Question

The points $\left( a,1 \right),\left( b,2 \right)$ and $\left( c,3 \right)$ are collinear. Which of the following is true?
(a) $c-b=c-a$
(b) $c-b=b-a$
(c) $c-a=a-b$
(d) $c-a=b-a$

Answer 1

Hint: In this question, we will take the general form of slope intercept on line and put given points on that line and solve to get the required condition.
Complete step-by-step answer:
Since all the given three points $\left( a,1 \right),\left( b,2 \right)$ and $\left( c,3 \right)$ are collinear, therefore there will be a line which will pass through all three points. Let this line on which all these three points will lie be $y=mx+d$, when, m is slope of line and d is y-intercept.
Now, $\left( a,1 \right)$ lies on this line, so it this point will satisfy this equation, therefore,
$1=ma+d$
Subtracting ma from both sides, we get,
$d=1-ma.........(i)$
Also, $\left( b,2 \right)$ lies on this line, so it will also satisfy this equation.
Therefore, $2=mb+d$
Putting value of equation (i) here we get,
$2=mb+1-ma$
Subtracting 1 from both sides, we get,
$mb-ma=1$
Taking m common, we get,
$m\left( b-a \right)=1$
Dividing b-a from both sides, we get,
$m=\dfrac{1}{b-a}.........(ii)$
Now, $\left( c,3 \right)$ also lies on the line, so this will also satisfy the equation of line.
Therefore, $3=mc+d$
Putting value of equation (i) and (ii) here,
$\begin{aligned} & 3=\left( \dfrac{1}{b-a} \right)c+1-ma \\\ & \Rightarrow 3=\dfrac{c}{b-a}+1-ma \\\ \end{aligned}$
Putting value of (ii) again, we get,

& 3=\dfrac{c}{b-a}+1-\dfrac{1\times a}{b-a} \\\ & \Rightarrow 3=\dfrac{c-a}{b-a}+1 \\\ \end{aligned}$$ Taking LCM, we get, $\begin{aligned} & 3=\dfrac{c-a+b-a}{b-a} \\\ & \Rightarrow 3=\dfrac{c+b-2a}{b-a} \\\ \end{aligned}$ Multiplying both sides of equation with b-a, we get, $\begin{aligned} & 3\left(b -a \right)=c+b-2a \\\ & \Rightarrow 3b-3a=c+b-2a \\\ \end{aligned}$ Adding $3a-b$, both sides of the equation, we get, $\begin{aligned} & 3b-b=c-2a+3a \\\ & \Rightarrow 2b=c+a \\\ \end{aligned}$ Subtracting b+a from both sides we get, $\begin{aligned} & 2b-b-a=c+a-b-a \\\ & \Rightarrow b-a=c-b \\\ & \Rightarrow c-b=b-a \\\ \end{aligned}$ Hence, the correct answer is option (b). Note: In this question, avoid using the distance formula of two-dimensional geometry as here we have 3 unknown terms which will make calculation difficult.