Let (x1, y1) be the point of intersection of the lines represented by the given equation.
Shifting the origin at (x1, y1), we have x = X + x1 and y = Y + y1 … (i)
Putting the values of x and y in the given equation, we obtain
a (X + x1)2 + 2h (X + x1) (Y + y1) + b (Y + y1)2 + 2g (X + x1) + 2f (Y + y1) + c = 0
aX2 + 2hXY + bY2 + 2 (ax1 + hy1 + g) X + 2(hx1 + by1 + f) + ax21 + 2hx1y1 + by21 + 2gx1 + 2fy1 + c = 0.
This equation represents a pair of straight lines passing through the new origin (X = 0, Y = 0). So, it must be a homogeneous equation in X and Y i.e., it should not contain terms involving X, Y and constant terms.
∴ ax1 + hy1 + g = 0 … (ii)
hx1 + by1 + f = 0 … (iii)
And, ax21 + 2hx1y1 + by21 + 2gx1 + 2fy1 + c = 0 … (iv)
Solving (ii) and (iii), we get
\(\frac{{{x}_{1}}}{hf-bg}=\frac{{{y}_{1}}}{gh-af}=\frac{1}{ab-{{h}^{2}}}\).
\({{x}_{1}}=\frac{hf-bg}{ab-{{h}^{2}}}\) and \({{y}_{1}}=\frac{gh-af}{ab-{{h}^{2}}}\).
Thus, the two lines intersect at \(\left( \frac{hf-bg}{ab-{{h}^{2}}},\,\frac{gh-af}{ab-{{h}^{2}}} \right)\).
REMARKS: If the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of straight lines, then abc + 2fgh – af2 – bg2 – ch2 = 0.
2fgh = af2 + bg2 + ch2 – abc.
∴ \(\frac{hf-bg}{ab-{{h}^{2}}}=\sqrt{{{\left( \frac{hf-bg}{ab-{{h}^{2}}} \right)}^{2}}}\).
= \(\sqrt{\frac{\left( {{h}^{2}}{{f}^{2}}+{{b}^{2}}{{g}^{2}}-2bfgh \right)}{{{\left( ab-{{h}^{2}} \right)}^{2}}}}\).
= \(\sqrt{\frac{{{h}^{2}}{{f}^{2}}+{{b}^{2}}{{g}^{2}}-ab{{f}^{2}}-{{b}^{2}}{{g}^{2}}-bc{{h}^{2}}+a{{b}^{2}}c}{{{\left( ab-{{h}^{2}} \right)}^{2}}}}\).
= \(\sqrt{\frac{\left( {{h}^{2}}-ab \right){{f}^{2}}-bc\left( {{h}^{2}}-ab \right)}{{{\left( {{h}^{2}}-ab \right)}^{2}}}}\).
= \(\sqrt{\frac{{{f}^{2}}-bc}{{{h}^{2}}-ab}}\).
Similarly,
\(\frac{hg-af}{ab-{{h}^{2}}}=\sqrt{\frac{{{g}^{2}}-ac}{{{h}^{2}}-ab}}\).
Thus, the coordinates of the point of intersection of the lines represented by ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 are \(\left( \sqrt{\frac{{{f}^{2}}-bc}{{{h}^{2}}-ab}},\,\sqrt{\frac{{{g}^{2}}-ac}{{{h}^{2}}-ab}} \right)\).