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Mathematics

The line 2x + y - 4 = 0 divides the line segment joining A(2, -2) and B(3, 7) in the ratio:

  1. 2 : 3

  2. 2 : 5

  3. 2 : 7

  4. 2 : 9

Section Formula

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Answer

Let the required point be P(x, y) which divides A(2, -2) and B(3, 7) in the ratio k : 1.

The line 2x + y - 4 = 0 divides the line segment joining A(2, -2) and B(3, 7) in the ratio:Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Substituting values, we get :

⇒(x,y)=(k×3+1×2k+1,k×7+1×(−2)k+1)⇒(x,y)=(3k+2k+1,7k−2k+1)\Rightarrow (x, y) = \Big(\dfrac{k \times 3 + 1 \times 2}{k + 1}, \dfrac{k \times 7 + 1 \times (-2)}{k + 1}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{3k + 2}{k + 1}, \dfrac{7k - 2}{k + 1}\Big)

Since P lies on the line 2x + y - 4 = 0, substituting the values of x and y:

⇒2(3k+2k+1)+(7k−2k+1)−4=0⇒6k+4+7k−2k+1−4=0⇒13k+2k+1=4⇒13k+2=4(k+1)⇒13k+2=4k+4⇒9k=2⇒k=29⇒k:1=29:1=2:9.\Rightarrow 2\Big(\dfrac{3k + 2}{k + 1}\Big) + \Big(\dfrac{7k - 2}{k + 1}\Big) - 4 = 0 \\[1em] \Rightarrow \dfrac{6k + 4 + 7k - 2}{k + 1} - 4 = 0 \\[1em] \Rightarrow \dfrac{13k + 2}{k + 1} = 4 \\[1em] \Rightarrow 13k + 2 = 4(k + 1) \\[1em] \Rightarrow 13k + 2 = 4k + 4 \\[1em] \Rightarrow 9k = 2 \\[1em] \Rightarrow k = \dfrac{2}{9} \\[1em] \Rightarrow k : 1 = \dfrac{2}{9} : 1 = 2 : 9.

Hence, Option 4 is the correct option.

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