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Question: If points \(\left( 5,5 \right),\left( 10,k \right)\) and \(\left( -5,1 \right)\) are collinear, then...

If points (5,5),(10,k)\left( 5,5 \right),\left( 10,k \right) and (−5,1)\left( -5,1 \right) are collinear, then k=k=

A) 3

B) 5

C) 7

D) 9

Explanation

Solution

We can either use the formula of area of triangle as 12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]=0\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 or the formula which is given by $$\dfrac{1}{2}\left| \begin{matrix}

{{x}{1}} & {{y}{1}} & 1 \\

{{x}{2}} & ,{{y}{2}} & 1 \\

{{x}{3}} & {{y}{3}} & 1 \\

\end{matrix} \right|=0$$. We have kept the area of the triangle equal to zero because the three points are collinear, that is, these lie on the same line which results in not forming any triangle. This means the area should be 0. By this we will further solve the question to get the value of the variable which is kk here.

Complete step-by-step answer:
We use the trick which says that the area of the triangle can be carried out to find whether the points are collinear. The condition which is already given is that the points are collinear.

This means that the area of the triangle is zero. So now we will use the formula to find the area of the triangle. So, area of triangle is,

{{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & \,{{y}_{2}} & 1 \\\ {{x}_{3}} & {{y}_{3}} & 1 \\\ \end{matrix} \right|=0$$ Now substitute the values. We by substituting the points we can have, $$\begin{aligned} & \dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & \,{{y}_{2}} & 1 \\\ {{x}_{3}} & {{y}_{3}} & 1 \\\ \end{matrix} \right|=0 $$

\Rightarrow \dfrac{1}{2}\left| \begin{matrix}

5 & 5 & 1 \\

10 & ,k & 1 \\

-5 & 1 & 1 \\

\end{matrix} \right|=0 \\

\end{aligned}$$

⇒12[5(k−1)−5(10+5)+1(10+5k)]=0\Rightarrow \dfrac{1}{2}\left[ 5\left( k-1 \right)-5\left( 10+5 \right)+1\left( 10+5k \right) \right]=0

⇒12[5k−5−50−25+10+5k]=0\Rightarrow \dfrac{1}{2}\left[ 5k-5-50-25+10+5k \right]=0

⇒12[10k−5−50−25+10]=0\Rightarrow \dfrac{1}{2}\left[ 10k-5-50-25+10 \right]=0

⇒12[10k−70]=0\Rightarrow \dfrac{1}{2}\left[ 10k-70 \right]=0

⇒10k−70=0\Rightarrow 10k-70=0

Therefore the value of k is carried out as,

10k=7010k=70

⇒k=7010\Rightarrow k=\dfrac{70}{10}

⇒k=7\Rightarrow k=7

Hence the correct option is (C).

Note:

Alternatively we can find the slope by performing the following steps. The points are collinear if the slope of any pairs of two points out of three points are equal. Here it means that if we consider (5,5)\left( 5,5 \right) as A point, (10,k)\left( 10,k \right) as B point and (−5,1)\left( -5,1 \right) as C point also if we take the pairs AB, BC and CA together then Slope of AB = Slope of BC = Slope of CA

Now we find the slope of AB first with the points A (5,5)\left( 5,5 \right) and B (10,k)\left( 10,k \right). By using the formula m=y−y1y2−y1m=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}} and m=x−x1x2−x1m=\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}} we will the points and substitute the third point because the points are linear so the third point will satisfy the equation. Thus we will find the value of k. By using the formula m=y−y1y2−y1m=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}} we have m=y−5k−5m=\dfrac{y-5}{k-5}. And using the formula m=x−x1x2−x1m=\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}} we have m=x−510−5m=\dfrac{x-5}{10-5}. Equating m=y−5k−5m=\dfrac{y-5}{k-5} and m=x−510−5m=\dfrac{x-5}{10-5} we get

y−5k−5=x−510−5\dfrac{y-5}{k-5}=\dfrac{x-5}{10-5}

⇒y−5k−5=x−55\Rightarrow \dfrac{y-5}{k-5}=\dfrac{x-5}{5}

⇒5(y−5)=(x−5)(k−5)\Rightarrow 5\left( y-5 \right)=\left( x-5 \right)\left( k-5 \right)

Now we will substitute the third point (−5,1)\left( -5,1 \right) in equation 5(y−5)=(x−5)(k−5)5\left( y-5 \right)=\left( x-5 \right)\left( k-5 \right). Therefore, we get

5(y−5)=(x−5)(k−5)5\left( y-5 \right)=\left( x-5 \right)\left( k-5 \right)

⇒5((1)−5)=((−5)−5)(k−5)\Rightarrow 5\left( \left( 1 \right)-5 \right)=\left( \left( -5 \right)-5 \right)\left( k-5 \right)

⇒5(−4)=(−10)(k−5)\Rightarrow 5\left( -4 \right)=\left( -10 \right)\left( k-5 \right)

⇒−20=−10k+50\Rightarrow -20=-10k+50

⇒10k=50+20\Rightarrow 10k=50+20

⇒10k=70\Rightarrow 10k=70

⇒k=7\Rightarrow k=7

Hence, the correct option is (C).

We need to take off the second term which is y1{{y}_{1}} should be negative and then proceed to the respective calculation. While solving the question we need to take care about performing multiplications otherwise, our answer can be wrong.