JEE Main 2020 · 9 January · morning · Single correct · +4 / −1

Let C be the centroid of the triangle with vertices (3, –1), (1, 3) and (2, 4). Let P be the point of intersection of the lines x + 3y – 1 = 0 and 3x – y + 1 = 0. Then the line passing through the points C and P also passes through the point :

  1. (a) (–9, –7)
  2. (b) (9, 7)
  3. (c) (7, 6)
  4. (d) (–9, –6)
Answer and solution

Answer (d) (–9, –6)

Solution

Maths sir · 4 steps

Find C and P, write the line through them, then test each option in that one equation.

  1. Centroid C. Average the vertices: C = ((3+1+2)/3, (-1+3+4)/3) = (2, 2).
  2. Intersection P. From 3x - y + 1 = 0 you get y = 3x + 1. Put this in x + 3y - 1 = 0: 10x + 2 = 0, so x = -1/5 and y = 2/5. So P = (-1/5, 2/5).
  3. Line CP. Slope = (2 - 2/5)/(2 + 1/5) = (8/5)/(11/5) = 8/11. Then y - 2 = 8/11(x - 2) gives 11y - 22 = 8x - 16.

    8x - 11y + 6 = 0

    Point8x - 11y + 6On the line?
    (a) (-9, -7)-72 + 77 + 6 = 11No
    (b) (9, 7)72 - 77 + 6 = 1No
    (c) (7, 6)56 - 66 + 6 = -4No
    (d) (-9, -6)-72 + 66 + 6 = 0Yes
  4. So. Only (-9, -6) makes the expression zero, so the line through C and P passes through it. The answer is (d).

The idea to keep. To check whether a point lies on a line, put the line in the form ax + by + c = 0 and substitute the point. It lies on the line only if you get exactly zero.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Straight Lines

Asked about 130 times in JEE Main since 2020. This question is from 2020.