Soalan 11:
Penyelesaian secara lukisan berskala dan / atau vektor tidak diterima.
Rajah 11 menunjukkan garis lurus AEC dan garis lurus DB.
Diberi bahawa AE : EC = 1 : 2.
(a) Cari
(i) koordinat A,
(ii) nilai p, jika luas Δ ABC ialah 48 unit2.
[4 markah]
(b) Titik F berada pada satah Cartes yang sama. Garis lurus DF dan garis lurus DB adalah berserenjang antara satu sama lain pada titik D.
Cari persamaan garis lurus DF dalam bentuk pintasan. [3 markah]
(c) Titik M bergerak dengan keadaan jaraknya dari titik F adalah sentiasa 2/3 kali jaraknya dari titik C. Titik F ialah pintasan-y bagi garis lurus DF.
Cari persamaan lokus M. [3 markah]
Jawapan:
(a)(i)
$$ \begin{array}{rlrl} (-4,-2) & =\left(\frac{(2)(x)+(1)(6)}{1+2},\right. & \left.\frac{(2)(y)+(1)(0)}{1+2}\right) \\ \frac{2 x+6}{3} & =-4, & \frac{2 y+0}{3} & =-2 \\ 2 x+6 & =-12, & 2 y & =-6 \\ 2 x & =-18, & y & =-3 \\ x & =-9 & , & \end{array} $$
$$ \therefore A(-9,-3) $$
(a)(ii)
$$ A(-9,-3), B(p, 5), C(6,0) $$
$$ \begin{aligned} \Delta A B C & =48 \text { unit }^2 \\ \frac{1}{2}\left|\begin{array}{cccc} 6 & p & -9 & 6 \\ 0 & 5 & -3 & 0 \end{array}\right| & =48 \\ \frac{1}{2}|[(6)(5)+(p)(-3)+(-9)(0)]-| & =48 \\ {[(0)(p)+(5)(-9)+(-3)(6)] } & =48 \\ \frac{1}{2}|(30-3 p)-(-63)| & =48 \\ |(30-3 p)+63| & =48(2) \\ |93-3 p| & =96 \end{aligned} $$
$$ \begin{aligned} 93-3 p & =96 \\ -3 p & =96-93 \\ -3 p & =3 \\ p & =\frac{3}{-3} \\ p & =-1 \end{aligned} $$
$$ \begin{aligned} 93-3 p & =-96 \\ -3 p & =-96-93 \\ -3 p & =-189 \\ p & =-\frac{189}{-3} \\ p & =63 \end{aligned} $$
$$ p<0, \therefore p=-1 $$
(b)
$$ \begin{aligned} & B(-1,5), D(-3,11) \\ & m_{B D}=\frac{11-5}{-3-(-1)} \\ &=-3 \end{aligned} $$
$$ \begin{aligned} B D \perp D F \rightarrow m_{B D} \times m_{D F} & =-1 \\ -3 \times m_{D F} & =-1 \\ m_{D F} & =\frac{1}{3} \end{aligned} $$
$$ \begin{aligned} &\text { Persamaan } D F \text { : }\\ &\begin{aligned} y-11 & =\frac{1}{3}[x-(-3)] \\ y & =\frac{1}{3}(x+3)+11 \\ y & =\frac{1}{3} x+12 \\ {\left[y-\frac{1}{3} x\right.} & =12] \div 12 \\ \frac{y}{12}-\frac{x}{36} & =1 \end{aligned} \end{aligned} $$
(c)
$$ \begin{aligned} & F \text { ialah } \text {pintasan y: } F(0,12)\\ &M(x, y), C(-6,0), F(0,12) \end{aligned} $$
$$ \begin{aligned} M F & =\frac{2}{3} M C \\ \sqrt{(x-0)^2+(y-12)^2} & =\frac{2}{3} \sqrt{(x-6)^2+(y-0)^2} \\ {\left[\sqrt{(x-0)^2+(y-12)^2}\right]^2 } & =\left[\frac{2}{3} \sqrt{(x-6)^2+(y-0)^2}\right]^2 \\ x^2+y^2-24 y+144 & =\frac{4}{9}\left(x^2-12 x+36+y^2\right) \\ 9\left(x^2+y^2-24 y+144\right) & =4\left(x^2-12 x+36+y^2\right) \\ 9 x^2+9 y^2-216 y+1296 & =4 x^2-48 x+144+4 y^2 \\ 5 x^2+5 y^2+48 x-216 y+1152 & =0 \end{aligned} $$
Penyelesaian secara lukisan berskala dan / atau vektor tidak diterima.
Rajah 11 menunjukkan garis lurus AEC dan garis lurus DB.
Diberi bahawa AE : EC = 1 : 2.
(a) Cari
(i) koordinat A,
(ii) nilai p, jika luas Δ ABC ialah 48 unit2.
[4 markah]
(b) Titik F berada pada satah Cartes yang sama. Garis lurus DF dan garis lurus DB adalah berserenjang antara satu sama lain pada titik D.
Cari persamaan garis lurus DF dalam bentuk pintasan. [3 markah]
(c) Titik M bergerak dengan keadaan jaraknya dari titik F adalah sentiasa 2/3 kali jaraknya dari titik C. Titik F ialah pintasan-y bagi garis lurus DF.
Cari persamaan lokus M. [3 markah]
Jawapan:
(a)(i)
$$ \begin{array}{rlrl} (-4,-2) & =\left(\frac{(2)(x)+(1)(6)}{1+2},\right. & \left.\frac{(2)(y)+(1)(0)}{1+2}\right) \\ \frac{2 x+6}{3} & =-4, & \frac{2 y+0}{3} & =-2 \\ 2 x+6 & =-12, & 2 y & =-6 \\ 2 x & =-18, & y & =-3 \\ x & =-9 & , & \end{array} $$
$$ \therefore A(-9,-3) $$
(a)(ii)
$$ A(-9,-3), B(p, 5), C(6,0) $$
$$ \begin{aligned} \Delta A B C & =48 \text { unit }^2 \\ \frac{1}{2}\left|\begin{array}{cccc} 6 & p & -9 & 6 \\ 0 & 5 & -3 & 0 \end{array}\right| & =48 \\ \frac{1}{2}|[(6)(5)+(p)(-3)+(-9)(0)]-| & =48 \\ {[(0)(p)+(5)(-9)+(-3)(6)] } & =48 \\ \frac{1}{2}|(30-3 p)-(-63)| & =48 \\ |(30-3 p)+63| & =48(2) \\ |93-3 p| & =96 \end{aligned} $$
$$ \begin{aligned} 93-3 p & =96 \\ -3 p & =96-93 \\ -3 p & =3 \\ p & =\frac{3}{-3} \\ p & =-1 \end{aligned} $$
$$ \begin{aligned} 93-3 p & =-96 \\ -3 p & =-96-93 \\ -3 p & =-189 \\ p & =-\frac{189}{-3} \\ p & =63 \end{aligned} $$
$$ p<0, \therefore p=-1 $$
(b)
$$ \begin{aligned} & B(-1,5), D(-3,11) \\ & m_{B D}=\frac{11-5}{-3-(-1)} \\ &=-3 \end{aligned} $$
$$ \begin{aligned} B D \perp D F \rightarrow m_{B D} \times m_{D F} & =-1 \\ -3 \times m_{D F} & =-1 \\ m_{D F} & =\frac{1}{3} \end{aligned} $$
$$ \begin{aligned} &\text { Persamaan } D F \text { : }\\ &\begin{aligned} y-11 & =\frac{1}{3}[x-(-3)] \\ y & =\frac{1}{3}(x+3)+11 \\ y & =\frac{1}{3} x+12 \\ {\left[y-\frac{1}{3} x\right.} & =12] \div 12 \\ \frac{y}{12}-\frac{x}{36} & =1 \end{aligned} \end{aligned} $$
(c)
$$ \begin{aligned} & F \text { ialah } \text {pintasan y: } F(0,12)\\ &M(x, y), C(-6,0), F(0,12) \end{aligned} $$
$$ \begin{aligned} M F & =\frac{2}{3} M C \\ \sqrt{(x-0)^2+(y-12)^2} & =\frac{2}{3} \sqrt{(x-6)^2+(y-0)^2} \\ {\left[\sqrt{(x-0)^2+(y-12)^2}\right]^2 } & =\left[\frac{2}{3} \sqrt{(x-6)^2+(y-0)^2}\right]^2 \\ x^2+y^2-24 y+144 & =\frac{4}{9}\left(x^2-12 x+36+y^2\right) \\ 9\left(x^2+y^2-24 y+144\right) & =4\left(x^2-12 x+36+y^2\right) \\ 9 x^2+9 y^2-216 y+1296 & =4 x^2-48 x+144+4 y^2 \\ 5 x^2+5 y^2+48 x-216 y+1152 & =0 \end{aligned} $$