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Joint Equation of Pair of Lines

 

Joint Equation of Pair of Lines Joining the Origin and the Points of Intersection of a Line and a Curve

If the line lx + my + n = 0, (n ≠ 0) i.e. the line not passing through origin) cuts the curve ax2 + by2 + 2gx + 2fy + c = 0 at two points A and B, then the joint equation of straight lines passing through A and B and the origin is given by homogenizing the equation of the curve by the equation of the line. i.e.

Origin and the Points of Intersection of a Line and a Curve

ax2 + 2hxy + by2 + (2gx + 2fy) (kx+my/–n) + c (lx+my/–n)2 = 0.

is the equation of the lines OA and OB.

Illustration:

Prove that the straight lines joining the origin to the points of intersection of the straight line hx + ky = 2hk and the curve (x – k)2 + (y – y)2 = c2 are at right angles if h2 + k2 = c2.

Solution:

Making the equation of the curve homogeneous with the help of that of the line, we get

x2 + y2 –2(kx + hy) (hx+ky/2hk) + (h2 + k2 – c2) (hx+ky/2hk) = 0

or 4h2k2x2 + 4h2k2y2 – 4h2x(hx+ky) – 4h2ky(hx + ky) + (h2 + k2 – c)(h2x2 + k2y2 + 2hxy) = 0.

This is the equation of the pair of lines joining the origin to the points of intersection of the given line and the curve. They will be at right angles if

Coefficient of x2 + coefficient of y2 = 0 i.e.

(h2 + k2) (h2 + k2 – c2) = 0 ⇒ h2 + k2 = c2  (since h2 + k2 ≠ 0).

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