Question

Find the value of $k$ if $A\left( 8,1 \right),B\left( k,-4 \right),C\left( 2,-5 \right)$ are collinear:
A.2
B.3
C.4
D.5

Answer 1

Hint: If three points are collinear then they must lie on the same plane. Then the area of the triangle formed by those three points is zero. The area of triangle formed by the points $\left( {{x}_{1,}}{{y}_{1}} \right),\left( {{x}_{2,}}{{y}_{2}} \right),\left( {{x}_{3,}}{{y}_{3}} \right)$ is $\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$. So these three points will be collinear if $\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]=0$. Use this result to solve such problems.

Complete step-by-step answer:
We have to find the value of $k$ if the points $A\left( 8,1 \right),B\left( k,-4 \right),C\left( 2,-5 \right)$ are collinear.
Here three points are given. We know that three points are collinear if they lie in the same line. So the area of the triangle formed by those three points is zero.
Also we know that the area of triangle formed by the points $\left( {{x}_{1,}}{{y}_{1}} \right),\left( {{x}_{2,}}{{y}_{2}} \right),\left( {{x}_{3,}}{{y}_{3}} \right)$ is $\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$.
Hence the points $\left( {{x}_{1,}}{{y}_{1}} \right),\left( {{x}_{2,}}{{y}_{2}} \right),\left( {{x}_{3,}}{{y}_{3}} \right)$ will be collinear if $\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]=0$.
Comparing the given points $A\left( 8,1 \right),B\left( k,-4 \right),C\left( 2,-5 \right)$ with the general points $\left( {{x}_{1,}}{{y}_{1}} \right),\left( {{x}_{2,}}{{y}_{2}} \right),\left( {{x}_{3,}}{{y}_{3}} \right)$we get ${{x}_{1}}=8,{{y}_{1}}=1;{{x}_{2}}=k,{{y}_{2}}=-4;{{x}_{3}}=2,{{y}_{3}}=-5$
Putting the values we get $\dfrac{1}{2}\left[ 8\left( -4-\left( -5 \right) \right)+k\left( -5-1 \right)+2\left( 1-\left( -4 \right) \right) \right]=0$.
Multiplying both side of the equation by$2$ we get $\left[ 8\left( -4-\left( -5 \right) \right)+k\left( -5-1 \right)+2\left( 1-\left( -4 \right) \right) \right]=0$,
Since $-\left( -x \right)=x$ so we have $\left[ 8\left( -4+5 \right)+k\left( -5-1 \right)+2\left( 1+4 \right) \right]=0$.
Calculating the portion lies in a round bracket we get $\left[ 8\left( 1 \right)+k\left( -6 \right)+2\left( 5 \right) \right]=0$.
Further simplifying the equation $8-6k+10=0$.
Adding the integers of the left side we get $18-6k=0$.
Add both sides of the equation by $-18$ we get $-18+18-6k=0-18$.
Further simplifying $-6k=-18$.
Cancelling negative signs from both sides we get $6k=18$.
Divide both sides of the above equation we have $\dfrac{6}{6}k=\dfrac{18}{6}$.
Simplifying we get $k=3$.
Therefore the value of $k$ is $3$.
Hence the correct answer is option (B).

Note: In this problem we have to use the result that the area of the triangle formed by three collinear points is zero. So while calculating students must be aware of small mistakes. In this problem while simplifying the terms inside brackets we need to go step by step to avoid mistakes. We can consider a line with endpoints A and C and point B on it. Then for these points to be collinear, they must satisfy the criterion given by length AB + length BC = length CA.