Question

If $A(1,{{p}^{2}}),$ B(0,1) and C(p,0) are the coordinates of three points, then find the value of p for which the area of the triangle is minimum.
A. $-\dfrac{1}{\sqrt{3}}$
B. $\dfrac{1}{\sqrt{3}}$
C. $\dfrac{1}{\sqrt{2}}$ or $-\dfrac{1}{\sqrt{3}}$
D. None

Answer 1

To solve this question first we will find the area of triangle ABC whose coordinates are $A(1,{{p}^{2}}),$ B(0,1) and C(p,0) by determinant method as area of triangle whose coordinates are $A({{x}_{1}},{{y}_{1}}),$ $B({{x}_{2}},{{y}_{2}}),$and $C({{x}_{3}},{{y}_{3}})$, is equals to $\dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & {{y}_{2}} & 1 \\\ {{x}_{3}} & {{y}_{3}} & 1 \\\ \end{matrix} \right|$, then only minimum possible value for area of triangle will be zero, so we will put $\dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & {{y}_{2}} & 1 \\\ {{x}_{3}} & {{y}_{3}} & 1 \\\ \end{matrix} \right|=0$ and then we will see which value of p satisfies the condition.

Complete step by step answer:

Now, in this question we are provided by three points of the triangle ABC which are $A(1,{{p}^{2}}),$ B(0,1) and C(p,0).
Now, there various methods of finding area of triangle in various situations, we will use formula for finding triangle which is equals to $\dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & {{y}_{2}} & 1 \\\ {{x}_{3}} & {{y}_{3}} & 1 \\\ \end{matrix} \right|$ , where term ahead $\dfrac{1}{2}$ represents the determinant of matrix $\left( \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & {{y}_{2}} & 1 \\\ {{x}_{3}} & {{y}_{3}} & 1 \\\ \end{matrix} \right)$.
Now, value of determinant \left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & {{z}_{1}} \\\ {{x}_{2}} & {{y}_{2}} & {{z}_{2}} \\\ {{x}_{3}} & {{y}_{3}} & {{z}_{3}} \\\ \end{matrix} \right|=\left\\{ {{x}_{1}}({{z}_{3}}{{y}_{2}}-{{y}_{3}}{{z}_{2}})-{{x}_{2}}({{z}_{3}}{{y}_{1}}-{{y}_{3}}{{z}_{1}})+{{x}_{3}}({{z}_{2}}{{y}_{1}}-{{y}_{2}}{{z}_{1}}) \right\\}
So, area of triangle ABC $=\dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & {{y}_{2}} & 1 \\\ {{x}_{3}} & {{y}_{3}} & 1 \\\ \end{matrix} \right|$
Value of determinant =\left\\{ {{x}_{1}}(1\cdot {{y}_{2}}-{{y}_{3}}\cdot 1)-{{x}_{2}}(1\cdot {{y}_{1}}-{{y}_{3}}\cdot 1)+{{x}_{3}}(1\cdot {{y}_{1}}-{{y}_{2}}\cdot 1) \right\\}
Now, from question we can compare that, ${{x}_{1}}=1,{{x}_{2}}=0,{{x}_{3}}=p$ and ${{y}_{1}}={{p}^{2}},{{y}_{2}}=1,{{y}_{3}}=0$
So, =\left\\{ 1\cdot (1\cdot 1-0\cdot 1)-0\cdot (1\cdot {{p}^{2}}-0\cdot 1)+p\cdot (1\cdot {{p}^{2}}-1\cdot 1) \right\\}
On simplifying, we get
$={{p}^{3}}-p+1$
Now, we know that area is a scalar quantity and also area cannot be negative.
So, only minimum possible value for triangle ABC is 0
So, $\dfrac {1}{2}{{p}^{3}}-p+1=0$
Now, if $p=\dfrac{1}{\sqrt{3}}$ ,
${{\left( \dfrac{1}{\sqrt{3}} \right)}^{3}}-\dfrac{1}{\sqrt{3}}+1\ne 0$
So, option ( a ) is not correct.
If $p=-\dfrac{1}{\sqrt{3}}$
${{\left( -\dfrac{1}{\sqrt{3}} \right)}^{3}}+\dfrac{1}{\sqrt{3}}+1\ne 0$
So, option ( b ) is also incorrect.
If $p=\dfrac{1}{\sqrt{2}}$
${{\left( \dfrac{1}{\sqrt{2}} \right)}^{3}}+\dfrac{1}{\sqrt{2}}+1\ne 0$
So, option ( c ) is also incorrect.
Therefore, none of the options is correct.

So, the correct answer is “Option D”.

Note: We chose the determinant formula for finding area of triangle because we had the coordinates points. While solving the determinant always remember to put correct values from the questions else the solution will become incorrect and answer so. Try to discard options by putting them in equations and check which option satisfies the equation.