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Question: Show that the points \(\left( 3,3 \right),\left( h,0 \right)\) and \(\left( 0,k \right)\) are collin...

Show that the points (3,3),(h,0)\left( 3,3 \right),\left( h,0 \right) and (0,k)\left( 0,k \right) are collinear, if 1h+1k=13\dfrac{1}{h}+\dfrac{1}{k}=\dfrac{1}{3}

Explanation

Solution

To show that the given points are collinear if the condition is satisfied we will use Slope formula. Firstly we will find the Slope of three pairs of points and put them all equal to each other. Then we will check that the terms we wanted are in which of the two among the three of them and take them. Finally we will solve them and show the condition as true.

Complete step by step answer:
It is given to us that for the points (3,3),(h,0)\left( 3,3 \right),\left( h,0 \right) and (0,k)\left( 0,k \right) below condition should satisfy,
1h+1k=13\dfrac{1}{h}+\dfrac{1}{k}=\dfrac{1}{3}
So we will start by finding the slope of three pair of points which are:
(3,3),(h,0)\left( 3,3 \right),\left( h,0 \right)……(1)\left( 1 \right)
(h,0),(0,k)\left( h,0 \right),\left( 0,k \right)……(2)\left( 2 \right)
(0,k),(3,3)\left( 0,k \right),\left( 3,3 \right)….(3)\left( 3 \right)
Now we will find the slope of all the equation above as follows:
Slope =y2−y1x2−x1=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}
For equation (1) we have,
⇒0−3h−3 ⇒−3h−3 \begin{aligned} & \Rightarrow \dfrac{0-3}{h-3} \\\ & \Rightarrow \dfrac{-3}{h-3} \\\ \end{aligned}
Slope of equation (1) is −3h−3\dfrac{-3}{h-3}….(4)\left( 4 \right)
For equation (2) we have,
⇒k−00−h ⇒k−h \begin{aligned} & \Rightarrow \dfrac{k-0}{0-h} \\\ & \Rightarrow \dfrac{k}{-h} \\\ \end{aligned}
Slope of equation (1) is k−h\dfrac{k}{-h}….(5)\left( 5 \right)
For equation (3) we have,
⇒3−k3−0 ⇒3−k3 \begin{aligned} & \Rightarrow \dfrac{3-k}{3-0} \\\ & \Rightarrow \dfrac{3-k}{3} \\\ \end{aligned}
Slope of equation (1) is 3−k3\dfrac{3-k}{3}….(6)\left( 6 \right)
Now as we know for three points to be collinear at-least slope of two pairs should be equal so we will put equation (4), (5) and (6) equal as follows:
−3h−3=k−h=3−k3\dfrac{-3}{h-3}=\dfrac{k}{-h}=\dfrac{3-k}{3}
Now taking first two terms we will simplify them as follows:
⇒−3h−3=k−h ⇒−3×−h=k(h−3) ⇒3h=hk−3k ⇒3h+3k=hk \begin{aligned} & \Rightarrow \dfrac{-3}{h-3}=\dfrac{k}{-h} \\\ & \Rightarrow -3\times -h=k\left( h-3 \right) \\\ & \Rightarrow 3h=hk-3k \\\ & \Rightarrow 3h+3k=hk \\\ \end{aligned}
Now divide both sides by 3hk3hk as follows:
⇒3h3hk+3k3hk=hk3hk ⇒1k+1h=13 \begin{aligned} & \Rightarrow \dfrac{3h}{3hk}+\dfrac{3k}{3hk}=\dfrac{hk}{3hk} \\\ & \Rightarrow \dfrac{1}{k}+\dfrac{1}{h}=\dfrac{1}{3} \\\ \end{aligned}
Hence proved
Hence points (3,3),(h,0)\left( 3,3 \right),\left( h,0 \right) and (0,k)\left( 0,k \right) are collinear, if 1h+1k=13\dfrac{1}{h}+\dfrac{1}{k}=\dfrac{1}{3}

Note: If two or more points lie on the same straight line they are known as collinear. There are two methods to find whether the points are collinear or not: the Slope Formula method and Area of Triangle method. The Slope Formula method states that if the slope of two pairs of the points is same then the points are collinear. Area of triangle method states that if the area of the triangle formed by the three points is zero the points are collinear.