Sec 4 A Math: Coordinate Geometry, practice questions & worked solutions

(a)

$$\begin{aligned} \text{Midpoint of } PQ &= \left(\frac{1+9}{2},\,\frac{-2+10}{2}\right)=(5,4)\\ \text{Gradient of } PQ &= \frac{-2-10}{1-9} = \frac{3}{2}\\ \text{Gradient of perpendicular bisector} &= -\frac{2}{3}\\ y-4 &= -\frac{2}{3}(x-5)\\ y &= -\frac{2}{3}x+\frac{22}{3} \end{aligned}$$

(b)

$$\begin{aligned} -\frac{2}{3}x+\frac{22}{3} &= 8x-10\\ -\frac{26}{3}x &= -\frac{52}{3}\\ x &= 2\\ R(2,6)& \end{aligned}$$

(c)

$$\begin{aligned} \text{Area} &= \frac{1}{2}\left|\begin{array}{cccc} 9 & 2 & 1 & 9\\ 10 & 6 & -2 & 10 \end{array}\right|\\ &= \frac{1}{2}\Big[(54-4+10)-(20+6-18)\Big]\\ &= 26\ \text{unit}^2 \end{aligned}$$

(a)

$$\begin{aligned} \text{Grad}_{PQ} &= -\frac{1}{\text{Grad}_{QR}}\\[2pt] \frac{(k-2)-0}{4k-(-6)} &= -\frac{4k-8}{(k-2)-5}\\[2pt] \frac{k-2}{4k+6} &= \frac{-4k+8}{k-7}\\[2pt] (k-2)(k-7) &= (-4k+8)(4k+6)\\ k^2-9k+14 &= -16k^2+8k+48\\ 17k^2-17k-34 &= 0\\ k^2-k-2 &= 0\\ (k-2)(k+1) &= 0\\ k=2\ \text{(N.A.)} \;&\text{or}\; k=-1 \end{aligned}$$

(b)

$$\begin{aligned} \text{Midpoint of } PR &= \left(\frac{-6+8}{2},\,\frac{0+5}{2}\right)=\left(1,\,\frac{5}{2}\right)\\[2pt] \text{Grad}_{PR} &= \frac{0-5}{-6-8}=\frac{5}{14} \end{aligned}$$ Eqn. of $\perp$ bisector of $PR$: $$\begin{aligned} y-\frac{5}{2} &= -\frac{14}{5}(x-1)\\ y &= -\frac{14}{5}x+\frac{53}{10} \end{aligned}$$ $$S\!\left(0,\,\frac{53}{10}\right)$$

(c)

$$\begin{aligned} \text{Area of } PQRS &= \frac{1}{2}\left|\begin{array}{ccccc} -6 & -4 & 8 & 0 & -6\\ 0 & -3 & 5 & 5.3 & 0 \end{array}\right|\\[4pt] &= \frac{1}{2}\left|(18-20+42.4)-(-24-31.8)\right|\\[2pt] &= 48.1\ \text{units}^2 \end{aligned}$$

(a)

Since $A$ is on the $y$-axis, $$A(0, 3)$$ Since $BC$ is parallel to the $y$-axis, and $x$-coordinate of $B$ is 4, $x$-coordinate of $C = 4$. $$\begin{aligned} y &= 2(4) + 3\\ &= 11\\ C&(4, 11) \end{aligned}$$ $$\begin{aligned} \sqrt{4^2 + (h - 3)^2} &= 11 - h\\ 16 + (h - 3)^2 &= (11 - h)^2\\ 16 + h^2 - 6h + 9 &= 121 - 22h + h^2\\ 16h &= 96\\ h &= 6\\ B&(4, 6) \end{aligned}$$

(b)

$$\begin{aligned} \text{Gradient of } BD &= -\frac{1}{2}\\ y &= -\frac{1}{2}x + c\\ 6 &= -\frac{1}{2}(4) + c\\ c &= 8 \end{aligned}$$ Equation of $BD$: $y = -\dfrac{1}{2}x + 8$

Since $CD$ is parallel to the $x$-axis, $y$-coordinate of $D = 11$. $$\begin{aligned} 11 &= -\frac{1}{2}x + 8\\ 3 &= -\frac{1}{2}x\\ x &= -6\\ D&(-6, 11) \end{aligned}$$

(c)

$$\begin{aligned} \text{Area} &= \frac{1}{2}\left| \begin{matrix} 0 & 4 & 4 & -6 & 0\\ 3 & 6 & 11 & 11 & 3 \end{matrix} \right|\\ &= \frac{1}{2}\left[(0 + 44 + 44 - 18) - (12 + 24 - 66 + 0)\right]\\ &= 50\ \text{units}^2 \end{aligned}$$

(a)

$$\begin{aligned} \text{gradient of } AB &= -2\\ \text{gradient of } AC\times\text{gradient of } AB &= -1\\ \text{gradient of } AC &= \frac{1}{2} \end{aligned}$$ Equation of line $AB$: $y=-2x+c_1$.

subt $(8,0)$, $c_1=16$.

Equation of line $AB$: $y=-2x+16$.

Equation of line $AC$: $y=\dfrac{1}{2}x+c_2$.

subt $(-6,-2)$, $c_2=1$.

Equation of line $AC$: $y=\dfrac{1}{2}x+1$. $$\begin{aligned} -2x+16 &= \frac{1}{2}x+1\\ x &= 6\\ y &= -2(6)+16 = 4 \end{aligned}$$ $A(6,4)$.

(b)

Since quadrilateral $ACNM$ has only one pair of parallel sides $MN$ and $AC$ by midpoint theorem. Therefore, it is a trapezium.

(c)

$$\begin{aligned} N &= \left(\frac{-6+8}{2},\,\frac{-2+0}{2}\right)=(1,-1)\\ M &= \left(\frac{6+8}{2},\,\frac{4+0}{2}\right)=(7,2)\\ \text{Area} &= \frac{1}{2}\left|\begin{array}{ccccc} 6 & -6 & 1 & 7 & 6\\ 4 & -2 & -1 & 2 & 4 \end{array}\right|\\ &= \frac{1}{2}\Big[(-12+6+2+28)-(12-7-2-24)\Big]\\ &= \frac{45}{2} \end{aligned}$$

(a)

$$\begin{aligned} \text{Gradient of } PR &= \frac{4-2}{5-9} = -\frac{1}{2}\\ \text{Gradient of } PQ &= 2\\ \text{Equation of } PQ: y-4 &= 2(x-5)\\ y &= 2x-6 \text{----- (1)} \end{aligned}$$

(b)

$$\begin{aligned} \text{Gradient of } SQ &= -\frac{1}{2}\\ \text{Equation of } SQ: y-11 &= -\frac{1}{2}(x+1.5)\\ y &= -\frac{1}{2}x+\frac{41}{4} \text{----- (2)}\\ (1)=(2): 2x-6 &= -\frac{1}{2}x+\frac{41}{4}\\ \frac{5}{2}x &= \frac{65}{4} \;\Rightarrow\; x=6.5 \end{aligned}$$ Sub $x=6.5$ into (1): $y=2(6.5)-6=7$. Coordinates of $Q$ are $(6.5,7)$.

(c)

$$\begin{aligned} \text{Gradient of } QR &= \frac{7-2}{6.5-9} = -2\\ \text{Equation of } QR: y-2 &= -2(x-9)\\ y &= -2x+20 \end{aligned}$$ Let the coordinates of $T$ be $(a,\,-2a+20)$. $$\begin{aligned} QT &= \sqrt{500}\ \text{units}\\ (a-6.5)^2+(-2a+20-7)^2 &= 500\\ (a-6.5)^2+(13-2a)^2 &= 500\\ a^2-13a+\frac{169}{4}+169-52a+4a^2-500 &= 0\\ 20a^2-260a-1155 &= 0\\ 4a^2-52a-231 &= 0\\ (2a-33)(2a+7) &= 0\\ a=16.5 \;&\text{or}\; a=-3.5\ (\text{rejected as } a>0) \end{aligned}$$ $y$-coordinate of $T$ is $-2(16.5)+20=-13$. Coordinates of $T$ are $(16.5,-13)$.

Alternative method

$$\begin{aligned} QR &= \sqrt{(6.5-9)^2+(7-2)^2}\\ &= \sqrt{\frac{125}{4}}\ \text{units}\\[3pt] \frac{QR}{QT} &= \frac{\frac{1}{2}\sqrt{125}}{2\sqrt{125}}=\frac{1}{4}\\[3pt] \frac{x-6.5}{2.5} &= 4 \Rightarrow x=16.5\\[3pt] \frac{7-y}{5} &= 4 \Rightarrow y=-13 \end{aligned}$$ Therefore, $T(16.5,-13)$.

(a)

$$\begin{aligned} \text{Midpoint of } PQ &= \left(\frac{a + 6}{2},\ \frac{4 + (-2)}{2}\right) = \left(\frac{a + 6}{2},\ 1\right)\\ \text{Gradient of } PQ &= \frac{4 - (-2)}{a - 6} = \frac{6}{a - 6}\\ \text{Gradient of Perpendicular Bisector } PQ &= -\frac{a - 6}{6} = \frac{6 - a}{6}\\ y - 1 &= \frac{6 - a}{6}\left(x - \frac{a + 6}{2}\right) \end{aligned}$$

(b)

$$\begin{aligned} 1 - 1 &= \frac{6 - a}{6}\left(2 - \frac{a + 6}{2}\right)\\ 0 &= \left(\frac{6 - a}{6}\right)\left(\frac{4 - a - 6}{2}\right)\\ 0 &= (6 - a)(-2 - a)\\ a &= 6 \ \text{or}\ a = -2 \end{aligned}$$

(c)

$$\begin{aligned} \text{Midpoint of } PQ &= \left(\frac{6 + 6}{2},\ 1\right) = (6, 1)\\ \frac{6 + x}{2} &= 6\\ \frac{-2 + y}{2} &= 1\\ S &= (6, 4) \end{aligned}$$ $$\text{Area} = 12 \text{ units}^2$$

(a)

Let $D$ be $(d, 0)$ $$\begin{aligned} \text{length } AB &= \sqrt{(6 - 0)^2 + (24 - 6)^2}\\ &= \sqrt{360}\ \text{units}\\ \sqrt{(6 - 0)^2 + (0 - d)^2} &= \sqrt{360}\\ d^2 &= 324\\ d &= 18 \ \text{or} \ -18\\ &\ (\text{rejected because } d > 0)\\ D &= (18, 0) \end{aligned}$$ Point $D$ is $(18, 0)$.

(b)

$$\begin{aligned} \text{midpoint of } BD &= \left(\frac{6 + 18}{2}, \frac{24 + 0}{2}\right)\\ &= (12, 12) \end{aligned}$$ Let $C$ be $(a, b)$. $$\text{midpoint of } AC = \left(\frac{0 + a}{2}, \frac{6 + b}{2}\right)$$ Since it is a rhombus, midpoint $BD$ is the same as midpoint $AC$. $$\begin{aligned} \left(\frac{0 + a}{2}, \frac{6 + b}{2}\right) &= (12, 12)\\ (a, b) &= (24, 18) \end{aligned}$$ Point $C$ is $(24, 18)$.

(c)

$$\begin{aligned} \text{Area of triangle } AOD &= \frac{1}{2}(6)(18)\\ &= 54\ \text{units}^2\\ \text{Area of rhombus } ABCD &= \frac{1}{2}\sqrt{(-12)^2 + 24^2}\sqrt{24^2 + (12)^2}\\ &= 360\ \text{units}^2\\ \text{Ratio} &= 54 : 360\\ &= 3 : 20 \end{aligned}$$

Alternative Method

$$\begin{aligned} \text{Area of triangle } AOD &= \frac{1}{2}(6)(18)\\ &= 54\ \text{units}^2 \end{aligned}$$ Area of rhombus $ABCD$ $$\begin{aligned} &= \frac{1}{2}\left| \begin{matrix} 18 & 24 & 6 & 0 & 18\\ 0 & 18 & 24 & 6 & 0 \end{matrix} \right|\\ &= \frac{1}{2}\left[(324 + 576 + 36) - (108 + 108)\right]\\ &= 360\ \text{units}^2\\ \text{Ratio} &= 54 : 360\\ &= 3 : 20 \end{aligned}$$

(a)

Let $B = (x, y)$ $$\begin{aligned} \text{Distance } AB &= \sqrt{(x - 6)^2 + (y - 8)^2} = 15\\ (x - 6)^2 + (y - 8)^2 &= 225 \text{------ (1)} \end{aligned}$$ Grad $AB = \dfrac{4}{3}$

Equation of $AB$: $y = \dfrac{4}{3}x$ :: (2) $$\begin{aligned} \text{Solving, } (x - 6)^2 + \left(\frac{4}{3}x - 8\right)^2 &= 225\\ \frac{25}{9}x^2 - \frac{100}{3}x - 125 &= 0\\ x^2 - 12x - 45 &= 0\\ (x + 3)(x - 15) &= 0\\ x = -3 \ &\text{or}\ x = 15 \ (\text{rej})\\ y &= -4\\ \therefore B &= (-3, -4) \end{aligned}$$

Or

$$\begin{aligned} \text{Distance } OA &= \sqrt{6^2 + 8^2} = 10\text{ units}\\ \text{Distance } OB &= 5\text{ units} \end{aligned}$$ Let $B = (x, y)$

By ratio theorem, $$\frac{5(6) + 10x}{15} = 0, \qquad \frac{5(8) + 10y}{15} = 0$$ $$\therefore B = (-3, -4)$$

(b)

Grad $BC = -\dfrac{3}{4}$ $$\begin{aligned} y &= -\frac{3}{4}x + c\\ -4 &= -\frac{3}{4}(-3) + c \implies c = -\frac{25}{4} \end{aligned}$$ Coordinates of $C = \left(0, -\dfrac{25}{4}\right)$

Let $D = (x, 0)$ $$\begin{aligned} \frac{\frac{25}{4}}{x} &= \frac{4}{3}\\ x &= \frac{75}{16} \end{aligned}$$ Coordinates of $D = \left(\dfrac{75}{16}, 0\right)$

(c)

$$\begin{aligned} \text{Area} &= \frac{1}{2}\left| \begin{matrix} 6 & -3 & 0 & \frac{75}{16} & 6\\ 8 & -4 & -\frac{25}{4} & 0 & 8 \end{matrix} \right|\\ &= \frac{1}{2}\left| -24 + \frac{75}{4} + \frac{75}{2} - \left(-24 - \frac{1875}{64}\right) \right|\\ &= \frac{5475}{128} \text{ or } 42\frac{99}{128} \text{ or } 42.8\text{ units}^2 \end{aligned}$$

or

$$\begin{aligned} \text{Area} &= \frac{1}{2}(AB + CD)(BC)\\ &= \frac{1}{2}\left(15 + \frac{125}{16}\right)\left(\frac{15}{4}\right) = \frac{5475}{128} \end{aligned}$$