Question

What will be ratio in which the line 3x + y – 11 = 0 divides the line segment joining the points (0, -1) and (-3, -4)?

(a) 1:2 (internally)

(b) 1:2 (externally)

(c) 2:1 (externally)

(d) 2:1 (internally)

I had been asked this question by my school principal while I was bunking the class.

This is a very interesting question from Geometry in division Coordinate Geometry of Mathematics – Class 10

Answer 1

Correct choice is (b) 1:2 (externally)

The explanation is: Let the ratio in which the line 3x + y – 11 = 0 divides the line segment joining the points (0, -1) and (-3, -4) be k:1.

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(0, -1) and B(-3, -4) and the ratio is k:1.

∴ x = \(\frac {k(-3)+1(0)}{k+1} = \frac {-3k}{k+1}\)

y = \(\frac {k(-4)+1(-1)}{k+1} = \frac {-4k-1}{k+1}\)

Since, the point \((\frac {-3k}{k+1}, \frac {-4k-1}{k+1} )\) lies on the line 3x+y-11 = 0.

3 \((\frac {-3k}{k+1} + \frac {-4k-1}{k+1} )\)-11 = 0

3(-3k) + (-4k – 1) – 11(k + 1) = 0

-9k – 4k – 1 – 11k – 11 = 0

-24k – 12 = 0

-24k = 12

k = \(\frac {12}{-24} = \frac {-1}{2}\)

The ratio is 1:2 (externally).