Question

Using the section formula, prove that the three points (-4, 6), (2, 4) and (14, 0) are collinear.

Answer 1

HINT: In this question, we will first let any one point divide the line that joins the other two points in k: 1 ratio.
If a real value of k is obtained, then it would be proved that all the three points are collinear.
The formula that will be required to solve this question is as follows
If point P (x, y) lies on line segment AB and satisfies AP: PB=m: n, AP: PB=m: n, then we say that P divides AB internally in the ratio m: n. The point of division has the coordinates
$P=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)$

Complete step-by-step answer:

As mentioned in the question, we have to find whether the given three points are collinear or not.
Now, let point (2, 4) be dividing the line joining the other two points in k: 1 ratio.
On applying the section formula that is given in the hint, we get

& P=\left( \dfrac{k(14)+(-4)}{k+1},\dfrac{k(6)+0}{k+1} \right) \\\ & (x,y)=\left( \dfrac{14k-4}{k+1},\dfrac{6k}{k+1} \right) \\\ \end{aligned}$$ Now, we can write the equation of the line that joins (-4, 6) and (14, 0) as follows $$\begin{aligned} & \dfrac{y-0}{x-14}=\dfrac{0-6}{14-(-4)} \\\ & \dfrac{y}{x-14}=\dfrac{-6}{18} \\\ & \dfrac{y}{x-14}=\dfrac{-1}{3} \\\ & 3y=-x+14 \\\ \end{aligned}$$ Now, on putting the coordinates of the point P in this equation, we get $$\begin{aligned} & 3y=-x+14 \\\ & 3\dfrac{6k}{k+1}=\dfrac{4-14k}{k+1}+14 \\\ & 18k=4-14k+14k+14 \\\ & 18k=18 \\\ & k=1 \\\ \end{aligned}$$ Now, as we get a real value of k, hence, the three points lie on the same line that is, all the points are collinear. NOTE: Another method of solving this question, other than section formula, is that we can try to find the area of the triangle with the given coordinates as the vertices of a triangle and if the area of this triangle comes out to be zero, then the three points would be collinear.