Question

If a variable straight line passing through the point of intersection of the lines x - 2y + 3 = 0 and 2x - y - 1 = 0 intersects the X and Y axes at A and B respectively, then the equation of the locus of a point which divides the segment AB in the ratio -2 : 3 is:

• A. \(14x^2 + 3xy - 15y^2 = 0\)
• B. \(xy = 14x + 15y\)
• C. \(x^2 + xy - y^2 = 0\)
• D. \(14x + 3xy - 15y = 0\)

Hint:

To find the locus of a point, eliminate the parameter (in this case, m) using the given conditions.

Answer 1

The Correct Option is D

Step 1: Find the point of intersection of the given lines.
The given lines are:
x - 2y + 3 = 0 ...(1)
2x - y - 1 = 0 ...(2)
Multiply equation (1) by 2:
2x - 4y + 6 = 0 ...(3)
Subtract equation (2) from equation (3):
(2x - 4y + 6) - (2x - y - 1) = 0
-3y + 7 = 0
y = \(\frac{7}{3}\)
Substitute y = \(\frac{7}{3}\) in equation (1):
x - 2(\(\frac{7}{3}\)) + 3 = 0
x - \(\frac{14}{3}\) + \(\frac{9}{3}\) = 0
x - \(\frac{5}{3}\) = 0
x = \(\frac{5}{3}\)
The point of intersection is (\(\frac{5}{3}\), \(\frac{7}{3}\)).

Step 2: Let the equation of the line passing through the intersection point be.
The equation of the line passing through (\(\frac{5}{3}\), \(\frac{7}{3}\)) is:
y - \(\frac{7}{3}\) = m(x - \(\frac{5}{3}\))
3y - 7 = m(3x - 5)
3y - 7 = 3mx - 5m
3mx - 3y + 7 - 5m = 0 ...(4)

Step 3: Find the coordinates of A and B.
For point A (x-intercept), put y = 0 in equation (4):
3mx + 7 - 5m = 0
x = \(\frac{5m - 7}{3m}\)
So, A = (\(\frac{5m - 7}{3m}\), 0)
For point B (y-intercept), put x = 0 in equation (4):
-3y + 7 - 5m = 0
y = \(\frac{7 - 5m}{3}\)
So, B = (0, \(\frac{7 - 5m}{3}\))

Step 4: Let the dividing point be (h, k).
Given that (h, k) divides AB in the ratio -2 : 3.
Using section formula:
h = \(\frac{3(\frac{5m - 7}{3m}) + (-2)(0)}{3 - 2} = \frac{5m - 7}{m}\)
k = \(\frac{3(0) + (-2)(\frac{7 - 5m}{3})}{3 - 2} = \frac{-14 + 10m}{3}\)

Step 5: Eliminate m to find the locus.
From h = \(\frac{5m - 7}{m}\), we get hm = 5m - 7, so m(h - 5) = -7, and m = \(\frac{-7}{h - 5} = \frac{7}{5 - h}\).
From k = \(\frac{-14 + 10m}{3}\), we get 3k = -14 + 10m, so 10m = 3k + 14, and m = \(\frac{3k + 14}{10}\).
Equating the two expressions for m:
\(\frac{7}{5 - h} = \frac{3k + 14}{10}\)
70 = (5 - h)(3k + 14)
70 = 15k + 70 - 3hk - 14h
0 = 15k - 3hk - 14h
14h + 3hk - 15k = 0
Replace (h, k) with (x, y):
14x + 3xy - 15y = 0
Therefore, the equation of the locus is 14x + 3xy - 15y = 0.