Calculate the coordinates of E.
About this question
Question 3.3 of Mathematics Grade 12 Paper 2, NSC Examination (May/June) (DBE, May/June 2026). It tests the Term 2 ATP topic Analytical Geometry.
- TermTerm 2
- ATP topicAnalytical Geometry
- PaperPaper 2
- Marks3 marks for the method (formula and substitution) and the correct final answer
AS intersects the \(y\)-axis at \(C(0\,;\,6)\). \(B(-4\,;\,8)\) is the midpoint of AC. F is the point \((0\,;\,-6)\). LF is drawn parallel to AS and cuts the \(x\)-axis at G. E is a point on LF such that the equation of AE is \(y = -\frac{7}{2}x - 18\). AE cuts the \(x\)-axis at T. SF is drawn.
Diagram: A lies in the second quadrant above B, with line AS sloping down to the right through B and C to S in the first quadrant. Line LF runs parallel to AS below it, from L (upper left, beyond G) through G on the negative \(x\)-axis and E (third quadrant) to F on the negative \(y\)-axis. Line AE runs steeply down from A through T on the negative \(x\)-axis (between G and O) to E. Segment SF joins S to F.
E lies on AE and LF, so equate:
\[-\frac{1}{2}x - 6 = -\frac{7}{2}x - 18\]\[-x - 12 = -7x - 36\]\[6x = -24\]\[x = -4\]\[y = -\frac{1}{2}(-4) - 6 = -4\]\(E(-4\,;\,-4)\)
This question also appeared in
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More Questions On Analytical Geometry
At T, \(x = 0\): \(y = 10\), so \(T(0\,;\,10)\)
\[RT^2 = (0 - (-5))^2 + (10 - 0)^2\]\[RT^2 = 25 + 100 = 125\]\(RT = \sqrt{125} = 5\sqrt{5}\) units
so \(C\hat{S}F = 90^\circ\)
\(\therefore\) CF is a diameter (chord subtends a right angle at the circle)
Centre = midpoint of CF = \(O(0\,;\,0)\) and \(r = 6\)
\(x^2 + y^2 = 36\)
The tangent is perpendicular to the radius at the point of contact.
\(C\hat{E}M = 90^\circ\)
Centre \(M(-6\,;\,4)\), radius \(AM = q + 1 = 5\)
\((x + 6)^2 + (y - 4)^2 = 25\)