Question

on a separate sheet of paper, draw and label a figure to represent the relationship.
points ( p(-2, 1) ), ( q(1, -3) ), ( r(1, 2) ) and ( s(-1, -3) ) are coplanar. point ( t ) is collinear with ( overleftrightarrow{pq} ) and ( overleftrightarrow{rs} ).
what are the approximate coordinates of point ( t )?

a) ( (0.2, 0) )

b) ( (0, -2) )

c) ( (-0.8, -0.5) )

d) ( (-0.3, -1.25) )

Explanation:

Step 1: Find the equation of line \( \overleftrightarrow{PQ} \)

First, find the slope of \( \overleftrightarrow{PQ} \) using points \( P(-2, 1) \) and \( Q(1, -3) \). The slope formula is \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
So, \( m_{PQ} = \frac{-3 - 1}{1 - (-2)} = \frac{-4}{3} \).
Using point - slope form \( y - y_1 = m(x - x_1) \) with point \( P(-2, 1) \):
\( y - 1=\frac{-4}{3}(x + 2) \)
\( y=\frac{-4}{3}x-\frac{8}{3}+1=\frac{-4}{3}x-\frac{5}{3} \)

Step 2: Find the equation of line \( \overleftrightarrow{RS} \)

Points \( R(1, 2) \) and \( S(-1, -3) \). The slope \( m_{RS}=\frac{-3 - 2}{-1 - 1}=\frac{-5}{-2}=\frac{5}{2} \)
Using point - slope form with point \( R(1, 2) \):
\( y - 2=\frac{5}{2}(x - 1) \)
\( y=\frac{5}{2}x-\frac{5}{2}+2=\frac{5}{2}x-\frac{1}{2} \)

Step 3: Find the intersection point \( T \) (solve the system of equations)

We have the system:
\(

$$\begin{cases}y = \frac{-4}{3}x-\frac{5}{3}\\y=\frac{5}{2}x-\frac{1}{2}\end{cases}$$

\)
Set \( \frac{-4}{3}x-\frac{5}{3}=\frac{5}{2}x-\frac{1}{2} \)
Multiply through by 6 to clear fractions:
\( 6\times(\frac{-4}{3}x-\frac{5}{3})=6\times(\frac{5}{2}x-\frac{1}{2}) \)
\( - 8x-10 = 15x - 3 \)
\( -8x-15x=-3 + 10 \)
\( -23x = 7 \)
\( x=-\frac{7}{23}\approx - 0.3 \)
Substitute \( x =-\frac{7}{23}\) into \( y=\frac{5}{2}x-\frac{1}{2} \)
\( y=\frac{5}{2}\times(-\frac{7}{23})-\frac{1}{2}=\frac{-35}{46}-\frac{23}{46}=\frac{-58}{46}\approx - 1.26\approx - 1.25 \)

Answer:

D) \((-0.3,-1.25)\)