Blog: Find the equation of a line that passes through point (0,2) and makes 45 degrees to the line 2x+3y+6=0.

Saturday, February 26, 2011

Find the equation of a line that passes through point (0,2) and makes 45 degrees to the line 2x+3y+6=0.

Hello!

As you probably know, a normal (orthogonal) vector for a line with the equation $\displaystyle{a}{x}+{b}{y}+{c}={0}$ is $\displaystyle\lt{a},{b}\gt.$ So for the given line it is $\displaystyle\lt{2},{3}\gt.$ Denote the equation in question as $\displaystyle{y}={m}{x}+{b},$ which is the same as $\displaystyle{m}{x}-{y}+{b}={0},$ its orthogonal vector is $\displaystyle\lt{m},-{1}\gt.$

For the angle between the lines to be 45 degrees, the angle between normal vectors must be $\displaystyle\pm{45}$ degrees. Take the dot product of these two vectors: from the one hand, it is $\displaystyle{2}{m}-{3},$ from the other it is

$\displaystyle{\left|<{m},-{1}>\right|}\cdot{\left|<{2},{3}>\right|}\cdot{\cos{{\left({45}\right)}}}=\sqrt{{{{m}}^{{2}}+{1}}}\cdot\sqrt{{{{2}}^{{2}}+{{3}}^{{2}}}}\cdot\frac{\sqrt{{{2}}}}{{2}}=\pm{\left({2}{m}-{3}\right)}.$

Square this equation and obtain

$\displaystyle{\left({{m}}^{{2}}+{1}\right)}\cdot\frac{{13}}{{2}}={{\left({2}{m}-{3}\right)}}^{{2}}={4}{{m}}^{{2}}-{12}{m}+{9},$ or

$\displaystyle{13}{{m}}^{{2}}+{13}={8}{{m}}^{{2}}-{24}{m}+{18},$ or

$\displaystyle{5}{{m}}^{{2}}+{24}{m}-{5}={0}.$

The roots of this quadratic equation is

$\displaystyle\frac{{-{12}\pm\sqrt{{{144}+{25}}}}}{{5}}=\frac{{-{12}\pm{13}}}{{5}},$ i.e. $\displaystyle{m}_{{1}}=-{5}$ and $\displaystyle{m}_{{2}}=\frac{{1}}{{5}}.$

Now we have to find the corresponding b's. The point (0, 2) satisfies the equation $\displaystyle{y}={m}{x}+{b},$ so $\displaystyle{b}={y}_{{0}}-{m}{x}_{{0}}={2}-{m}\cdot{0}={2}.$ Thus $\displaystyle{b}_{{1}}={b}_{{2}}={2}.$

This way, the answer is: $\displaystyle{y}=-{5}{x}+{2}$ and $\displaystyle{y}=\frac{{1}}{{5}}{x}+{2}.$

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Saturday, February 26, 2011

Find the equation of a line that passes through point (0,2) and makes 45 degrees to the line 2x+3y+6=0.

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