Question #93665
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
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1
Expert's answer
2019-09-03T11:02:30-0400

1. Find the coordinates of L:


x=13+112=242=12x=\frac{13+11}{2}=\frac{24}{2}=12y=3+132=102=5y=\frac{-3+13}{2}=\frac{10}{2}=5


Find the slope of the line QL:


mQL=y2y1x2x1=52123=39=1/3m_{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+by=mx+b ) and solve for b:


2=133+b2=\frac{1}{3}\cdot3+b2=1+b2=1+bb=1b=1

The equation of the line QL is: y=13x+1y=\frac{1}{3}x+1.


Find the slope of the line PR:


mPR=39134=129=43m_{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=434+b9=-\frac{4}{3}\cdot 4+b9=163+b9=-\frac{16}{3}+bb=433b=\frac{43}{3}

The equation of the line PR is: y=43x+433y=-\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:


{y=13x+1y=43x+433\begin{cases} y=\frac{1}{3}x+1 \\ y=-\frac{4}{3}x+\frac{43}{3} \end{cases}13x+1=43x+433\frac{1}{3}x+1=-\frac{4}{3}x+\frac{43}{3}53x=403\frac{5}{3}x=\frac{40}{3}x=403:53=8x=\frac{40}{3}:\frac{5}{3}=8y=83+1=113y=\frac{8}{3}+1=\frac{11}{3}

Hence N(8,113)N(8,\frac{11}{3}) - the point of intersection of QL and PR.


3. Find the slope of the line RS:


mRS=13(3)1113=162=8m_{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:


mMQ=1mRS=18m_{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=183+b2=\frac{1}{8}\cdot{3}+bb=238b=2-\frac{3}{8}b=138b=\frac{13}{8}

The equation of the line MQ is: y=18x+138y=\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:


{y=18x+138y=43x+433\begin{cases} y=\frac{1}{8}x+\frac{13}{8} \\ y=-\frac{4}{3}x+\frac{43}{3} \end{cases}

{8y=x+133y=4x+43\begin{cases} 8y=x+13 \\ 3y=-4x+43 \end{cases}{x=8y133y=4x+43\begin{cases}x=8y-13 \\ 3y=-4x+43 \end{cases}

3y=4(8x13)+433y=-4(8x-13)+43

3y=32x+52+433y=-32x+52+43

35y=9535y=95

y=197y=\frac{19}{7}

x=819713=617x=8\cdot\frac{19}{7}-13=\frac{61}{7}

The coordinates of M are (617,197)(\frac{61}{7},\frac{19}{7})

Then the length of QM:


QM=(6173)2+(1972)2QM=\sqrt{(\frac{61}{7}-3)^2+(\frac{19}{7}-2)^2}


QM=(407)2+(57)2=162549=5657QM=\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=x1(y2y3)+x2(y3y1)+x3(y1y2)2A=\lvert \frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\rvert

A=4(2113)+3(1139)+8(92)2A=\lvert\frac{4(2-\frac{11}{3})+3(\frac{11}{3}-9)+8(9-2)}{2}\rvert

A=20316+562=503A=\lvert\frac{-\frac{20}{3}-16+56}{2}\rvert=\frac{50}{3}




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