Mathematics — Standard (041) — CBSE Class 10 board question

Let the point $\left(\dfrac{8}{5}, y\right)$ divide the line segment joining $A(1, 2)$ and $B(2, 3)$ in the ratio $k : 1$.

By the section formula, the x-coordinate is:

$$\frac{8}{5} = \frac{k(2) + 1(1)}{k + 1} = \frac{2k + 1}{k + 1}$$

$$8(k + 1) = 5(2k + 1)$$

$$8k + 8 = 10k + 5$$

$$2k = 3 \implies k = \frac{3}{2}$$

So the ratio is $\dfrac{3}{2} : 1 = \mathbf{3 : 2}$.

Now, using the section formula for y-coordinate:

$$y = \frac{k(3) + 1(2)}{k + 1} = \frac{3\left(\frac{3}{2}\right) + 2}{\frac{3}{2} + 1} = \frac{\frac{9}{2} + 2}{\frac{5}{2}} = \frac{\frac{13}{2}}{\frac{5}{2}} = \mathbf{\frac{13}{5}}$$

The ratio is 3 : 2 and $y = \dfrac{13}{5}$.

Source: Chapter 7, Section 7.3 — Section Formula

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• Key formula used: Section formula — if a point divides $(x_1,y_1)$ to $(x_2,y_2)$ in ratio $k:1$, then $x = \dfrac{kx_2 + x_1}{k+1}$.
• Always assume ratio $k:1$ (not $m:n$) to reduce unknowns to one.
• First find $k$ using the given x-coordinate, then substitute that $k$ to find $y$.
• Examiners award marks for: setting up the equation (1 mark), finding the ratio (1 mark), and finding $y$ (1 mark). Show all steps clearly.