Point of Intersection of Pair of Straight Lines

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 $$\left( \sqrt{\frac{{{f}^{2}}-bc}{{{h}^{2}}-ab}},\,\sqrt{\frac{{{g}^{2}}-ac}{{{h}^{2}}-ab}} \right)$$.