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
x₁/
(hf - bg) = y₁/ (gh - af) = 1/ (ab - h²)
⇒ x₁ = (hf - bg)/ (ab - h²) and y₁ = (gh - af) /(ab - h²)
Thus,
the two lines intersect at [(hf - bg)/ (ab - h²), (gh - af) / (ab - h²)]
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
∴ (hf - bg)/ (ab - h²) = √ [[(hf - bg)/ (ab - h²)]²
=
√ [(h²f² + b²g² - 2bfgh)/ (ab - h²)²]
=
√ [(h²f² + b²g² - abf² - b²g² - bch² + ab²c)/ (ab - h²)²]
=
√ (h² - ab) f² - bc (h² - ab)/ (h² - ab)²
=
√ (f² - bc)/ (h² - ab)
Similarly,
(hg
– af)/ (ab - h²) = √ (g² - ac)/ (h² - ab)
Thus,
the coordinates of the point of intersection of the lines represented by ax2
+ 2hxy + by2 + 2gx + 2fy + c = 0 are {√ (f² - bc)/ (h² -
ab), √ (g² - ac)/ (h² - ab)}
No comments:
Post a Comment