Coordinate Geometry Past Year

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A) Straight lines

1. The point R\, divides the line joining the points P\left(3,2\right) and Q\left(5,8\right) in the ratio 3:4\,. Find the equation of the line passing through R\, and perpendicular to PQ\,.

  • y=-\frac{1}{3}x+\frac{41}{7} \mbox{ or } 7x+21y-123=0
    • \begin{align} & \mbox{Coordinates of } R = \left(\frac{3\left(5\right)+4\left(3\right)}{3+4}, \frac{3\left(8\right)+4\left(2\right)}{3+4}\right)\\ & =\left(\frac{27}{7},\frac{32}{7}\right)\\ & m_{PQ} = \frac{8-2}{5-3}=3 \\ & l_{1} \perp PQ \\ & \therefore : m_{1}m_{PQ} =-1 \\ & m_{1}=-\frac{1}{3}\\ & l_{1}: m_{1}=-\frac{1}{3},\left(\frac{27}{7},\frac{32}{7}\right):\\ & y-\frac{32}{7} =-\frac{1}{3}\left(x-\frac{27}{7}\right) \\ & y-\frac{32}{7} =-\frac{1}{3}+\frac{27}{21}\\ & y=-\frac{1}{3}x+\frac{41}{7} \\ \end{align}


2. The straight line l_{1}\, which passes through the points A\left(4,0\right) and B\left(2,4\right) intersect the y\,-axis at the point P\,. The straight line l_{2}\, is perpendicular to l_{1}\, and passes through B\,. If l_{2}\, intersects the x\,-axis and y\,-axis at the point Q\, and R\, respectively, show that PR:QR=\sqrt{5}:3.

    • \begin{align} & l_{1} : \frac{y-0}{x-4}=\frac{4-0}{2-4}\\ & \qquad \frac{y}{x-4}=-2\\ & \qquad y =-2x+8 \\ & \mbox{at }P,x=0\\ & \therefore y =8, \therefore P\left(0,8\right)\\ & l_{2}\perp l_{1}, \therefore m_{2}m_{1}=-1\\ & \therefore m_{2} =\frac{1}{2} \\ & l_{2}:m=\frac{1}{2}, B\left(2,4\right) \\ & y-4=\frac{1}{2}\left(x-2\right)\\ & y=\frac{1}{2}x+3 \\ & \mbox{at }Q,y=0\\ & \therefore y =-6, \therefore Q\left(-6,0\right)\\ & \mbox{at }R,x=0\\ & \therefore y =3, \therefore R\left(0,3\right)\\ & \frac{PR}{QR}=\frac{\sqrt{\left(0-0\right)^{2}+\left(8-3\right)^{2}}}{\sqrt{\left(0+6\right)^{2}+\left(3+0\right)^{2}}} \\ & \qquad =\frac{5}{\sqrt{45}}=\frac{5}{3\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{5\sqrt{5}}{3\left(5\right)}=\frac{\sqrt{5}}{3}\\ & \therefore PR:QR =\sqrt{5}:3 \\ \end{align}


3. Given two parallel line l_{1}\, and l_{2}\,, passing through \left(5,0\right) and \left(-5,0\right) intersect 4x+3y=25\, respectively at P\, and Q\,. If PQ\, is equal to 5\, units, find the possible slopes of l_{1}\, and l_{2}\,.

  • m=\frac{4}{7} \mbox { or }m=-\frac{4}{13}
    • \begin{align} & \mbox{Let } \mbox{gradient/slope } = m \\ & l_{1} : y-0=m\left(x-5\right) \\ & \qquad y=mx-5m \\ & l_{2} : y-0=m\left(x+5\right) \\ & \qquad y=mx+5m \\ \end{align}
    • \begin{align} & \mbox{To find } P : \\ & y=mx-5m \frac{\qquad}{}(1)\\ & 4x+3y=25 \frac{\qquad}{}(2)\\ & (1)\to(2) : 4x+3\left(mx-5m\right)=25\\ & 4x+3mx-15m=25 \\ & x\left(4+3m\right)=25+15m \\ & x=\frac{5\left(5+3m\right)}{4+3m}\\ & \therefore y=m\left[\frac{5\left(5+3m\right)}{4+3m}\right]-5m \\ & \qquad =\frac{5m\left(5+3m\right)-5m\left(4+3m\right)}{4+3m} \\ & \qquad =\frac{5m\left(5+3m-4-3m\right)}{4+3m} \\ & \qquad =\frac{5m}{4+3m} \\ & \therefore P\left(\frac{5\left(5+3m\right)}{4+3m},\frac{5m}{4+3m}\right)\\ \end{align}
    • \begin{align} & \mbox{To find } Q : \\ & y=mx+5m \frac{\qquad}{}(3)\\ & 4x+3y=25 \frac{\qquad}{}(2)\\ & (1)\to(2) : 4x+3\left(mx+5m\right)=25\\ & 4x+3mx+15m=25 \\ & x\left(4+3m\right)=25-15m \\ & x=\frac{5\left(5-3m\right)}{4+3m}\\ & \therefore y=m\left[\frac{5\left(5-3m\right)}{4+3m}\right]+5m \\ & \qquad =\frac{5m\left(5-3m\right)+5m\left(4+3m\right)}{4+3m} \\ & \qquad =\frac{5m\left(5-3m+4+3m\right)}{4+3m} \\ & \qquad =\frac{45m}{4+3m} \\ & \therefore Q\left(\frac{5\left(5-3m\right)}{4+3m},\frac{45m}{4+3m}\right)\\ \end{align}
    • \begin{align} & PQ =5 \\ & \sqrt{\left[\frac{5\left(5+3m\right)}{4+3m}-\frac{5\left(5-3m\right)}{4+3m}\right]^{2}+\left[\frac{5m}{4+3m}-\frac{45m}{4+3m}\right]^{2}}=5 \\ & \left(\frac{30m}{4+3m}\right)^{2}+\left(\frac{-40m}{4+3m}\right)^{2}=25\\ & 900m^{2}+1600m^{2}=25\left(4m+3\right)^{2} \\ & 2500m^{2} = 25\left(4m+3\right)^{2} \\ & 50m = \pm 5\left(4m+3\right) \\ & 10m = \pm \left(4m+3\right) \\ & 10m=4+3m \mbox { or } 10m=-4-3m \\ & 7m=4 \qquad \qquad \quad 13m=-4 \\ & m=\frac{4}{7} \mbox { or }m=-\frac{4}{13} \\ \end{align}


4. The vertices of a quadrilateral are A\left(1,4\right),B\left(9,5\right),C\left(5,-2\right) and D\left(-3,-3\right) . Show that this quadrilateral is a rhombus.

    • \begin{align} & AB = \sqrt{\left(9-1\right)^{2}+\left(5-4\right)^{2}}=\sqrt{65}\mbox{ unit} \\ & BC = \sqrt{\left(9-5\right)^{2}+\left(5+2\right)^{2}}=\sqrt{65}\mbox{ unit} \\ & CD = \sqrt{\left(5+3\right)^{2}+\left(-2+3\right)^{2}}=\sqrt{65}\mbox{ unit} \\ & DA = \sqrt{\left(1+3\right)^{2}+\left(4+3\right)^{2}}=\sqrt{65}\mbox{ unit} \\ & AB=BC=CD=DA, \therefore ABCD \mbox{ is a rhombus}\\ \end{align}

The point P\, divides AB\, internally in the ratio 1:4\,, and the point Q\, divides AD\, internally in the ratio 3:2\,. Determine the coordinates of point P\, and point Q\,, and find the equation of PQ\,.

  • P\left(\frac{13}{5},\frac{21}{5} \right), Q\left(-\frac{7}{5},-\frac{1}{5} \right), 55x-50y+67=0
    • \begin{align} & P=\left(\frac{1\left(9\right)+4\left(1\right)}{1+4},\frac{1\left(5\right)+4\left(4\right)}{1+4}\right)\\ & \quad =\left(\frac{13}{5},\frac{21}{5} \right)\\ & Q=\left(\frac{3\left(-3\right)+2\left(1\right)}{3+2},\frac{3\left(-3\right)+2\left(4\right)}{3+2}\right)\\ & \quad =\left(-\frac{7}{5},-\frac{1}{5} \right)\\ & \mbox{Equation of } PQ :\\ & \frac{y-\left(-\frac{1}{5}\right)}{x-\left(-\frac{7}{5}\right)}=\frac{\frac{21}{5}-\left(-\frac{1}{5}\right)}{\frac{13}{5}-\left(-\frac{7}{5}\right)}\\ & \frac{y+\frac{1}{5}}{x+\frac{7}{5}}=\frac{11}{10}\\ & \frac{5y+1}{5x+7}=\frac{11}{10}\\ & 50y+10=55x+77 \\ & 55x-50y+67=0 \\ \end{align}

Find the distance from C\, to PQ\, and the area of triangle CPQ\,.

  • d=\frac{2}{5}\sqrt{221}\mbox{ unit},17.68 \mbox{ unit}^{2}
    • \begin{align} & C\left(5,-2\right), 55x-50y+67=0 : \\ & d=\frac{\left|55\left(5\right)-50\left(-2\right)+67\right|}{\sqrt{55^{2}+\left(-50\right)^{2}}}\\ & \quad =\frac{442}{\sqrt{5525}}=\frac{442}{5\sqrt{221}}=\frac{442\sqrt{221}}{5\left(221\right)}=\frac{2}{5}\sqrt{221}\\ & PQ =\sqrt{\left[\frac{13}{5}-\left(-\frac{7}{5}\right)\right]^{2} + \left[\frac{21}{5}-\left(-\frac{1}{5}\right)\right]^{2}} \\ & \quad =\sqrt{\frac{884}{25}}=\frac{2}{5}\sqrt{221}\mbox{ unit} \\ & \mbox{Area of triangle } CPQ =\frac{1}{2}d\left(PQ\right) \\ & =\frac{1}{2}\left(\frac{2}{5}\sqrt{221}\right)\left(\frac{2}{5}\sqrt{221}\right)\\ & = 17.68 \mbox{ unit}^{2}\\ \end{align}


5. Given that PQRS\, is a parallelogram where P\left(0,9\right),Q\left(2,-5\right),R\left(7,0\right) and S\left(a,b\right) are points on the plane. Find the value of a\, and b\, . Find the shortest distance from P\, to QR\, and the area of the parallelogram PQRS\,.

  • a=5, b=14, d=8\sqrt{2}\mbox{ unit}, 80\mbox{ unit}^{2}
    • \begin{align} & \mbox{Midpoint of } PR = \mbox{Midpoint of } QS \\ & \left(\frac{7}{2},\frac{9}{2}\right)= \left(\frac{a+2}{2},\frac{b-5}{2}\right)\\ & \therefore a+2=7 \qquad b-5=9 \\ & a=5, b=14\\ & \mbox{Equation of QR }: \\ & \frac{y-0}{x-7}=\frac{-5-0}{2-7} \\ & \frac{y}{x-7}=1 \\ & y=x-7 \\ & x-y-7=0 \\ & P\left(0,9\right), x-y-7=0 : \\ & d=\frac{\left|0-9-7\right|}{\sqrt{1^{2}+\left(-1\right)^{2}}}\\ & \quad = \frac{16}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\ & \quad =8\sqrt{2}\mbox{ unit} \\ & QR =\sqrt{\left(7-2\right)^{2}+\left(0+5\right)^{2}}\\ & \qquad =\sqrt{50}=5\sqrt{2}\mbox{ unit}\\ & \mbox{Area of }PQRS = \left(QR\right)d\\ & \qquad\qquad \qquad \quad =5\sqrt{2}\left(8\sqrt{2}\right)\\ & \qquad\qquad \qquad \quad =80\mbox{ unit}^{2}\\ \end{align}


6. Find the equation of both straight lines that are inclined at an angle of 45^{\circ} with straight line 2x+y-3=0\, and passing through a point \left(1,-4\right).

  • y=-\frac{1}{3}x-\frac{11}{3}, \mbox{ and } y=3x-7
    • \begin{align} & 2x+y-3 =0 \\ & \therefore m_{1} =-2 \\ & \mbox{Let gradient of line} =m_{2}\\ & \therefore \left|\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right|=\tan 45^{\circ} \\ & \left|\frac{m_{2}-\left(-2\right)}{1+\left(-2\right)m_{2}}\right|=1 \\ & \frac{m_{2}+2}{1-2m_{2}} =\pm 1 \\ & \therefore m_{2}+2 = 1-2m_{2} \mbox{ or } m_{2}+2 = 2m_{2}-1 \\ & 3m_{2}=-1 \qquad \qquad \qquad 3 = m_{2} \\ & \therefore m_{2}=-\frac{1}{3}, m_{2} =3\\ & \mbox{When }m_{2}=-\frac{1}{3},\\ & l_{2} : y+4=-\frac{1}{3}\left(x-1\right)\\ & y=-\frac{1}{3}x-\frac{11}{3} \\ & \mbox{When }m_{2}=3,\\ & l_{2} : y+4=3\left(x-1\right)\\ & y=3x-7 \\ & \therefore y=-\frac{1}{3}x-\frac{11}{3}, \mbox{ and } y=3x-7 \\ \end{align}


7. The equation of two parallel straight lines l_{1}\, and l_{2}\, are 3x+4y+12=0\, and 3x+4y-3=0\,. Find the distance between l_{1}\, and l_{2}\,. The straight line l_{3}\, is parallel to l_{1}\, and l_{2}\, and the distance between l_{1}\, and l_{3}\, is the same as the distance between l_{2}\, and l_{3}\,. Find the equation of l_{3}\,.

  • 3 \mbox{ unit};6x+8y+9 =0\,
    • \begin{align} & l_{1} : \mbox{When } x=0, y=-3 \\ & \therefore \left(0,-3\right) \mbox{ is on } l_{1} \\ & \left(0,-3\right), 3x+4y-3=0 :\\ & d=\frac{\left|3\left(0\right)+4\left(-3\right)-3\right|}{\sqrt{3^{2}+4^{2}}}\\ & \quad = \frac{15}{5}=3\\ & \therefore \mbox{Distance between } l_{1} \mbox{ and } l_{2} =3 \mbox{ unit}\\ & \mbox{Let point on }l_{3}: P\left(x,y\right)\\ & d_{1}=d_{2} \\ & \frac{\left|3x+4y+12\right|}{\sqrt{3^{2}+4^{2}}}= \frac{\left|3x+4y-3\right|}{\sqrt{3^{2}+4^{2}}}\\ & \therefore 3x+4y+12=3x+4y-3 \mbox{ or } 3x+4y+12=-3x-4y+3 \\ & 12=-3 \mbox{(rejected)} \qquad \qquad \qquad \quad 6x+8y+9 =0 \\ & \therefore l_{3} : 6x+8y+9 =0 \\ \end{align}


8. The points A\, and B\, are located on the line 5x-12y=6\,. The perpendicular distances of A\, and B\, respectively from the line 3x-4y=-2\, are both 4\, units. A\, lies in the same side of 3x-4y=-2\, as the origin, and B\, lies on the other side. Find the coordinates of A\, and B\,.

  • A\left(12,\frac{9}{2}\right), B\left(-18, -8\right)
    • \begin{align} & \mbox{Let point on }5x-12y=6 \mbox{ to be } \left(a,b\right) \\ & \therefore 5a-12b=6 \\ & \left(a,b\right), 3x-4y+2=0:\\ & d=4 \\ & \frac{\left|3a-4b+2\right|}{\sqrt{3^{2}+\left(-4\right)^{2}}} =4 \\ & \left|3a-4b+2\right| = 20 \\ & 3a-4b+2 = \pm 20 \\ & \therefore 3a-4b+2 = 20 \mbox{ or } 3a-4b+2 = - 20 \\ & 3a-4b = 18\frac{\qquad}{}(1) \mbox{ or } 3a-4b = - 22\frac{\qquad}{}(2) \\ & \\ & 3a-4b = 18\frac{\qquad}{}(1) \\ & 5a-12b=6 \frac{\qquad}{}(3) \\ & (1)\times (3): 9a-12b=54\frac{\qquad}{}(5) \\ & (5)-(3) : 4a =48, a=12, \therefore b= \frac{9}{2} \\ & \\ & 3a-2b = -22\frac{\qquad}{}(2) \\ & 5a-12b=6 \frac{\qquad}{}(3) \\ & (2)\times (3): 9a-12b=-66\frac{\qquad}{}(6) \\ & (6)-(3) : 4a =-72, a=-18, \therefore b=-8 \\ & \\ & \therefore \mbox{ the points are } \left(12, \frac{9}{2}\right), \left(-18, -8\right)\\ \end{align}
    • \begin{align} & \mbox{For } \left(0,0\right), ah+bk+c = 3\left(0\right)-4\left(0\right)+2 =2 \\ & \mbox{For } \left(12,\frac{9}{2}\right), ah+bk+c = 3\left(12\right)-4\left(\frac{9}{2}\right)+2 =20 \\ & \mbox{For } \left(-18,-8\right), ah+bk+c = 3\left(-18\right)-4\left(-8\right)+2 =-20 \\ & \left(12,\frac{9}{2}\right) \mbox{ have same sign of } ah+bk+c \mbox{ with origin} \\ & \left(-18,-8\right) \mbox{ have opposite sign of } ah+bk+c \mbox{ with origin} \\ & \therefore A\left(12,\frac{9}{2}\right), B\left(-18, -8\right)\\ \end{align}

B) Curves

9. The coordinates of point A\, and B\, are \left(3,0\right) and \left(0,4\right) respectively. The point P\, moves such that AP=3PB\,. Show that locus of P\, is a circle and find the centre and radius of this circle.

  • \mbox{Center } \left(-\frac{3}{8},-\frac{9}{2}\right), \mbox{radius } =\frac{15}{8}
    • \begin{align} & \mbox{Let } P\left(x,y\right) \\ & AP =3PB \\ & \sqrt{\left(x-3\right)^{2}+y^{2}}=3\sqrt{x^{2}+\left(y-4\right)^{2}}\\ & \left(x-3\right)^{2}+y^{2} = 9\left[x^{2}+\left(y-4\right)^{2}\right]\\ & x^{2}-6x+9+y^{2} = 9\left[x^{2}+y^{2}-8y+16\right]\\ & x^{2}-6x+9+y^{2} = 9x^{2}+9y^{2}-72y+144\\ & 8x^{2}+8y^{2}+6x-72y+135=0\\ & \mbox{This is an equation of a circle }\frac{\qquad}{}\mbox{(shown)}\\ & x^{2}+^{2}+\frac{3}{4}x-9y+\frac{135}{8}=0\\ & \left(x+\frac{3}{8}\right)^{2}+\left(y+\frac{9}{2}\right)^{2}=-\frac{135}{8}+\left(\frac{3}{8}\right)^{2}+\left(\frac{9}{2}\right)^{2}\\ & \left(x+\frac{3}{8}\right)^{2}+\left(y+\frac{9}{2}\right)^{2}=\frac{225}{64} \\ & \left(x+\frac{3}{8}\right)^{2}+\left(y+\frac{9}{2}\right)^{2}=\left(\frac{15}{8}\right)^{2} \\ & \mbox{Center } \left(-\frac{3}{8},-\frac{9}{2}\right), \mbox{radius } =\frac{15}{8} \end{align}


10. Find the perpendicular distance from the centre of the circle x^{2}+y^{2}-8x+2y+8=0\, to the straight line 3x+4y=28\,. Hence, find the shortest distance between the circle and the straight line.


11. Find the equation of the circle which touches the line 5x+y=3\, at the point \left(-2,-7\right) and has a centre which lies on the line x-2y=19\,. Hence, find the shortest distance between the circle and the straight line.


12. The point A\, has the coordinate \left(8,1\right) and the point B\, has the coordinates \left(7,0\right). Find the equation of the circle passing through A\, and B\,, and its tangent at the point B\, has the equation 3x-4y-21=0\,. Find the equation of the tangent parallel with the tangent at point B\,.


13. Show that the centre of the circles passing through the points \left(3,2\right) and \left(6,3\right) lie on the line 3x+y=16\,. Two of the circles above touch the line x+2y=3\,. Find the equations of the two circles. Determine the points on the line x+2y=2\, where the two circles pass.


14. Show that x^{2}+y^{2}-2ax-2by+c=0\, is the equation of the circle with centre \left(a,b\right) and radius \sqrt{a^{2}+b^{2}-c}.

    • \begin{align} & x^{2}+y^{2}-2ax-2by+c=0 \\ & \left(x-a\right)^{2}-a^{2}+\left(y-b\right)^{2}-b^{2}+c=0 \\ & \left(x-a\right)^{2}+\left(y-b\right)^{2}=a^{2}+b^{2}-c \\ & \therefore \mbox{Equation of circle with center } \left(a,b\right) \mbox{ and radius } \sqrt{a^{2}+b^{2}-c}\frac{\quad}{}\mbox{(shown)}\\ \end{align}

MT-CG-PY-14.pngMedia:MT-CG-PY-14.dia

The above figure shows three circles C_{1}, C_{2}\, and C_{3}\, touching one another, where their centers lie on a straight line. If C_{1}\, and C_{2}\, have equations x^{2}+y^{2}-10x-4y+28=0\, and x^{2}+y^{2}-16x+4y+52=0\, respectively, find equation of C_{3}\,.


15. Sketch the graph of 100x^{2}+36y^{2}=225\,. Find the area of the quadrilateral formed by joining the point of intersection of this curve with the coordinate axis.


16. The sum of the distance of the point P\, from the point \left(4,0\right) and the distance P\, from the origin is 8\, units. Show that the locus of P\, is the ellipse \frac{\left(x-2\right)^{2}}{16}+\frac{y^{2}}{12}=1 and sketch the ellipse.


17. A curve has a parametric equations x=t\left(t+2\right), y=2\left(t+1\right). Find the Cartesian equation of this curve. Sketch the curve.


18. a)The point P\, moves such that its distance from the x\,-axis is equal to its distance from the point A\left(0,2\right). Find the equation of the locus of P\,.

  • x^{2}=4\left(1-y\right)
    • \begin{align} & \mbox{Let }P\left(x,y\right)\\ & AP =d \\ & \sqrt{x^{2}+\left(y-2\right)^{2}}=\left|y\right|\\ & x^{2}+y^{2}-4y+4=y^{2} \\ & x^{2}=4-4y\\ & x^{2}=4\left(1-y\right) \end{align}

b) The point Q\, lie on the curve xy=16\, and the point O\, is the origin. Find the equation of the locus of the mid-point of OQ\, .

  • xy=4\,
    • \begin{align} & \mbox{Let }P\left(x,y\right), Q\left(a,\frac{16}{a}\right)\\ & P =\mbox{Midpoint } OQ \\ & \left(x,y\right) = \left(\frac{a}{2},\frac{16}{2a}\right)\\ & \therefore x = \frac{a}{2}, y =\frac{8}{a}\frac{\quad}{}(2) \\ & a=2x\frac{\quad}{}(1) \\ & (1)\to(2) : y=\frac{8}{2x} \\ & y=\frac{4}{x} \\ & \therefore xy=4 \\ \end{align}

19. The point Q\, moves such that the length of the tangent from Q\, to the circle x^{2}+y^{2}+4x+8y+9=0\, is equal to the distance of Q\, from the origin. Determine the locus of Q\,.

20. The straight line y=mx-2\, intersect with the curve y^{2}=4ax\, at two different points P\left(x_{1},y_{1}\right) and Q\left(x_{2},y_{2}\right). Show that

a) m \neq 0 and m<-\frac{1}{2}

b) x_{1}+x_{2}=\frac{4\left(m+1\right)}{m^{2}}

c) y_{1}+y_{2}=\frac{4}{m}

If the point O\, is the origin and point T\, is a point such that OPTQ\, is a parallelogram, prove that when m\, changes, the equation of locus P\, is y^{2}+4y=4x\, .

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