Question

In what ratio does the line x-y-2 = 0 divide the linesegment joining the points A(3, -1) and B(8, 9) ?

Answer 1

Let the line $x-y–2=0$ divide the line segment joining the points A(3,-1) and B(8,9) in the ratio k:1 at P

Then, by section formula the coordinates of P are

$x=(\frac{m{x}_2+n{x}_1}{m+n},y=\frac{m{y}_2+n{y}_1}{m+n})$

P = $x=\frac{8k+3}{k+1},y=\frac{9k-1}{k+1}$

Since, P lies on the line $x-y-2=0$, we have.

$\left(\frac{8k+3}{k+1}\right)-\left(\frac{9k-1}{k+1}\right)-2=0$

⇒ $8k+3–9k+1–2k-2=0$

⇒$8k–9k-2k+3+1–2=0$

⇒$-3k+2=0$

⇒ $-3k=-2$

⇒$k=\frac{2}{3}$

So, the required ratio is $\frac{2}{3}:1$, which is equal to 2 : 3