Question

Determine the ratio in which the line 3x + 4y - 9 = 0 divides the line segment joining the points (1,3) and (2,7).

Answer 1

Let the line 3x + 4y - 9 = 0 divides the line segment joining the points (1,3) and (2,7) in the ratio k:1

then the coordinate of the point is $( \frac{2k+1}{k+1}, \frac{7k+3}{k+1})$.

Since this the intersection point, it also lies on the line 3x + 4y - 9 = 0.Thus

$3(\frac{2k+1}{k+1}) + 4(\frac{7k+3}{k+1}) - 9 = 0\\ \\3(2k+1)+4(7k+3)-9(k+1)=0\\ \\6k+3+28k+12-9k-9=0\\ \\25k+6=0\\ \\25k=-6\\ \\k=- \frac{6}{25}$

Since k is negative, the line 3x + 4y - 9 = 0 divides the line segment joining the points (1,3) and (2,7) externally in the ratio $\boxed{6:25}$

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let the line 3x + 4y - 9 = 0 divides the lines segment joining A(1, 3) and B(2, 7) in ratio k : 1 at point P.

Therefore, Coordinates of P = (2k + 1/k + 1, 7k + 3/k + 1)

Since, P lies on the line 3x + 4y - 9 = 0

So, Coordinates of P satisfies 3x + 4y - 9 = 0

⇒ 3(2k + 1/k + 1) + 4(7k + 3/k + 1) - 9 = 0

⇒ 6k + 3 + 28k + 12 - 9k - 9 = 0

⇒ 25k + 6 = 0

⇒ k = - 6/25

Line 3x + 4y - 9 = 0 divides the line segment joining the points A and B in 6 : 25 externally.