Question

select the correct answer from each drop - down menu.
$overleftrightarrow{cd}$ is perpendicular to $overleftrightarrow{ab}$ and passes through point $c(5, 12)$.
if the coordinates of $a$ and $b$ are $(-10, -3)$ and $(7, 14)$, respectively, the $x$-intercept of $overleftrightarrow{cd}$ is $circ$. the point $circ$ lies on $overleftrightarrow{cd}$.

Explanation:

Step1: Find slope of \( \overleftrightarrow{AB} \)

Using slope formula \( m=\frac{y_2 - y_1}{x_2 - x_1} \) with \( A(-10,-3) \) and \( B(7,14) \), we get \( m_{AB}=\frac{14 - (-3)}{7 - (-10)}=\frac{17}{17} = 1 \).

Step2: Find slope of \( \overleftrightarrow{CD} \)

Since \( \overleftrightarrow{CD}\perp\overleftrightarrow{AB} \), \( m_{CD}=-\frac{1}{m_{AB}}=- 1 \).

Step3: Find equation of \( \overleftrightarrow{CD} \)

Using point - slope form \( y - y_1=m(x - x_1) \) with \( m = - 1 \) and \( C(5,12) \), we have \( y-12=-1(x - 5) \), which simplifies to \( y=-x + 17 \).

Step4: Find x - intercept of \( \overleftrightarrow{CD} \)

Set \( y = 0 \) in \( y=-x + 17 \), then \( 0=-x + 17\Rightarrow x = 17 \).

Step5: Check which point lies on \( \overleftrightarrow{CD} \)

Substitute the x and y coordinates of the candidate point into \( y=-x + 17 \) and check if the equation is satisfied.

Answer:

First, find the slope of \( \overleftrightarrow{AB} \). The slope formula is \( m = \frac{y_2 - y_1}{x_2 - x_1} \). For points \( A(-10, -3) \) and \( B(7, 14) \), the slope of \( \overleftrightarrow{AB} \) is \( \frac{14 - (-3)}{7 - (-10)} = \frac{17}{17} = 1 \). Since \( \overleftrightarrow{CD} \) is perpendicular to \( \overleftrightarrow{AB} \), the slope of \( \overleftrightarrow{CD} \) is the negative reciprocal, which is \( -1 \).

Now, use the point - slope form \( y - y_1 = m(x - x_1) \) to find the equation of \( \overleftrightarrow{CD} \). We know that \( m=-1 \) and the point \( C(5,12) \). Substituting these values into the point - slope form:

\( y - 12=-1(x - 5) \)

Simplify the equation:

\( y - 12=-x + 5 \)

\( y=-x + 17 \)

To find the x - intercept, set \( y = 0 \):

\( 0=-x + 17 \)

\( x = 17 \)

So the x - intercept of \( \overleftrightarrow{CD} \) is \( 17 \).

For the second part, we can check which point lies on the line \( y=-x + 17 \). Let's assume some common points (since the original problem's drop - down options are not shown, but following the calculation of the line equation \( y=-x + 17 \), we can test points. For example, if we test the point \( (17,0) \) (the x - intercept), when \( x = 17 \), \( y=-17 + 17=0 \), so \( (17,0) \) lies on \( \overleftrightarrow{CD} \). If we test another point, say \( (10,7) \), \( y=-10 + 17 = 7 \), so \( (10,7) \) also lies on the line. (The specific point would depend on the drop - down options, but the process is to substitute the x and y values of the point into the equation \( y=-x + 17 \) and check if the equation holds.)