Question

A, B, C and D are four points on the same straight line. The distances between successive points are equal such that AB = BC = CD. If A is (1,-3), C is (4, a) and D is (b, 5), then the values of a and b are :

• A. $a=\frac{7}{3}andb=\frac{11}{2}$
• B. $a=\frac{4}{3}andb=\frac{11}{2}$
• C. $a=\frac{4}{3}andb=\frac{7}{2}$
• D. $a=\frac{7}{3}andb=\frac{7}{2}$

Answer 1

The correct option is A $a=\frac{7}{3}andb=\frac{11}{2}$


We know that AB = BC = CD. Thus the points B and C trisect AB.
$\therefore$ B and C are the points of trisection of AD.

The coordinates of the point that divides a line segment, joining the points $(x_1,y_1)and(x_2,y_2)$ in the ratio m : n are

$(\frac{n\times x_1+m\times x_2}{m+n},\frac{n\times y_1+m\times y_2}{n+m})$


Now, C (4, a) divides A (1, -3) and D (b, 5) internally in the ratio 2 : 1.

Applying section formula, we get,

$4=\frac{2\left(b\right)+1\left(1\right)}{2+1}$

$\Rightarrow 12=2b+1$

$\Rightarrow b=\frac{11}{2}$

and

$a=\frac{2\left(5\right)+1(-3)}{2+1}$

$\Rightarrow a=\frac{10-3}{3}$

$\Rightarrow a=\frac{7}{3}$