Coordinate Geometry Part1

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Exercise 1

1. $P\left(3,2\right)$ , $Q\left(-1,1\right)$ , $R\left(2,-2\right)$ . Prove that $PQR\,$ is an isosceles triangle.

- $\begin{align} & PQ = \sqrt{\left(3+1\right)^{2}+\left(2-1\right)^{2}}=\sqrt{17}\\ & QR = \sqrt{\left(-1-2\right)^{2}+\left(1+2\right)^{2}}=\sqrt{18}\\ & PR = \sqrt{\left(3-2\right)^{2}+\left(2+2\right)^{2}}=\sqrt{17}\\ & \mbox{Since} PQ=PR, \therefore PQR \mbox{ is an isosceles triangle.} \end{align}$

2. $A\left(-2,3\right)$ , $B\left(2,q\right)$ . Find possible values of $q\,$ if $AB=5\,$ units.

$q=0 \mbox{ or } q=6\,$
- $\begin{align} & AB =5 \\ & \sqrt{\left(2+2\right)^{2}+\left(q-3\right)^{2}}=5\\ & 16 + q^{2}-6q + 9 = 25 \\ & q^{2}-6q = 0 \\ & q \left(q-6\right)=0\\ & q=0 \mbox{ or } q=6 \end{align}$

3. $A\left(2,-1\right)$ , $B\left(4,0\right)$ , $C\left(8,2\right)$ , $D\left(-2,-2\right)$ , $E\left(-4,-4\right)$ , $F\left(f,-3\right)$

a) Prove that $A\,$ , $B\,$ and $C\,$ are collinear

- $\begin{align} & m_{AB} =\frac{0-\left(-1\right)}{4-2}=\frac{1}{2}\\ & m_{BC} =\frac{2-0}{8-4}=\frac{1}{2}\\ & \mbox{Since } AB\parallel BC, \mbox{ and } B \mbox{ is common point, }\\ & \therefore A,B \mbox{ and } C \mbox{ are collinear.}\frac{\quad}{}\mbox{(Proved)} \end{align}$

b) Determine whether the following are collinear

i) $A\,$ , $B\,$ and $E\,$
$\mbox{Yes}\,$
- $\begin{align} & m_{AB} =\frac{0-\left(-1\right)}{4-2}=\frac{1}{2}\\ & m_{BE} =\frac{0-\left(-4\right)}{4-\left(-4\right)}=\frac{1}{2}\\ & \mbox{Since } AB\parallel BE, \mbox{ and } B \mbox{ is common point, }\\ & \therefore A,B \mbox{ and } E \mbox{ are collinear.} \end{align}$
ii) $B\,$ , $D\,$ and $E\,$
$\mbox{No}\,$
- $\begin{align} & m_{BD} =\frac{0-\left(-2\right)}{4-\left(-2\right)}=\frac{1}{3}\\ & m_{DE} =\frac{-2-\left(-4\right)}{-2-\left(-4\right)}=1\\ & \mbox{Since } BD \mbox{ is not parallel to } DE,\therefore B,D \mbox{ and } E \mbox{ are not collinear.} \end{align}$

c) Given that $A\,$ , $D\,$ and $F\,$ are collinear. Find value of $f\,$ .

$f=-6\,$
- $\begin{align} & \mbox{Since }A, D, F \mbox{ are collinear} \therefore AD\parallel DF \\ & \therefore m_{AD} = m_{DF}\\ & \frac{-1-\left(-2\right)}{2-\left(-2\right)} =\frac{-3-\left(-2\right)}{f-\left(-2\right)}\\ & \frac{1}{4} =\frac{-1}{f+2}\\ & f+2=-4 \\ & f=-6 \\ \end{align}$

4. $P\left(2,1\right)$ , $Q\left(7,0\right)$ , $R\left(-3,5\right)$ Find coordinates of $S\,$ if

a) $PQRS\,$ is a parallelogram

$S\left(-8,6\right)$
- Media:MT-CG-E1-4a.dia
- $\begin{align} & \mbox{Let } S\left(a,b\right) \\ & \mbox{midpoint of } PR=\mbox{midpoint of } QS \\ & \left(\frac{2-3}{2}, \frac{1+5}{2}\right) = \left(\frac{a+7}{2}, \frac{b+0}{2}\right)\\ & \therefore -1=a+7 \qquad 6=b \\ & a=-8, b=6 \\ &\therefore S\left(-8,6\right) \end{align}$

b) $PRQS\,$ is a parallelogram

$S\left(12,-4\right)$
- Media:MT-CG-E1-4b.dia
- $\begin{align} & \mbox{Let } S\left(a,b\right) \\ & \mbox{midpoint of } PQ=\mbox{midpoint of } RS \\ & \left(\frac{2+7}{2}, \frac{1+0}{2}\right) = \left(\frac{a-3}{2}, \frac{b+5}{2}\right)\\ & \therefore 9=a-3 \qquad 1=b+5 \\ & a=12, b=-4 \\ &\therefore S\left(12,-4\right) \end{align}$

c) $PQSR\,$ is a parallelogram

$S\left(2,4\right)$
- Media:MT-CG-E1-4c.dia
- $\begin{align} & \mbox{Let } S\left(a,b\right) \\ & \mbox{midpoint of } PS=\mbox{midpoint of } QR \\ & \left(\frac{a+2}{2}, \frac{b+1}{2}\right) = \left(\frac{7-3}{2}, \frac{0+5}{2}\right)\\ & \therefore a+2=4 \qquad b+1=5 \\ & a=2, b=4 \\ &\therefore S\left(2,4\right) \end{align}$

5. $A\left(-4,-1\right)$ , $B\left(5,5\right)$ . Find coordinates of $C\,$ if

a) $C\,$ is between $A\,$ and $B\,$ . $C\,$ divides $AB\,$ in the ratio $2:3\,$

$C\left(-\frac{2}{5},\frac{7}{5}\right)$
- $\begin{align} \mbox{Coordinates of } C &= \left(\frac{2\left(5\right)+3\left(-4\right)}{2+3}, \frac{2\left(5\right)+3\left(-1\right)}{2+3}\right)\\ & =\left(-\frac{2}{5},\frac{7}{5}\right) \end{align}$
- Check : File:MT-CG-E1-5a.dia
- $\triangle x= \frac{18}{5} :\frac{27}{5}=2:3 ; \triangle y= \frac{12}{5} :\frac{18}{5}=2:3$

b) $C\,$ divides $BA\,$ internally in the ratio $7:2\,$

$C\left(-2,\frac{1}{3}\right)$
- $\begin{align} \mbox{Coordinates of } C &= \left(\frac{7\left(-4\right)+2\left(5\right)}{7+2}, \frac{7\left(-1\right)+2\left(5\right)}{7+2}\right)\\ & =\left(-2,\frac{1}{3}\right) \end{align}$
- Check : File:MT-CG-E1-5b.dia
- $\triangle x= 7 :2 ; \triangle y= \frac{14}{3} :\frac{4}{3}$

c) $C\,$ divides $AB\,$ externally in the ratio $2:1\,$

$C\left(14,11\right)$
- $\begin{align} \mbox{Coordinates of } C &= \left(\frac{2\left(5\right)-1\left(-4\right)}{2-1}, \frac{2\left(5\right)-1\left(-1\right)}{2-1}\right)\\ & =\left(14,11\right) \end{align}$
- Check : File:MT-CG-E1-5c.dia
- $\triangle x= 18 :9 ; \triangle y= 12 :6$

d) $B\,$ divides $AC\,$ internally in the ratio $2:3\,$

$C\left(\frac{37}{2},11\right)$
- $\begin{align} & \mbox{Let } C\left(a,b\right) \\ & \left(5, 5\right) = \left(\frac{2\left(a\right)+3\left(-4\right)}{2+3}, \frac{2\left(b\right)+3\left(-1\right)}{2+3}\right)\\ & \therefore 5=\frac{2a-12}{5} \qquad 5=\frac{2b-3}{5} \\ & a=\frac{37}{2}, b=14 \\ &\therefore C\left(\frac{37}{2},14\right) \end{align}$
- Check : File:MT-CG-E1-5d.dia
- $\triangle x= 9 :\frac{27}{2} = 2 :3; \triangle y= 6 :9$

e) $A\,$ divides $BC\,$ in the ratio $-2:1\,$

$C\left(\frac{1}{2},2\right)$
- $\begin{align} & \mbox{Let } C\left(a,b\right) \\ & \left(-4, -1\right) = \left(\frac{-2\left(a\right)+1\left(5\right)}{-2+1}, \frac{-2\left(b\right)+1\left(5\right)}{-2+1}\right)\\ & \therefore -4=\frac{-2a+5}{-1} \qquad -1=\frac{-2b+5}{-1} \\ & a=\frac{1}{2}, b=2 \\ &\therefore C\left(\frac{1}{2},2\right) \end{align}$
- Check : File:MT-CG-E1-5e.dia
- $\triangle x= 9 : \frac{9}{2} ; \triangle y= 6 :3$

6) $A\left(0,4\right)$ , $B\left(3,0\right)$ . Find $C\,$ if

a) $AC=2CB\,$

b) $3BC=2CA\,$

c) $3AB=BC\,$

d) $2AC=5BC\,$

Exercise 2

1) Find equation of the line that

a) passes through $\left(-2,3\right)$ and has gradient $-3\,$

$y=-3x-3\,$
- $\begin{align} & l_{1} : m =-3, \left(-2,3\right):\\ & y-y_{1}=m\left(x-x_{1}\right)\\ & y-3= -3\left(x+2\right)\\ & y= -3x-6+3 \\ & y= -3x-3 \\ \end{align}$

b) passes through $\left(4,-2\right)$ and $\left(-5,7\right)$

$y=-x+2\,$
- $\begin{align} & l_{1} : \left(4,-2\right), \left(-5,7\right):\\ & \frac{y-\left(-2\right)}{x-4}=\frac{7-\left(-2\right)}{\left(-5\right)-4}\\ & \frac{y+2}{x-4}=-1\\ & y+2= -x+4 \\ & y= -x+2 \\ \end{align}$

c) parallel to $x\,$ -axis and passes through point $\left(-3,5\right)$

$y=5\,$
- Media:MT-CG-E2-1c.dia
- $\therefore y=5$

d) passes through $\left(0,2\right)$ and has gradient $\frac{2}{3}$

$y=\frac{2}{3}x+2\,$
- $\begin{align} & l_{1} : m =\frac{2}{3}, \left(0,2\right):\\ & y-y_{1}=m\left(x-x_{1}\right)\\ & y-2= \frac{2}{3}\left(x-0\right)\\ & y= \frac{2}{3}x+2 \\ \end{align}$

e) passes through $\left(-2,-7\right)$ and $\left(-1,8\right)$

$y=15x+23\,$
- $\begin{align} & l_{1} :\left(-2,-7\right), \left(-1,8\right):\\ & \frac{y-\left(-7\right)}{x-\left(-2\right)}=\frac{8-\left(-7\right)}{-1-\left(-2\right)}\\ & \frac{y+7}{x+2}=15\\ & y+7= 15x+30 \\ & y= 15x+23 \\ \end{align}$

f) passes through origin and $\left(4,-1\right)$

$y=-\frac{1}{4}x$
- $\begin{align} & l_{1} :\left(0,0\right), \left(4,-1\right):\\ & \frac{y-0}{x-0}=\frac{-1-0}{4-0}\\ & y=-\frac{1}{4}x\\ \end{align}$

g) passes through $\left(1,4\right)$ and $\left(-3,4\right)$

$y=4\,$
- Media:MT-CG-E2-1g.dia
- $\therefore y=4$
- OR
- $\begin{align} & l_{1} :\left(1,4\right), \left(-3,4\right):\\ & \frac{y-4}{x-1}=\frac{4-4}{-3-4}\\ & y-4=0\\ & y=4\\ \end{align}$

h) has $x\,$ -intercept $3\,$ and $y\,$ -intercept $5\,$

$5x+3y-15=0 \mbox { or } y=-\frac{5}{3}x+5$
- $\begin{align} & l_{1} :\left(3,0\right), \left(0,5\right):\\ & \frac{y-0}{x-3}=\frac{5-0}{0-3}\\ & -3y=5x-15\\ & 5x+3y-15=0 \end{align}$

2) Find equation of the line that

a) passes through $\left(-2,3\right)$ and is perpendicular with the line $3x+4y+1=0\,$

$4x-3y+17=0 \mbox{ or } y=\frac{4}{3}x+\frac{17}{3}$
- $\begin{align} & l_{1} : 3x+4y+1=0\\ & m_{1} = -\frac{3}{4} \\ & l_{2} \perp l_{1} \\ & \therefore m_{2} = \frac{1}{\left(-\frac{3}{4}\right)} = \frac{4}{3} \\ & l_{2} : m_{2}=\frac{4}{3}, \left(-2,3\right): \\ & y-3=\frac{4}{3}\left[x-\left(-2\right)\right]\\ & 3\left(y-3\right) = 4\left(x+2\right) \\ & 3y-9 =4x+8\\ & 4x-3y+17=0\\ \end{align}$

b) passes through $\left(6,5\right)$ and is parallel to the line joining $\left(1,3\right)$ & $\left(-1,6\right)$

$3x+2y-28=0 \mbox{ or } y=-\frac{3}{2}x+14$
- $\begin{align} & m_{1}=\frac{6-3}{-1-1}=-\frac{3}{2}\\ & l_{2} \parallel l_{1} \\ & \therefore m_{2} = m_{1}= -\frac{3}{2} \\ & l_{2} : m_{2}=-\frac{3}{2}, \left(6,5\right): \\ & y-5=-\frac{3}{2}\left(x-6\right)\\ & 2\left(y-5\right) = -3\left(x-6\right) \\ & 2y-10 =-3x+18\\ & 3x+2y-28=0\\ \end{align}$

3) a) Find the perpendicular bisector of the line that joins $A\left(-4,6\right)$ & $B\left(3,-5\right)$

$7x-11y+9=0 \mbox{ or } y=\frac{7}{11}x+\frac{9}{11}$
- Media:MT-CG-E2-3a.dia
- $\begin{align} & \mbox{Mid point of } AB =\left(\frac{-4+3}{2},\frac{6-5}{2}\right)\\ & \qquad \qquad \qquad \quad=\left(-\frac{1}{2},\frac{1}{2}\right)\\ & m_{AB} = \frac{6-\left(-5\right)}{-4-3}=-\frac{11}{7}\\ & l_{2} \perp AB, \therefore m_{2} =\frac{7}{11}\\ & l_{2} : m=\frac{7}{11}, \left(-\frac{1}{2},\frac{1}{2}\right): \\ & y-\frac{1}{2} = \frac{7}{11}\left(x+\frac{1}{2}\right)\\ & 11y-\frac{11}{2}=7x+\frac{7}{2}\\ & 7x-11y+9=0 \end{align}$

b) $P\left(2,5\right)$ & $Q\left(-1,4\right)$ . $D\,$ divides $PQ\,$ internally in the ratio $2:1\,$ . Find equation of the line that passes through $D\,$ and is perpendicular to $PQ\,$ .

$y=-3x+\frac{13}{3}\,$
- $\begin{align} & D =\left(\frac{2\left(-1\right)+2}{2+1},\frac{2\left(4\right)+5}{2+1}\right)\\ & \qquad \qquad=\left(0,\frac{13}{3}\right)\\ & m_{PQ} = \frac{5-4}{2-\left(-1\right)}=\frac{1}{3}\\ & l_{2} \perp PQ, \therefore m_{2} =-3\\ & l_{2} : m=-3, \left(0,\frac{13}{3}\right): \\ & y-\frac{13}{3} = -3\left(x-0\right)\\ & y=-3x+\frac{13}{3}\\ \end{align}$

4) $ABC\,$ is a triangle. $A\left(3,3\right)$ , $B\left(6,-3\right)$ , $C\left(-1,-2\right)$ . $M\,$ is the midpoint of $BC\,$ . Find the equation of

a) $AM\,$

$y=11x-30\,$
- $\begin{align} & M =\left(\frac{6-1}{2},\frac{-3-2}{2}\right)=\left(\frac{5}{2},-\frac{5}{2}\right)\\ & AM : A\left(3,3\right), M\left(\frac{5}{2},-\frac{5}{2}\right) :\\ & \frac{y-3}{x-3}=\frac{-\frac{5}{2}-3}{\frac{5}{2}-3}\\ & \frac{y-3}{x-3}=11\\ & y-3=11x-33 \\ & y=11x-30\\ \end{align}$

b) the line that passes through $A\,$ and is parallel with $BC\,$

$x+7y-24=0 \mbox{ or }y=-\frac{1}{7}x+\frac{17}{3}$
- $\begin{align} & m_{BC} = \frac{-3-\left(-2\right)}{6-\left(-1\right)}=-\frac{1}{7}\\ & l_{1} : m=-\frac{1}{7},A\left(3,3\right):\\ & y-3=-\frac{1}{7}\left(x-3\right)\\ & 7y-21=3-x \\ & x+7y-24=0 \end{align}$

c) the line that passes through $B\,$ and is perpendicular with $AC\,$

$4x+5y-9=0 \mbox{ or }y=-\frac{4}{5}x+\frac{9}{5}$
- $\begin{align} & m_{AC} = \frac{-2-3}{-1-3}=\frac{5}{4}\\ & \therefore m_{2} = -\frac{4}{5}\\ & l_{2} : m=-\frac{4}{5},B\left(6,-3\right):\\ & y+3=-\frac{4}{5}\left(x-6\right)\\ & 5y+15=-4x+24 \\ & 4x+5y-9=0 \end{align}$

5) The three sides of a triangle have the equations $3x+4y-17=0\,$ , $x-2y+1=0\,$ & $3x-y+8=0\,$ Find the coordinates of all its vertexes.

$\left(3,2\right),\left(-1,5\right),\left(-3,-1\right)$
- $\begin{align} & 3x+4y-17=0\frac{\qquad}{}(1)\\ & x-2y+1=0\frac{\qquad}{}(2)\\ & 3x-y+8=0\frac{\qquad}{}(3)\\ & (1) \mbox{ and } (2) : \\ & 3x+4y-17=0\frac{\qquad}{}(1)\\ & (2)\times2 : 2x-4y+2=0\frac{\qquad}{}(3)\\ & (1)+(2) : 5x-15=0 \\ & x=3 , \therefore y= 2 \\ & (1) \mbox{ and } (3) : \\ & 3x+4y-17=0\frac{\qquad}{}(1)\\ & 3x-y+8=0\frac{\qquad}{}(3)\\ & (1)-(3) : 5y -25=0 \\ & y=5, x=-1 \\ & (2) \mbox{ and } (3) : \\ & x-2y+1=0\frac{\qquad}{}(2)\\ & 3x-y+8=0\frac{\qquad}{}(3)\\ & (2)\times 3 : 3x-6y+3=0\frac{\qquad}{}(4)\\ & (3)-(4):5y+5=0 \\ & y=-1, x=-3 \\ & \therefore \mbox{ The coordinates of the vertexes are } \left(3,2\right),\left(-1,5\right),\left(-3,-1\right) \end{align}$

6) Point $A\,$ lies on the line $4-7x\,$ and distance $AB\,$ , where $B\,$ is $\left(-3,0\right)$ is $5\,$ . Find all possible coordinates of $A\,$ .

$\left(0,4\right), \left(1,-3\right)$
- $\begin{align} & \mbox{Let }A \left(a,4-7a\right) \\ & AB = 5 \\ & \sqrt{\left(a+3\right)^{2}+\left(4-7a\right)^{2}} =5 \\ & a^{2}+6a+9+16-56a+49a^{2}=25\\ & 50a^{2}-50a = 0\\ & 50a\left(a-1\right) = 0 \\ & a =0, a=1 \\ & \therefore A\left(0,4\right) \mbox{ or } \left(1,-3\right) \end{align}$

7) $ABC\,$ is a triangle. $A\left(-1,-3\right)$ , $B\left(-8,-4\right)$ , $C\left(-9,3\right)$ . The perpendicular bisector of $AB\,$ and the perpendicular bisector of $BC\,$ meet at $D\,$ . Find coordinates of $D\,$ . Show that $D\,$ also lies on the perpendicular bisector of $AC\,$ .

$D\left(-5,0\right)$
- $\begin{align} & \mbox{Midpoint }AB =\left(\frac{-1-8}{2}+\frac{-3-4}{2}\right)=\left(-\frac{9}{2},-\frac{7}{2}\right)\\ & m_{AB} =\frac{-4-\left(-3\right)}{-8-\left(-1\right)}=\frac{1}{7} \\ & l_{1} \perp m_{AB} \\ & \therefore m_{1} = -7 \\ & l_{1} : y+\frac{7}{2} =-7\left(x+\frac{9}{2}\right) \\ & y =-7x -35 \\ & \mbox{Midpoint }BC =\left(\frac{-8-9}{2}+\frac{-4+3}{2}\right)=\left(-\frac{17}{2},-\frac{1}{2}\right)\\ & m_{BC} =\frac{-4-3}{-8-\left(-9\right)}=-7 \\ & l_{2} \perp m_{BC} \\ & \therefore m_{2} = \frac{1}{7} \\ & l_{2} : y+\frac{1}{2} =\frac{1}{7}\left(x+\frac{17}{2}\right) \\ & 7y+\frac{7}{2}= x+\frac{17}{2}\\ & x-7y+5=0 \\ & \mbox{to find } D :\\ & y =-7x -35\frac{\qquad}{}(1) \\ & x-7y+5=0 \\ & x=7y-5=0 \frac{\qquad}{}(2) \\ & (2)\to(1) :\\ & y =-7\left(7y-5\right)-35 \\ & y=-49y+35-35 \\ & y =0, \therefore x=-5 \\ & \therefore D\left(-5,0\right)\\ \end{align}$
- $\begin{align} & \mbox{Midpoint }AC =\left(\frac{-1-9}{2}+\frac{-3+3}{2}\right)=\left(-5,0\right)\\ & m_{AC} =\frac{3-\left(-3\right)}{-9-\left(-1\right)}=-\frac{3}{4} \\ & l_{3} \perp m_{AC} \\ & \therefore m_{3} = \frac{4}{3} \\ & l_{3} : y =\frac{4}{3}\left(x+5\right) \\ & 3y= 4x+20\\ & 4x-3y+20=0 \\ & \mbox{Substituting } D\left(-5,0\right) \\ & \mbox{R.H.S. }= 4\left(-5\right)+20 = 0 = \mbox{L.H.S. }\\ & \therefore D \mbox{ is also lies on the perpendicular bisector of } AC \frac{\qquad}{}\mbox{(Proved)} \end{align}$

Exercise 3

1. Find the acute angle between the following lines

a) $y=x-1\,$ and $y=-2x+5\,$

$\theta =71.6^{\circ}\,$
- $\begin{align} m_{1} &=1,m_{2}=-2 \\ \tan \theta &= \left|\frac{1-\left(-2\right)}{1+\left(1\right)\left(-2\right)}\right|=3\\ \therefore \theta & = 71.6^{\circ} \\ \end{align}$

b) $x+4y-7=0\,$ and $2x+y-1=0\,$

$\theta =49.4^{\circ}\,$
- $\begin{align} m_{1} &=-\frac{1}{4},m_{2}=-2 \\ \tan \theta &= \left|\frac{-\frac{1}{4}-\left(-2\right)}{1+\left(-\frac{1}{4}\right)\left(-2\right)}\right|=\frac{\left(\frac{7}{4}\right)}{\left(\frac{3}{2}\right)}=\frac{7}{6}\\ \therefore \theta & = 49.4^{\circ} \\ \end{align}$

c) $3x+2y=6\,$ and $2x-3y=6\,$

$\theta =90.0^{\circ}\,$
- $\begin{align} m_{1} &=-\frac{3}{2},m_{2}=\frac{2}{3} \\ \tan \theta &= \left|\frac{-\frac{3}{2}-\frac{13}{6}}{1+\left(-\frac{3}{2}\right)\left(\frac{2}{3}\right)}\right|=\left|\frac{-\frac{13}{6}}{0}\right|=\infty\\ \therefore \theta & = 90.0^{\circ} \\ \end{align}$
- OR
- $\begin{align} & m_{1}m_{2} =\left(-\frac{3}{2}\right)\left(\frac{2}{3}\right)=-1 \\ & \therefore l_{1} \perp l_{2} \\ & \therefore \theta = 45.0^{\circ} \\ \end{align}$

d) $y=x+1\,$ and $y=4\,$

$\theta =45.0^{\circ}\,$
- $\begin{align} m_{1} &=1,m_{2}=0 \\ \tan \theta &= \left|\frac{1-0}{1+0}\right|=1\\ \therefore \theta & = 45.0^{\circ} \\ \end{align}$

2. Find the equation of the lines

a) which passes through $\left(-2,7\right)$ and makes an angle $30^{\circ}\,$ with $x\,$ -axis.
$y=\frac{\sqrt{3}}{3}x+\frac{2\sqrt{3}+21}{3} \mbox{ and } y=\frac{\sqrt{3}}{3}x+\frac{21-2\sqrt{3}}{3}$
- $\begin{align} & m = \pm \tan 30^{\circ}=\pm\frac{1}{\sqrt{3}}=\pm\frac{\sqrt{3}}{3}\\ \end{align}$
- $\begin{array}{ll} m=\frac{\sqrt{3}}{3}, \left(2,-7\right): \qquad\qquad& m=-\frac{\sqrt{3}}{3}, \left(2,-7\right):\\ y-7=\frac{\sqrt{3}}{3}\left(x+2\right) & y-7=-\frac{\sqrt{3}}{3}\left(x+2\right) \\ y=\frac{\sqrt{3}}{3}x+\frac{2\sqrt{3}+21}{3} & y=\frac{\sqrt{3}}{3}x+\frac{21-2\sqrt{3}}{3}\\ \end{array}$
- $\therefore y=\frac{\sqrt{3}}{3}x+\frac{2\sqrt{3}+21}{3} \mbox{ and } y=\frac{\sqrt{3}}{3}x+\frac{21-2\sqrt{3}}{3}$
b) which passes through $\left(2,-1\right)$ and makes an angle $45^{\circ}\,$ with the line $3x+2y-4=0\,$ .
$y=-\frac{1}{5}x-\frac{3}{5}\mbox{ and } y=5x-11$
- $\begin{align} & 3x+2y-4=0 \\ & m_{1} = -\frac{3}{2} \\ &\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}=\pm\tan 45^{\circ}\\ &\frac{m_{1}+\frac{3}{2}}{1-\frac{3}{2}m_{1}}=\pm 1\\ \end{align}$
- $\begin{array}{lll} \therefore \frac{m_{1}-\frac{3}{2}}{1+\frac{3}{2}m_{1}}=1 &\mbox{ or } &\frac{m_{1}-\frac{3}{2}}{1+\frac{3}{2}m_{1}}=\pm -1\\ m_{1}+\frac{3}{2} = 1-\frac{3}{2}m_{1} & & m_{1}+\frac{3}{2} = \frac{3}{2}m_{1}-1\\ \frac{5}{2}m_{1}=-\frac{1}{2} & & \frac{5}{2}=\frac{1}{2}m_{1} \\ m=-\frac{1}{5} & & m_{1}=5 \\ m=-\frac{1}{5},\left(2,-1\right) & & m=5,\left(2,-1\right)\\ y+1 =-\frac{1}{5}\left(x-2\right) & & y+1 =5\left(x-2\right)\\ \therefore y =-5x-11 & & y=-\frac{1}{5}x-\frac{3}{5} \end{array}$
- $y=-\frac{1}{5}x-\frac{3}{5}\mbox{ and } y=5x-11$

3. Find the perpendicular distance of the following points with the given line

a) $\left(3,-2\right);4x-3y-11=0$

$\frac{7}{5}\mbox{ units}$
- $\begin{align} & \left(3,-2\right),4x-3y-11=0: \\ & d=\frac{\left|4\left(3\right)-3\left(-2\right)-11\right|}{\sqrt{4^{2}+\left(-3\right)^{2}}}\\ & =\frac{7}{5}\mbox{ units}\\ \end{align}$

b) $\left(0,0\right);y=2x+3$

$\frac{3\sqrt{5}}{5}\mbox{ units}$
- $\begin{align} & \left(0,0\right),2x-y+3=0: \\ & d=\frac{\left|2\left(0\right)-1\left(0\right)+3\right|}{\sqrt{2^{2}+\left(-1\right)^{2}}}\\ & =\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{5}\mbox{ units}\\ \end{align}$

c) $\left(2,1\right);y=\frac{2}{7}x-1$

$\frac{10\sqrt{53}}{53}\mbox{ units}$
- $\begin{align} & y=\frac{2}{7}x-1 \\ & 7y=2x-7 \\ & \left(2,1\right),2x-7y-7=0: \\ & d=\frac{\left|2\left(2\right)-7\left(1\right)-7\right|}{\sqrt{2^{2}+\left(-7\right)^{2}}}\\ & =\frac{10}{\sqrt{53}}=\frac{10\sqrt{53}}{53}\mbox{ units}\\ \end{align}$

d) $\left(-11,5\right);x+2y=7$

$\frac{10\sqrt{53}}{53}\mbox{ units}$
- $\begin{align} & \left(-11,5\right),x+2y-7=0: \\ & d=\frac{\left|1\left(-11\right)+2\left(5\right)-7\right|}{\sqrt{1^{2}+2^{2}}}\\ & =\frac{8}{\sqrt{5}}=\frac{8\sqrt{5}}{5}\mbox{ units}\\ \end{align}$

e) $\left(-5,4\right);x=-2$

$3\mbox{ units}\,$
- $\begin{align} & \left(-5,3\right),x+2=0: \\ & d=\frac{\left|1\left(-5\right)+0\left(3\right)+2\right|}{\sqrt{1^{2}+0^{2}}}\\ & =3\mbox{ units}\\ \end{align}$
- OR
- Media:MT-CG-E3-3e.dia
- $\therefore d=3\mbox{ units}$

4. Show the points $A\left(-5,2\right)$ and $B\left(1,0\right)$ lie on the same side of the line $y=-\frac{2}{3}x+\frac{5}{3}$ .

- $\begin{align} & y=-\frac{2}{3}x+\frac{5}{3} \\ & 3y = -2x +5 \\ & 2x+3y-5=0 \\ & A\left(-5,2\right) : ah+bk+c =2\left(-5\right)+3\left(2\right)-5 =-9 \\ & B\left(1,0\right) : ah+bk+c =2\left(1\right)+3\left(0\right)-5 =-3 \\ & A \mbox{ and } B \mbox{ has the same sign for } ah+bk+c\\ & \therefore \mbox{ they lie on the same side of the line }\frac{\qquad}{}\mbox{(shown)} \end{align}$

5. a) $A\left(2,2\right)$ , $B\left(-4,4\right)$ , $C\left(-3,-2\right)$ . Find shortest distance of $A\,$ to $BC\,$ . Hence, find the area of triangle $ABC\,$ .

$\frac{34\sqrt{37}}{37}\mbox{ units}; 17\mbox{ units}^{2}$
- Media:MT-CG-E3-5a.dia
- $\begin{align} & \mbox{equation }BC : \\ & \frac{y-4}{x+4}=\frac{-2-4}{-3+4}\\ & y-4=-6\left(x+4\right) \\ & y-4=-6x-24\\ & 6x+y+20=0 \\ & A\left(2,2\right), 6x+y+20=0:\\ & d= \frac{\left|6\left(2\right)+1\left(2\right)+20\right|}{\sqrt{6^{2}+1^{2}}}\\ & =\frac{34}{\sqrt{37}} =\frac{34\sqrt{37}}{37}\mbox{ units}\\ & BC = \sqrt{\left(-4+3\right)^{2}+\left(4+2\right)^{2}}=\sqrt{37}\\ & \mbox{Area of triangle } ABC \\ & = \frac{1}{2}d\left(BC\right)=\frac{1}{2}\left(\frac{34}{\sqrt{37}}\right)\sqrt{37}=17\mbox{ units}^{2} \end{align}$

b) $PQRS\,$ is a parallelogram. $P\left(-2,2\right)$ , $Q\left(-3,-2\right)$ , $R\left(3,0\right)$ , $S\,$ . Find coordinates of $S\,$ and shortest distance of $P\,$ to $QR\,$ . Hence, find the area of parallelogram $PQRS\,$ .

$S\left(4,4\right), d=\frac{11\sqrt{10}}{10}\mbox{ units}; 22\mbox{ units}^{2}$
- Media:MT-CG-E3-5b.dia
- $\begin{align} & \mbox{Let } S\left(a,b\right) \\ & \mbox{Mid point }PR = \mbox{Mid point }QS \\ & \left(\frac{-2+3}{2},\frac{2+0}{2}\right)=\left(\frac{-3+a}{2},\frac{-2+b}{2}\right) \\ & 1=-3+a \qquad 2=-2+b \\ & a=4 \qquad\qquad b=4 \\ & \therefore S\left(4,4\right) \\ & \mbox{equation }QR : \\ & \frac{y+2}{x+3}=\frac{0+2}{3+3}\\ & 3\left(y+2\right)=x+3 \\ & 3y+6=x+3\\ & x-3y-3=0 \\ & P\left(-2,2\right), x-3y-3=0:\\ & d= \frac{\left|1\left(-2\right)-3\left(2\right)-3\right|}{\sqrt{1^{2}+\left(-3\right)^{2}}}\\ & =\frac{\left|-11\right|}{\sqrt{10}} =\frac{11\sqrt{10}}{10}\mbox{ units}\\ & QR = \sqrt{\left(3+3\right)^{2}+\left(0+2\right)^{2}}=\sqrt{40}=2\sqrt{10}\\ & \mbox{Area of parallelogram } PQRS \\ & = d\left(QR\right)=\frac{11}{\sqrt{10}}\left(2\sqrt{10}\right)=22\mbox{ units}^{2} \end{align}$

6. Find the perpendicular distance between the following parallel lines

a) $y=2x-4\,$ and $y=2x-1\,$

$\frac{3}{5}\sqrt{2}$
- Media:MT-CG-E3-6a.dia
- $\begin{align} & l_{1} : y=2x-4 \\ & \mbox{When } x=0, y=-4\\ & \therefore \left(0,-4\right) \mbox{ lies on }l_{1} \\ & l_{2} : y=2x-1, \therefore 2x-y-1=0 \\ & \left(0,-4\right),2x-y-1=0 :\\ & d = \frac{\left|2\left(0\right)-1\left(-4\right)-1\right|}{\sqrt{2^{2}+\left(-1\right)^{2}}} \\ & \quad = \frac{3}{\sqrt{5}}=\frac{3}{5}\sqrt{2} \end{align}$

b) $3x-4y-7=0\,$ and $3x-4y+11=0\,$

$\frac{18}{5}$
- Media:MT-CG-E3-6a.dia
- $\begin{align} & l_{1} : 3x-4y-7=0 \\ & \mbox{When } x=0, y=-\frac{7}{4}\\ & \therefore \left(0,-\frac{7}{4}\right) \mbox{ lies on }l_{1} \\ & \left(0,-\frac{7}{4}\right),3x-4y+11=0 :\\ & d = \frac{\left|3\left(0\right)-4\left(-\frac{7}{4}\right)+11\right|}{\sqrt{3^{2}+\left(-4\right)^{2}}} \\ & \quad = \frac{18}{5} \end{align}$

Coordinate Geometry Part1

From StpmWiki

Contents

Exercise 1

Exercise 2

Exercise 3

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