### orthocentre

Question Sample Titled 'orthocentre'

Suppose ${k}$ is a non-zero real constant.. It is given that ${f{{\left({x}\right)}}}={4}{x}^{{2}}-{6}{k}{x}+{5}{k}^{{2}}$.

 (a) Does the graph of ${y}={f{{\left({x}\right)}}}$ have any ${x}$-intercept? Explain your answer. (2 marks) (b) Using the method of completing the square, express, in terms of ${k}$, the coordinates of the vertex of the graph of ${y}={f{{\left({x}\right)}}}$. (3 marks) (c) Let ${A}$ and ${B}$ be the vertices of the graph of ${y}={f{{\left({x}\right)}}}$ and ${y}=\dfrac{{11}}{{4}}{k}^{{2}}-{f{{\left({x}\right)}}}$ respectively in the same rectangular system. Denote the origin by ${O}$. Brody claims that ${B}$ is the orthocentre of $\triangle{O}{A}{B}$. Do you agree? Explain your answer. (3 marks)

 (a) $\Delta$ $={\left(-{6}{k}\right)}^{{2}}-{4}{\left({4}\right)}{\left({5}{k}^{{2}}\right)}$ 1M  $=-{44}{k}^{{2}}$  $<{0}$ ∴   The graph of ${y}$ $={f{{\left({x}\right)}}}$ has no ${x}$-intercept. 1A (b) ${f{{\left({x}\right)}}}$ $={4}{x}^{{2}}-{6}{k}{x}+{5}{k}^{{2}}$  $={4}{\left({x}^{{2}}-\dfrac{{3}}{{2}}{k}{x}\right)}+{5}{k}^{{2}}$  $={4}{\left({x}^{{2}}-\dfrac{{3}}{{2}}{k}{x}+{\left(\dfrac{{3}}{{4}}{k}\right)}^{{2}}-{\left(\dfrac{{3}}{{4}}{k}\right)}^{{2}}\right)}+{5}{k}^{{2}}$ 1M  $={4}{\left({x}-\dfrac{{3}}{{4}}{k}\right)}^{{2}}-{4}\times{\left(\dfrac{{3}}{{4}}{k}\right)}^{{2}}+{5}{k}^{{2}}$  $={4}{\left({x}-\dfrac{{3}}{{4}}{k}\right)}^{{2}}+\dfrac{{11}}{{4}}{k}^{{2}}$ 1A ∴   Vertex $={\left(\dfrac{{3}}{{4}}{k},\dfrac{{11}}{{4}}{k}^{{2}}\right)}$ 1A  (c) ${A}$ $={\left(\dfrac{{3}}{{4}}{k},\dfrac{{11}}{{4}}{k}^{{2}}\right)}$ For the graph of ${y}$ $=\dfrac{{11}}{{4}}{k}^{{2}}-{f{{\left({x}\right)}}}$, vertex$={\left(\dfrac{{3}}{{4}}{k},{0}\right)}$ ${B}$ $={\left(\dfrac{{3}}{{4}}{k},{0}\right)}$ 1A ∵  ${A}{B}$ is a vertical line and ${O}{B}$ is a horizontal line. ∴  ${A}{B}\bot{O}{B}$ ${A}{B}$ and ${O}{B}$ are two altitudes of $\triangle{O}{A}{B}$ with both passing through ${B}$. 1M ∴  ${B}$ is the orthocentre of $\triangle{O}{A}{B}$. 1A ∴   He is correct.

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