Question 65 of 70intermediate🔧 ApplyShort Answer4 marks

Find the value of pp so that the three lines 3x+y−2=03x + y - 2 = 0, px+2y−3=0px + 2y - 3 = 0 and 2x−y−3=02x - y - 3 = 0 may intersect at one point.

Correct Answer

p = 5

Exercise: Miscellaneous Exercise on Chapter 9 | Q: 8 | (Chapter: 22)
For More Understanding

Explanation

For three lines to intersect at one point (be concurrent), they must all pass through a common point. The strategy is to first find the intersection point of two lines, then substitute this point into the third line to find the unknown parameter p.

Solution Steps

  1. Step 1: Find the intersection point of lines (1) and (3). Given lines are: (1) 3x + y - 2 = 0 and (3) 2x - y - 3 = 0. Using cross-multiplication method: x/(-3-2) = y/(-4+9) = 1/(-3-2), which gives x = 1 and y = -1. So the point of intersection is (1, -1).

  2. Step 2: Since all three lines are concurrent, the point (1, -1) must satisfy equation (2): px + 2y - 3 = 0.

  3. Step 3: Substitute x = 1 and y = -1 into equation (2): p(1) + 2(-1) - 3 = 0, which gives p - 2 - 3 = 0, therefore p = 5.

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