Question

24
in the diagram, line ( overline{ab} ) and line ( overline{ef} ) are horizontal lines and ( overline{cb} ) is a vertical line segment. if ( fb:fc = 4:3 ), what are the coordinates of point ( d )?

a. ( (-4,-2) )
b. ( (-5,-3) )
c. ( (-6,-4) )
d. ( (-7,-5) )
e. ( (-8,6) )

Explanation:

Step1: Find the length of FB and FC

First, we find the vertical distance between \( C(-3, -1) \) and \( B \) (since \( \overline{CB} \) is vertical, the x - coordinate of \( B \) is the same as that of \( F \), and the y - coordinate of \( F \) is the same as that of \( D \) and \( E \)). The y - coordinate of \( C \) is \( - 1 \), and let's assume the y - coordinate of \( B \) is \( y_B \). The length of \( FC=\vert - 1-y_F\vert \), and \( FB = \vert y_B - y_F\vert \). But since \( AB \) and \( EF \) are horizontal, \( y \) - coordinate of \( A(-10,-8) \), \( B \), \( E \), \( D \), \( F \) are the same. Let's find the vertical distance between \( C(-3,-1) \) and the line \( AB \) (or \( EF \)). The vertical distance (change in y - coordinates) between \( C(-3,-1) \) and \( A(-10,-8) \) is \( \vert-1-(-8)\vert=7 \)? Wait, no. Wait, \( FB:FC = 4:3 \), and \( CB \) is vertical. Let's find the length of \( CB \). The y - coordinate of \( C \) is \( - 1 \), the y - coordinate of \( B \) is \( - 8 \) (since \( A \) is \( (-10,-8) \) and \( AB \) is horizontal, so \( B \) has the same y - coordinate as \( A \)). So the length of \( CB=\vert-1-(-8)\vert = 7 \). Let \( FB = 4x \) and \( FC = 3x \), then \( FB+FC=CB \), so \( 4x + 3x=7 \), \( 7x = 7 \), \( x = 1 \). So \( FB = 4 \) and \( FC = 3 \).

Step2: Find the x - coordinate of F

Since \( C(-3,-1) \) and \( \overline{CB} \) is vertical, the x - coordinate of \( F \) is the same as that of \( C \), so \( F=(-3,y_F) \). And since \( AB \) is horizontal with \( A(-10,-8) \), the y - coordinate of \( B \) is \( - 8 \), so the y - coordinate of \( F \) is also \( - 8+FB \)? Wait, no. Wait, \( FB \) is vertical? No, \( \overline{CB} \) is vertical, so \( FB \) is vertical? Wait, \( AB \) and \( EF \) are horizontal, so \( AB \) and \( EF \) are parallel, and \( CB \) is vertical, so \( FB \) is vertical (since \( B \) is on \( AB \) and \( F \) is on \( EF \), and \( AB\parallel EF \), \( CB\perp AB \), so \( CB\perp EF \), so \( FB \) is vertical). So the length of \( FB \) is the vertical distance between \( F \) and \( B \), and \( FC \) is the vertical distance between \( F \) and \( C \). Since \( C(-3,-1) \) and \( B \) has y - coordinate \( - 8 \) (because \( A(-10,-8) \) is on \( AB \), a horizontal line), then the vertical distance between \( C \) and \( B \) is \( \vert-1-(-8)\vert=7 \). Since \( FB:FC = 4:3 \), let \( FB = 4k \), \( FC = 3k \), then \( 4k + 3k=7 \), so \( k = 1 \), so \( FB = 4 \), \( FC = 3 \). So the y - coordinate of \( F \) is \( y_B+FB \)? Wait, \( B \) is at \( (x_B,-8) \), \( F \) is at \( (x_F,y_F) \), since \( CB \) is vertical, \( x_F=x_C=-3 \), and since \( FB \) is vertical, \( y_F=-8 + FB \)? Wait, no. If \( B \) is at \( (x_B,-8) \) and \( F \) is above \( B \) (since \( C \) is above \( F \)), then \( y_F=-8 + FB \). Since \( FC = 3 \), and \( C \) is at \( (-3,-1) \), then \( y_F=-1 - FC=-1 - 3=-4 \)? Wait, that's a better way. The y - coordinate of \( C \) is \( - 1 \), \( FC = 3 \), so moving down 3 units from \( C \) along the vertical line \( x=-3 \) gives \( F \)'s y - coordinate: \( - 1-3=-4 \). So \( F=(-3,-4) \).

Step3: Find the equation of line AC

We have points \( A(-10,-8) \) and \( C(-3,-1) \). The slope of line \( AC \) is \( m=\frac{-1-(-8)}{-3-(-10)}=\frac{7}{7} = 1 \). The equation of line \( AC \) using point - slope form \( y - y_1=m(x - x_1) \) with \( A(-10,-8) \) is \( y+8 = 1\times(x + 10) \), so \( y=x + 2 \).

Step4: Find the coordinates of D

Since \( D \) is on \( EF \), and \( EF \) is horizontal with \( y \) - coordinate equal to that of \( F \), which is \( - 4 \). So we substitute \( y=-4 \) into the equation of line \( AC \): \( - 4=x + 2 \), so \( x=-6 \). Wait, no, wait. Wait, earlier we found \( F \) has y - coordinate \( - 4 \)? Wait, let's re - calculate. The vertical distance between \( C(-3,-1) \) and \( B \) (since \( A \) is \( (-10,-8) \), \( B \) is on \( AB \) horizontal line, so \( B \) has \( y=-8 \)). The length of \( CB \) is \( \vert-1-(-8)\vert = 7 \). Since \( FB:FC = 4:3 \), then \( FC=\frac{3}{4 + 3}\times7 = 3 \), \( FB=\frac{4}{4 + 3}\times7 = 4 \). So the y - coordinate of \( F \) is \( y_C-FC=-1-3=-4 \) (since we move down from \( C \) to \( F \) by \( FC = 3 \) units). So \( F \) is at \( (-3,-4) \). Now, line \( AC \) goes from \( A(-10,-8) \) to \( C(-3,-1) \). The slope \( m=\frac{-1-(-8)}{-3-(-10)}=\frac{7}{7}=1 \). Equation of line \( AC \): \( y+8 = 1\times(x + 10)\Rightarrow y=x + 2 \). Now, point \( D \) is on \( EF \), which is a horizontal line with \( y=-4 \) (since \( F \) is on \( EF \) and \( F=(-3,-4) \)). So substitute \( y=-4 \) into \( y=x + 2 \): \( - 4=x + 2\Rightarrow x=-6 \). Wait, but the options are \( A. (-4,-2) \), \( B. (-5,-3) \), \( C. (-6,-4) \), \( D. (-7,-5) \), \( E. (-8,-6) \). Wait, maybe I made a mistake in the y - coordinate of \( F \). Wait, \( AB \) is horizontal with \( A(-10,-8) \), so \( B \) has \( y=-8 \). \( CB \) is vertical, so \( C(-3,-1) \) and \( B \) have the same x - coordinate? No, \( \overline{CB} \) is vertical, so \( C \) and \( B \) have the same x - coordinate? Wait, no, vertical lines have the same x - coordinate. So if \( \overline{CB} \) is vertical, then \( C \) and \( B \) have the same x - coordinate. So \( B=(-3,-8) \). Ah! That's my mistake. So \( C(-3,-1) \) and \( B(-3,-8) \), so the length of \( CB=\vert-1-(-8)\vert = 7 \). Then \( FB:FC = 4:3 \), and \( FB \) is the length from \( F \) to \( B \), \( FC \) is from \( F \) to \( C \). Since \( B(-3,-8) \) and \( C(-3,-1) \), \( F \) is on the vertical line \( x = - 3 \), between \( B \) and \( C \). So \( FB \) is the distance from \( F \) to \( B \), \( FC \) is from \( F \) to \( C \). Let \( F=(-3,y) \), then \( FB=\vert y-(-8)\vert=\vert y + 8\vert \), \( FC=\vert-1 - y\vert \). Since \( FB:FC = 4:3 \), and \( y \) is between \( - 8 \) and \( - 1 \) (because \( F \) is between \( B \) and \( C \)), so \( FB=y + 8 \), \( FC=-1 - y \). Then \( \frac{y + 8}{-1 - y}=\frac{4}{3} \), cross - multiply: \( 3(y + 8)=4(-1 - y)\Rightarrow3y+24=-4 - 4y\Rightarrow3y + 4y=-4 - 24\Rightarrow7y=-28\Rightarrow y=-4 \). So \( F=(-3,-4) \), which means \( EF \) is the horizontal line \( y=-4 \). Now, line \( AD \): since \( A(-10,-8) \) and \( D \) is on \( EF \) ( \( y=-4 \) ) and on line \( AC \). The slope of \( AC \) is \( \frac{-1-(-8)}{-3-(-10)}=\frac{7}{7}=1 \), so the equation is \( y+8 = 1\times(x + 10)\Rightarrow y=x + 2 \). Now, substitute \( y=-4 \) into \( y=x + 2 \): \( - 4=x + 2\Rightarrow x=-6 \). So \( D=(-6,-4) \).

Answer:

C. \((-6, -4)\)