Question

The value of m for which coordinates (3,5), (m,6) and $(\frac{1}{2}$, $\frac{15}{2})$ are collinear.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B

2

Let P(m, 6) divides the line segment AB joining A(3, 5), B( $\frac{1}{2}$, $\frac{15}{2})$ in the ratio k : 1.

Applying section formula, we get the co-ordinates of P;

$\left(\frac{\frac{1}{2}k+3\times 1}{k+1}\right)$, $\left(\frac{\frac{15}{2}k+5\times 1}{k+1}\right)$=

$(\frac{k+6}{2(k+1)}$, $\frac{15k+10}{2(k+1)})$

But P(m, 6) = P $(\frac{k+6}{2(k+1)}$, $\frac{15k+10}{2(k+1)})$ ⇒ m = $\frac{k+6}{2(k+1)}$ and also $\frac{15k+10}{2(k+1)}$ = 6

⇒ $\frac{15k+10}{2(k+1)}$ = 6 ⇒15k +10 = 12(k+1) ⇒ 15k +10 = 12k + 12

⇒ 15k - 12k = 12-10 ⇒ 3k = 2 ⇒ k = $\frac{2}{3}$

Putting k = $\frac{2}{3}$ in the equation m = $\frac{k+6}{2(k+1)}$

we get: m = $\frac{(\frac{2}{3}+6)}{2(\frac{2}{3}+1)}$ = $\frac{\left(\frac{2+18}{3}\right)}{2\left(\frac{2+3}{3}\right)}$ = $\frac{20}{3}$ × $\frac{3}{10}$ = $\frac{20}{10}$ ( ∵ k = $\frac{2}{3}$)

m = $\frac{10\times 2}{10}$ = 2

Required value of m is 2 ⇒ m = 2