### Find orthocentre of a triangle given three vertices

Question Sample Titled 'Find orthocentre of a triangle given three vertices'

Let ${O}$ be the origin. The coordinates of the points ${P}$ and ${Q}$ are ${\left({0},{54}\right)}$ and ${\left({88},{22}\right)}$ respectively. The ${x}$-coordinate of the orthocentre of $\triangle{O}{P}{Q}$ is

A
${8}$ .
B
${22}$ .
C
${15}$ .
D
${128}$ .

 Recall that the definition of orthocentre of a triangle is the intersection of the three (or any two) altitudes of that triangle. Let ${H}$ be the orthocenter of $\triangle{O}{P}{Q}$ . ${H}$ lies on the altitude passing through vertex ${Q}$ .  Note that ${O}{P}$ is a vertical line, the ${y}$-coordinate of ${H}$ is ${22}$ . Denote the altitude passing through ${P}$ be ${L}_{{1}}$ . The equation of ${L}_{{1}}$ is ${y}$ $={\left[-{1}\div{\left(\dfrac{{22}}{{88}}\right)}\right]}{x}+{54}$ ${L}_{{1}}\bot{O}{Q}$ ${y}$ $=-{4}{x}+{54}$ ∴   The orthocenter can be found by solving ${y}$ $={22}$ $\ldots{\left({1}\right)}$ ${L}_{{1}}:{y}$ $=-{4}{x}+{54}$ $\ldots{\left({2}\right)}$ Subsitute ${\left({1}\right)}$ into ${\left({2}\right)}$ : ${22}$ $=-{4}{x}+{54}$ ${x}$ $={8}$ ∴   The ${x}$-coordinate of the corresponding orthocentre is ${8}$ .

${x}$${y}$${O}$${22}$${y}={22}$${P}{\left({0},{54}\right)}$${Q}{\left({88},{22}\right)}$${H}$${L}_{{1}}$

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