Question

a) In what ratio, the line segment joining (-3, -1) and (-8, -9) is divided at the point (-5, 3)?

b) For what value of $a$, the points (a, 1), (1, -1) and (11, 4) are collinear?

[6 MARKS]

Answer 1

Part a): 2 Marks
Part b): 4 Marks

a)



Let P(-5,3) divide the line segment joining A(-3,-1) and B(-8,-9) in the ratio k:1.

Using section formula, we have $-5=\frac{1\times (-3)+k\times (-8)}{k+1}\Rightarrow -5k-5=-3-8k\Rightarrow k=\frac{2}{3}$
So, required ratio is 2 : 3

b)
Points are (a, 1), (1, -1) and (11, 4)
Let point (1, -1) divide the line joining (a, 1) and (11, 4) in the ratio k:1.
Using section formula, we have $1=\frac{1\times a+k\times 11}{k+1}......\left(i\right)-1=\frac{1\times 1+k\times 4}{k+1}.....\left(ii\right)$
From (ii), we get $k=-\frac{2}{5}$
Putting $k=-\frac{2}{5}$ in (i), we get
$1=\frac{1\times a+k\times 11}{k+1}\Rightarrow k+1=a+11k\Rightarrow 1-a=10k\Rightarrow 1-a=10(-\frac{2}{5})\Rightarrow 1-a=-4\Rightarrow a=5$