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In what ratio does the point M(p, -1) divide the line segment joining the points A(1, -3) and B(6, 2) ? Hence, find the value of p.

In what ratio does the point M(p, -1) divide the line segment joining the points A(1, -3) and B(6, 2) ? Hence, find the value of p. Chapterwise Revision, Concise Mathematics Solutions ICSE Class 10.

Section Formula

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Answer

Let the point M(p, -1) divide the line segment joining the points A(1, -3) and B(6, 2) in ratio k : 1.

By section formula,

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

Substituting value we get,

⇒−1=k×2+1×−3k+1⇒−1=2k−3k+1⇒−(k+1)=2k−3⇒−k−1=2k−3⇒2k+k=−1+3⇒3k=2⇒k=23.\Rightarrow -1 = \dfrac{k \times 2 + 1 \times -3}{k + 1} \\[1em] \Rightarrow -1 = \dfrac{2k - 3}{k + 1} \\[1em] \Rightarrow -(k + 1) = 2k - 3 \\[1em] \Rightarrow -k - 1 = 2k - 3 \\[1em] \Rightarrow 2k + k = -1 + 3 \\[1em] \Rightarrow 3k = 2 \\[1em] \Rightarrow k = \dfrac{2}{3}.

⇒ k : 1 = 23:1\dfrac{2}{3} : 1 = 2 : 3.

By section formula,

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

Substituting value we get,

⇒p=2×6+3×12+3⇒p=12+35⇒p=155⇒p=3.\Rightarrow p = \dfrac{2 \times 6 + 3 \times 1}{2 + 3} \\[1em] \Rightarrow p = \dfrac{12 + 3}{5} \\[1em] \Rightarrow p = \dfrac{15}{5} \\[1em] \Rightarrow p = 3.

Hence, ratio = 2 : 3 and p = 3.

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