Question 11:
Solution by scale drawing and / or vector are not accepted. Diagram 5 shows the triangle PQR on a Cartesian plane.
It is given that the gradient of the straight lines PQ, QR and PR is -3m, m and 3m respectively, such that m is a constant.
(a) Show that m = ½.
[2 marks]
(b) Find the coordinates of R.
[3 marks]
(c)(i) Find the area of Δ PQR.
(ii) Point U moves such that its distance from point R is always 5/2 times its distance from point Q.
Find the equation of the locus of U.
[5 marks]
Answer:
(a)
$$ \begin{aligned} & m_{P Q}=-3 m \\ & \frac{4-1}{-1-1}=-3 m \\ & -\frac{3}{2}=-3 m \\ & \frac{-\frac{3}{2}}{-3}=m \\ & \frac{1}{2}=m(\text { Shown }) \end{aligned} $$
(b)
$$ \begin{aligned} \ \text { Let }, R(x, y) & \\ m_{Q R} & =m \\ \frac{4-y}{-1-x} & =\frac{1}{2} \\ 2(4-y) & =1(-1-x) \\ 8-2 y & =-1-x \\ -2 y & =-1-x-8 \\ -2 y & =-9-x \ldots （1） \end{aligned} $$
$$ \begin{aligned} m_{P R} & =3 m \\ \frac{1-y}{1-x} & =3\left(\frac{1}{2}\right) \\ \frac{1-y}{1-x} & =\frac{3}{2} \\ 3(1-x) & =2(1-y) \\ 3-3 x-2 & =-2 y \\ 1-3 x & =-2 y \\ 3 x-1 & =2 y \ldots （2） \end{aligned} $$
$$ \begin{aligned} & (1)=(2) ; \\ & 3 x-1=9+x \\ & 3 x-x=9+1 \\ & 2 x=10 \\ & x=5 \end{aligned} $$
$$ \begin{aligned} &\text { From (1), }\\ &\begin{aligned} 2 y & =9+x \\ 2 y & =9+5 \\ 2 y & =14 \\ y & =7 \end{aligned}\\ &\therefore R(5,7) \end{aligned} $$
(c)(i)
$$ \begin{aligned} &\text { Area, } \triangle P Q R\\ &\begin{aligned} & =\frac{1}{2}\left|\begin{array}{cccc} 1 & -1 & 5 & 1 \\ 1 & 4 & 7 & 1 \end{array}\right| \\ & =\frac{1}{2}|[(1)(4)+(-1)(7)+(5)(1)]-[(1)(-1)+(4)(5)+(7)(1)]| \\ & =\frac{1}{2}|2-26| \\ & =\frac{1}{2}|-24| \\ & =\frac{1}{2}(24) \\ & =12 \text { unit }^2 \end{aligned} \end{aligned} $$
(c)(ii)
$$ \begin{aligned} R U=\frac{5}{2} Q U & \\ \sqrt{(x-5)^2+(y-7)^2} & =\frac{5}{2} \sqrt{\left[(x-(-1)]^2+(y-4)^2\right.} \\ \sqrt{x^2-10 x+25+y^2-14 y+49} & =\frac{5}{2} \sqrt{x^2+2 x+1+y^2-8 y+16} \\ x^2+y^2-10 x-14 y+74 & =\left(\frac{5}{2}\right)^2\left(x^2+y^2+2 x-8 y+17\right) \\ x^2+y^2-10 x-14 y+74 & =\frac{25}{4}\left(x^2+y^2+2 x-8 y+17\right) \\ 4 x^2+4 y^2-40 x-56 y+296 & =25 x^2+25 y^2+50 x-200 y+425 \\ 0 & =21 x^2+21 y^2+90 x-144 y+129 \\ 0 & =7 x^2+7 y^2+30 x-48 y+43 \end{aligned} $$
Solution by scale drawing and / or vector are not accepted. Diagram 5 shows the triangle PQR on a Cartesian plane.
It is given that the gradient of the straight lines PQ, QR and PR is -3m, m and 3m respectively, such that m is a constant.
(a) Show that m = ½.
[2 marks]
(b) Find the coordinates of R.
[3 marks]
(c)(i) Find the area of Δ PQR.
(ii) Point U moves such that its distance from point R is always 5/2 times its distance from point Q.
Find the equation of the locus of U.
[5 marks]
Answer:
(a)
$$ \begin{aligned} & m_{P Q}=-3 m \\ & \frac{4-1}{-1-1}=-3 m \\ & -\frac{3}{2}=-3 m \\ & \frac{-\frac{3}{2}}{-3}=m \\ & \frac{1}{2}=m(\text { Shown }) \end{aligned} $$
(b)
$$ \begin{aligned} \ \text { Let }, R(x, y) & \\ m_{Q R} & =m \\ \frac{4-y}{-1-x} & =\frac{1}{2} \\ 2(4-y) & =1(-1-x) \\ 8-2 y & =-1-x \\ -2 y & =-1-x-8 \\ -2 y & =-9-x \ldots （1） \end{aligned} $$
$$ \begin{aligned} m_{P R} & =3 m \\ \frac{1-y}{1-x} & =3\left(\frac{1}{2}\right) \\ \frac{1-y}{1-x} & =\frac{3}{2} \\ 3(1-x) & =2(1-y) \\ 3-3 x-2 & =-2 y \\ 1-3 x & =-2 y \\ 3 x-1 & =2 y \ldots （2） \end{aligned} $$
$$ \begin{aligned} & (1)=(2) ; \\ & 3 x-1=9+x \\ & 3 x-x=9+1 \\ & 2 x=10 \\ & x=5 \end{aligned} $$
$$ \begin{aligned} &\text { From (1), }\\ &\begin{aligned} 2 y & =9+x \\ 2 y & =9+5 \\ 2 y & =14 \\ y & =7 \end{aligned}\\ &\therefore R(5,7) \end{aligned} $$
(c)(i)
$$ \begin{aligned} &\text { Area, } \triangle P Q R\\ &\begin{aligned} & =\frac{1}{2}\left|\begin{array}{cccc} 1 & -1 & 5 & 1 \\ 1 & 4 & 7 & 1 \end{array}\right| \\ & =\frac{1}{2}|[(1)(4)+(-1)(7)+(5)(1)]-[(1)(-1)+(4)(5)+(7)(1)]| \\ & =\frac{1}{2}|2-26| \\ & =\frac{1}{2}|-24| \\ & =\frac{1}{2}(24) \\ & =12 \text { unit }^2 \end{aligned} \end{aligned} $$
(c)(ii)
$$ \begin{aligned} R U=\frac{5}{2} Q U & \\ \sqrt{(x-5)^2+(y-7)^2} & =\frac{5}{2} \sqrt{\left[(x-(-1)]^2+(y-4)^2\right.} \\ \sqrt{x^2-10 x+25+y^2-14 y+49} & =\frac{5}{2} \sqrt{x^2+2 x+1+y^2-8 y+16} \\ x^2+y^2-10 x-14 y+74 & =\left(\frac{5}{2}\right)^2\left(x^2+y^2+2 x-8 y+17\right) \\ x^2+y^2-10 x-14 y+74 & =\frac{25}{4}\left(x^2+y^2+2 x-8 y+17\right) \\ 4 x^2+4 y^2-40 x-56 y+296 & =25 x^2+25 y^2+50 x-200 y+425 \\ 0 & =21 x^2+21 y^2+90 x-144 y+129 \\ 0 & =7 x^2+7 y^2+30 x-48 y+43 \end{aligned} $$