Answer to Question #93665 in Analytic Geometry for belle

Question

P(4,9) ,Q(3,2),R(13,-3) and S(11,13) are four points in a plane and L is the mid point of RS . the line through Q perpendicular to RS meets PR at M find :
1 the equation of the lines QL and PR
2 the coordinates of the point of intersection N ,of QL and PR
3 the length of QM
4 the area of triangle PQN
'

Answer 1

1. Find the coordinates of L:

$$x=\frac{13+11}{2}=\frac{24}{2}=12$$ $$y=\frac{-3+13}{2}=\frac{10}{2}=5$$

Find the slope of the line QL:

$$m_{QL}=\frac{y_2-y_1}{x_2-x_1}=\frac{5-2}{12-3}=\frac{3}{9}=1/3$$

Substitute the slope and point Q in the slope intercept form of the equation ($y=mx+b$ ) and solve for b:

$$2=\frac{1}{3}\cdot3+b$$ $$2=1+b$$ $$b=1$$

The equation of the line QL is: $y=\frac{1}{3}x+1$.

Find the slope of the line PR:

$$m_{PR}=\frac{-3-9}{13-4}=\frac{-12}{9}=-\frac{4}{3}$$

Substitute the slope and point P in the slope intercept form of the equation and solve for b:

$$9=-\frac{4}{3}\cdot 4+b$$ $$9=-\frac{16}{3}+b$$ $$b=\frac{43}{3}$$

The equation of the line PR is: $y=-\frac{4}{3}x+\frac{43}{3}$.

2. If N is the point of intersection of QL and PR, then its coordinates satisfy the system of equations:

$$\begin{cases} y=\frac{1}{3}x+1 \\ y=-\frac{4}{3}x+\frac{43}{3} \end{cases}$$ $$\frac{1}{3}x+1=-\frac{4}{3}x+\frac{43}{3}$$ $$\frac{5}{3}x=\frac{40}{3}$$ $$x=\frac{40}{3}:\frac{5}{3}=8$$ $$y=\frac{8}{3}+1=\frac{11}{3}$$

Hence $N(8,\frac{11}{3})$ - the point of intersection of QL and PR.

3. Find the slope of the line RS:

$$m_{RS}=\frac{13-(-3)}{11-13}=\frac{16}{-2}=-8$$

Since the line MQ is perpendicular to the line RS,  their slopes have a particular relationship to each other:

$$m_{MQ}=-\frac{1}{m_{RS}}=\frac{1}{8}$$

Find the equation of the line MQ. Substitute the slope and point Q in the slope intercept form of the equation and solve for b:

$$2=\frac{1}{8}\cdot{3}+b$$ $$b=2-\frac{3}{8}$$ $$b=\frac{13}{8}$$

The equation of the line MQ is: $y=\frac{1}{8}x+\frac{13}{8}$

If N is the point of intersection of MQ and PR, then its coordinates satisfy the system of equations:

$$\begin{cases} y=\frac{1}{8}x+\frac{13}{8} \\ y=-\frac{4}{3}x+\frac{43}{3} \end{cases}$$

$$\begin{cases} 8y=x+13 \\ 3y=-4x+43 \end{cases}$$ $$\begin{cases}x=8y-13 \\ 3y=-4x+43 \end{cases}$$

$$3y=-4(8x-13)+43$$

$$3y=-32x+52+43$$

$$35y=95$$

$$y=\frac{19}{7}$$

$$x=8\cdot\frac{19}{7}-13=\frac{61}{7}$$

The coordinates of M are $(\frac{61}{7},\frac{19}{7})$

Then the length of QM:

$$QM=\sqrt{(\frac{61}{7}-3)^2+(\frac{19}{7}-2)^2}$$

$$QM=\sqrt{(\frac{40}{7})^2+(\frac{5}{7})^2}=\sqrt{\frac{1625}{49}}=\frac{5\sqrt{65}}{7}$$

4. Find the area of the triangle using the formula:

$$A=\lvert \frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\rvert$$

$$A=\lvert\frac{4(2-\frac{11}{3})+3(\frac{11}{3}-9)+8(9-2)}{2}\rvert$$

$$A=\lvert\frac{-\frac{20}{3}-16+56}{2}\rvert=\frac{50}{3}$$