Write down the coordinates of \(L\).
About this question
Question 4.1 of Mathematics Grade 12 Paper 2, NSC Examination (DBE, November 2024). It tests the Term 2 ATP topic Analytical Geometry.
- TermTerm 2
- ATP topicAnalytical Geometry
- PaperPaper 2
- Marks2 marks for the method (formula and substitution) and the correct final answer
\(M(1\,;\,3)\) is the centre of the circle. The circle cuts the \(x\)-axis at \(N\). \(ST\) is a tangent to the circle at \(P(3\,;\,9)\). \(R(d\,;\,1)\), with \(d \gt 0\), and \(L\) lie on the circle. \(O\) and \(V\) are the \(x\)-intercepts of \(PL\) and \(RL\) respectively.
Diagram: \(PL\) is a diameter passing through \(M\) and through the origin \(O\), with \(L\) in the third quadrant. Chords \(PR\) and \(RL\) are drawn, \(R\) lying just above the \(x\)-axis on the right. \(RL\) cuts the \(x\)-axis at \(V\), between \(O\) and \(N\). The tangent \(ST\) touches the circle at \(P\), with \(S\) near the \(y\)-axis and \(T\) to the right.
\(PL\) is a diameter, so \(M\) is the midpoint of \(PL\).
\[x_L = 2(1) - 3 = -1\,;\quad y_L = 2(3) - 9 = -3\]\(L(-1\,;\,-3)\)
More Questions On Analytical Geometry
\(F\) is the \(y\)-intercept of \(y = 2x - 2\), so \(F(0\,;\,-2)\).
\(F\) is the midpoint of \(AC\):
\[\left(\frac{x + 3}{2}\,;\,\frac{y + 4}{2}\right) = (0\,;\,-2)\]\[\frac{x + 3}{2} = 0 \Rightarrow x = -3\]\[\frac{y + 4}{2} = -2 \Rightarrow y = -8\]\(C(-3\,;\,-8)\)
(\(LF \parallel AS\))
\(F(0\,;\,-6)\) is the \(y\)-intercept, so \(c = -6\)
\(y = -\frac{1}{2}x - 6\)
\(\therefore PT \perp RT\)
\(S(-15\,;\,-2)\) lies on \(AB\) produced.
\[AS = \sqrt{(3 - (-15))^2 + (4 - (-2))^2} = \sqrt{360} = 6\sqrt{10}\]\[AB = \sqrt{(-9 - 3)^2 + (0 - 4)^2} = 4\sqrt{10}\]With \(D(1\,;\,0)\):
\[AD = \sqrt{(3 - 1)^2 + (4 - 0)^2} = 2\sqrt{5}\]\[\frac{\text{Area }\triangle ABD}{\text{Area }\triangle ASC} = \frac{\frac{1}{2}(AB)(AD)\sin\hat{A}}{\frac{1}{2}(AS)(AC)\sin\hat{A}}\]\[= \frac{(4\sqrt{10})(2\sqrt{5})}{(6\sqrt{10})(6\sqrt{5})}\]\[= \frac{8}{36}\]\(= \frac{2}{9}\)
\(S(-12\,;\,1)\) and \(T(0\,;\,10)\)
\[m_{ST} = \frac{10 - 1}{0 - (-12)} = \frac{9}{12} = \frac{3}{4}\]\(VR \perp ST\), so
\[m_{VR} = -\frac{4}{3}\]Substitute \(R(-5\,;\,0)\):
\[0 = -\frac{4}{3}(-5) + c\]\[c = -\frac{20}{3}\]\(y = -\frac{4}{3}x - \frac{20}{3}\)
Inclination of EM:
\[\tan(\text{inclination of } EM) = m_{EM} = -\frac{3}{4}\]\[\text{Inclination of } EM = 180^\circ - 36{,}87^\circ = 143{,}13^\circ\]\[\therefore E\hat{M}D = 180^\circ - 143{,}13^\circ = 36{,}87^\circ\]Inclination of CM:
\[m_{CM} = \frac{0 - (-5)}{3 - (-7)} = \frac{5}{10} = \frac{1}{2}\]\[\text{Inclination of } CM = 26{,}57^\circ\]\[\therefore T\hat{M}D = 26{,}57^\circ\]\[E\hat{M}T = 36{,}87^\circ + 26{,}57^\circ = 63{,}44^\circ\]\(EM = MT\) (radii), so \(E\hat{T}M = T\hat{E}M\)
\[E\hat{T}M = \frac{180^\circ - 63{,}44^\circ}{2}\]\(E\hat{T}M = 58{,}28^\circ\)