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Point P divides the line segment joining R(-1, 3) and S(9, 8) in the ratio k : 1. If P lies on the line x - y + 2 = 0, then the value of k is:

  1. 12\dfrac{1}{2}

  2. 13\dfrac{1}{3}

  3. 14\dfrac{1}{4}

  4. 23\dfrac{2}{3}

Section Formula

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Answer

Let point P be (x, y).

Point P divides the line segment joining R(-1, 3) and S(9, 8) in the ratio k : 1. If P lies on the line x - y + 2 = 0, then the value of k is: Reflection, RSA Mathematics Solutions ICSE Class 10.

Given,

m1 : m2 = k : 1

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×9+1×(−1)k+1,k×8+1×3k+1)=(9k−1k+1,8k+3k+1).\Rightarrow (x, y) = \Big(\dfrac{k \times 9 + 1 \times (-1)}{k + 1}, \dfrac{k \times 8 + 1 \times 3}{k + 1}\Big) \\[1em] = \Big(\dfrac{9k - 1}{k + 1}, \dfrac{8k + 3}{k + 1}\Big).

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

⇒9k−1k+1−8k+3k+1+2=0⇒9k−1−(8k+3)k+1+2=0⇒9k−1−8k−3k+1+2=0⇒k−4k+1+2=0⇒k−4+2(k+1)k+1=0⇒k−4+2k+2k+1=0⇒3k−2k+1=0⇒3k−2=0⇒3k=2⇒k=23.\Rightarrow \dfrac{9k - 1}{k + 1} - \dfrac{8k + 3}{k + 1} + 2 = 0 \\[1em] \Rightarrow \dfrac{9k - 1 - (8k + 3)}{k + 1} + 2 = 0 \\[1em] \Rightarrow \dfrac{9k - 1 - 8k - 3}{k + 1} + 2 = 0 \\[1em] \Rightarrow \dfrac{k - 4}{k + 1} + 2 = 0 \\[1em] \Rightarrow \dfrac{k - 4 + 2(k + 1)}{k + 1} = 0 \\[1em] \Rightarrow \dfrac{k - 4 + 2k + 2}{k + 1} = 0 \\[1em] \Rightarrow \dfrac{3k - 2}{k + 1} = 0 \\[1em] \Rightarrow 3k - 2 = 0 \\[1em] \Rightarrow 3k = 2 \\[1em] \Rightarrow k = \dfrac{2}{3}.

Hence, Option 4 is the correct option.

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