Question 11:
Solutions by scale drawing and / or vector is not accepted.
Diagram 11 shows the straight lines AEC and DB.
It is given that AE : EC = 1 : 2.
(a) Find
(i) the coordinates of A,
(ii) the value of p, if the area of Δ ABC is 48 unit2.
[4 marks]
(b) Point F lies on the same Cartesian plane. The straight lines DF and DB are perpendicular to each other at the point D.
Find the equation of the straight line DF in intercept form. [3 marks]
(c) Point M moves such that its distance from the point F is always 2/3 times its distance from the point C. The point F is the y-intercept of the straight line DF.
Find the equation of the locus of M. [3 marks]
Answer:
(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)
$$ \begin{aligned} \Delta A B C & =48 \text { unit }^2 \\ \frac{1}{2}\left|\begin{array}{cccc} 6 & p & -9 & 6 \\ 0 & 5 & -3 & 0 \end{array}\right| & =48 \\ \frac{1}{2}|[(6)(5)+(p)(-3)+(-9)(0)]-| & =48 \\ {[(0)(p)+(5)(-9)+(-3)(6)] } & =48 \\ \frac{1}{2}|(30-3 p)-(-63)| & =48 \\ |(30-3 p)+63| & =48(2) \\ |93-3 p| & =96 \end{aligned} $$
$$ \begin{aligned} 93-3 p & =96 \\ -3 p & =96-93 \\ -3 p & =3 \\ p & =\frac{3}{-3} \\ p & =-1 \end{aligned} $$
$$ \begin{aligned} 93-3 p & =-96 \\ -3 p & =-96-93 \\ -3 p & =-189 \\ p & =-\frac{189}{-3} \\ p & =63 \end{aligned} $$
$$ p<0, \therefore p=-1 $$
(b)
$$ \begin{aligned} & B(-1,5), D(-3,11) \\ & m_{B D}=\frac{11-5}{-3-(-1)} \\ &=-3 \end{aligned} $$
$$ \begin{aligned} B D \perp D F \rightarrow m_{B D} \times m_{D F} & =-1 \\ -3 \times m_{D F} & =-1 \\ m_{D F} & =\frac{1}{3} \end{aligned} $$
$$ \begin{aligned} & \text { Equation of } D F \text { : }\\ &\begin{aligned} y-11 & =\frac{1}{3}[x-(-3)] \\ y & =\frac{1}{3}(x+3)+11 \\ y & =\frac{1}{3} x+12 \\ {\left[y-\frac{1}{3} x\right.} & =12] \div 12 \\ \frac{y}{12}-\frac{x}{36} & =1 \end{aligned} \end{aligned} $$
(c)
$$ \begin{aligned} & F \text { is } y \text {-intercept: } F(0,12)\\ &M(x, y), C(-6,0), F(0,12) \end{aligned} $$
$$ \begin{aligned} M F & =\frac{2}{3} M C \\ \sqrt{(x-0)^2+(y-12)^2} & =\frac{2}{3} \sqrt{(x-6)^2+(y-0)^2} \\ {\left[\sqrt{(x-0)^2+(y-12)^2}\right]^2 } & =\left[\frac{2}{3} \sqrt{(x-6)^2+(y-0)^2}\right]^2 \\ x^2+y^2-24 y+144 & =\frac{4}{9}\left(x^2-12 x+36+y^2\right) \\ 9\left(x^2+y^2-24 y+144\right) & =4\left(x^2-12 x+36+y^2\right) \\ 9 x^2+9 y^2-216 y+1296 & =4 x^2-48 x+144+4 y^2 \\ 5 x^2+5 y^2+48 x-216 y+1152 & =0 \end{aligned} $$
Solutions by scale drawing and / or vector is not accepted.
Diagram 11 shows the straight lines AEC and DB.
It is given that AE : EC = 1 : 2.
(a) Find
(i) the coordinates of A,
(ii) the value of p, if the area of Δ ABC is 48 unit2.
[4 marks]
(b) Point F lies on the same Cartesian plane. The straight lines DF and DB are perpendicular to each other at the point D.
Find the equation of the straight line DF in intercept form. [3 marks]
(c) Point M moves such that its distance from the point F is always 2/3 times its distance from the point C. The point F is the y-intercept of the straight line DF.
Find the equation of the locus of M. [3 marks]
Answer:
(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)
$$ \begin{aligned} \Delta A B C & =48 \text { unit }^2 \\ \frac{1}{2}\left|\begin{array}{cccc} 6 & p & -9 & 6 \\ 0 & 5 & -3 & 0 \end{array}\right| & =48 \\ \frac{1}{2}|[(6)(5)+(p)(-3)+(-9)(0)]-| & =48 \\ {[(0)(p)+(5)(-9)+(-3)(6)] } & =48 \\ \frac{1}{2}|(30-3 p)-(-63)| & =48 \\ |(30-3 p)+63| & =48(2) \\ |93-3 p| & =96 \end{aligned} $$
$$ \begin{aligned} 93-3 p & =96 \\ -3 p & =96-93 \\ -3 p & =3 \\ p & =\frac{3}{-3} \\ p & =-1 \end{aligned} $$
$$ \begin{aligned} 93-3 p & =-96 \\ -3 p & =-96-93 \\ -3 p & =-189 \\ p & =-\frac{189}{-3} \\ p & =63 \end{aligned} $$
$$ p<0, \therefore p=-1 $$
(b)
$$ \begin{aligned} & B(-1,5), D(-3,11) \\ & m_{B D}=\frac{11-5}{-3-(-1)} \\ &=-3 \end{aligned} $$
$$ \begin{aligned} B D \perp D F \rightarrow m_{B D} \times m_{D F} & =-1 \\ -3 \times m_{D F} & =-1 \\ m_{D F} & =\frac{1}{3} \end{aligned} $$
$$ \begin{aligned} & \text { Equation of } D F \text { : }\\ &\begin{aligned} y-11 & =\frac{1}{3}[x-(-3)] \\ y & =\frac{1}{3}(x+3)+11 \\ y & =\frac{1}{3} x+12 \\ {\left[y-\frac{1}{3} x\right.} & =12] \div 12 \\ \frac{y}{12}-\frac{x}{36} & =1 \end{aligned} \end{aligned} $$
(c)
$$ \begin{aligned} & F \text { is } y \text {-intercept: } F(0,12)\\ &M(x, y), C(-6,0), F(0,12) \end{aligned} $$
$$ \begin{aligned} M F & =\frac{2}{3} M C \\ \sqrt{(x-0)^2+(y-12)^2} & =\frac{2}{3} \sqrt{(x-6)^2+(y-0)^2} \\ {\left[\sqrt{(x-0)^2+(y-12)^2}\right]^2 } & =\left[\frac{2}{3} \sqrt{(x-6)^2+(y-0)^2}\right]^2 \\ x^2+y^2-24 y+144 & =\frac{4}{9}\left(x^2-12 x+36+y^2\right) \\ 9\left(x^2+y^2-24 y+144\right) & =4\left(x^2-12 x+36+y^2\right) \\ 9 x^2+9 y^2-216 y+1296 & =4 x^2-48 x+144+4 y^2 \\ 5 x^2+5 y^2+48 x-216 y+1152 & =0 \end{aligned} $$