CD}$[/tex].



Answer :

To solve this problem step-by-step:

1. Find the slope of line [tex]\(AB\)[/tex]:
- The coordinates of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are [tex]\( A(-10, -3) \)[/tex] and [tex]\( B(7, 14) \)[/tex], respectively.
- 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].
- Applying the coordinates:
[tex]\[ m_{AB} = \frac{14 - (-3)}{7 - (-10)} = \frac{14 + 3}{7 + 10} = \frac{17}{17} = 1 \][/tex]
- Thus, the slope of line [tex]\(AB\)[/tex] is [tex]\(1.0\)[/tex].

2. Determine the slope of line [tex]\(CD\)[/tex] which is perpendicular to [tex]\(AB\)[/tex]:
- The slope of a line perpendicular to another is the negative reciprocal.
- Since the slope of [tex]\(AB\)[/tex] is [tex]\(1.0\)[/tex], the slope of [tex]\(CD\)[/tex] is:
[tex]\[ m_{CD} = -\frac{1}{1} = -1 \][/tex]
- Thus, the slope of line [tex]\(CD\)[/tex] is [tex]\(-1.0\)[/tex].

3. Formulate the equation of line [tex]\(CD\)[/tex]:
- The general form of the line equation is [tex]\( y = mx + c \)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(c\)[/tex] is the y-intercept.
- Using the slope of [tex]\(CD\)[/tex] ([tex]\( -1 \)[/tex]) and point [tex]\(C(5, 12)\)[/tex]:
[tex]\[ y = -x + c \][/tex]
- Substitute the point [tex]\(C\)[/tex] into the equation to solve for [tex]\(c\)[/tex]:
[tex]\[ 12 = -5 + c \quad \Rightarrow \quad c = 12 + 5 = 17 \][/tex]
- Thus, the equation of line [tex]\(CD\)[/tex] is:
[tex]\[ y = -x + 17 \][/tex]

4. Find the x-intercept of line [tex]\(CD\)[/tex]:
- To find the x-intercept, set [tex]\(y\)[/tex] to 0 and solve for [tex]\(x\)[/tex]:
[tex]\[ 0 = -x + 17 \quad \Rightarrow \quad x = 17 \][/tex]
- Therefore, the x-intercept of [tex]\( \overleftrightarrow{CD} \)[/tex] is [tex]\( 17 \)[/tex].

5. Find a point that lies on line [tex]\(CD\)[/tex]:
- From the previous step, the x-intercept gives us a point on the line when [tex]\( y = 0 \)[/tex].
- Hence, the point is [tex]\( (17, 0) \)[/tex].

To conclude, if [tex]$\overleftrightarrow{C D}$[/tex] is perpendicular to [tex]$\overleftrightarrow{A B}$[/tex] and passes through point [tex]\(C(5, 12)\)[/tex], then:
- The x-intercept of [tex]\( \overleftrightarrow{C D} \)[/tex] is [tex]\( \boxed{17} \)[/tex].
- The point [tex]\( \boxed{(17, 0)} \)[/tex] lies on [tex]\( \overleftrightarrow{C D} \)[/tex].

Final Answer:

- The x-intercept of [tex]\( \overleftrightarrow{C D} \)[/tex] is [tex]\( \boxed{17} \)[/tex].
- The point [tex]\( \boxed{(17, 0)} \)[/tex] lies on [tex]\( \overleftrightarrow{C D} \)[/tex].