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Find the ratio in which the line 2x + y = 4 divides the line segment joining the points P(2, -2) and Q(3, 7).

Section Formula

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Answer

Let ratio in which 2x + y = 4 divides the line segment joining the points P(2, -2) and Q(3, 7) be k : 1 at point (x, y).

By section formula,

x=m1x2+m2x1m1+m2x = \dfrac{m1x2 + m2x1}{m1 + m2}

Substituting values we get,

⇒x=k×3+1×2k+1⇒x(k+1)=3k+2⇒x=3k+2k+1.\Rightarrow x = \dfrac{k \times 3 + 1 \times 2}{k + 1} \\[1em] \Rightarrow x(k + 1) = 3k + 2 \\[1em] \Rightarrow x = \dfrac{3k + 2}{k + 1}.

y=m1y2+m2y1m1+m2y = \dfrac{m1y2 + m2y1}{m1 + m2}

Substituting values we get,

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

Substituting value of x and y in 2x + y = 4.

⇒2(3k+2k+1)+7k−2k+1=4⇒6k+4k+1+7k−2k+1=4⇒6k+4+7k−2k+1=4⇒13k+2=4k+4⇒9k=2⇒k=29.⇒k:1=29:1=2:9.\Rightarrow 2 \Big(\dfrac{3k + 2}{k + 1}\Big) + \dfrac{7k - 2}{k + 1} = 4 \\[1em] \Rightarrow \dfrac{6k + 4}{k + 1} + \dfrac{7k - 2}{k + 1} = 4 \\[1em] \Rightarrow \dfrac{6k + 4 + 7k - 2}{k + 1} = 4 \\[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, ratio in which the line 2x + y = 4 divides the line segment joining the points P(2, -2) and Q(3, 7) = 2 : 9.

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