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If ax^{2}+2hxy+by^{2}=0\, be two sides of a parallelogram and px+qy=1\, is one diagonal,prove that the other diagonal is y(bp-hq)=x(aq-hp)\,

Let P(x_{1},y_{1}),Q(x_{2},y_{2})\, be the points where the diagonal px+qy=1\, meets the pair of lines.

OR and PQ bisect each other at M(\alpha ,\beta )\,

\alpha ={\frac  {x_{1}+x_{2}}{2}},\beta ={\frac  {y_{1}+y_{2}}{2}}\,

Eliminating y from ax^{2}+2hxy+by^{2}=0,px+qy=1\,

ax^{2}+2hx({\frac  {1-px}{q}})+b({\frac  {1-px}{q}})^{2}=0\,


The roots of this quadratic equation are x_{1},x_{2}\,

where x_{1}+x_{2}=-{\frac  {2(hq-bp)}{aq^{2}-2hpq-np^{2}}},\alpha ={\frac  {bp-hq}{aq^{2}-2hpq+bp^{2}}}\,

Similarly by eliminating x from the given equations ,y_{1}+y_{2}=-{\frac  {2(hp-aq)}{aq^{2}-2hpq-np^{2}}},\beta ={\frac  {aq-hp}{aq^{2}-2hpq+bp^{2}}}\,

Now the equation to the line joining origin and M is \alpha y=\beta x\,

Hence y(bp-hq)=x(aq-hp)\,

Main Page:Geometry:Straight Lines-II