JEE Main 2021 (24 February Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: Let $a, b, c$ be in arithmetic progression. Let the centroid of the triangle with vertices $(a, c), (2, b)$ and $(a, b)$ be $(\frac{10}{3}, \frac{7}{3})$ . If $α, β$ are the roots of the equation $a x^{2} + b x + 1 = 0$ , then the value of $α^{2} + β^{2} - α β$ is:

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Look closely at the vertices

A (a, c)

B (2, b)

, and

C (a, b)

. Notice how the x-coordinates of

A

and

C

are identical, and the y-coordinates of

B

and

C

are also identical. This geometric configuration implies that the triangle is right-angled at

C

. While not strictly necessary for the centroid formula, visualizing this helps you understand the structure of the problem.

Hint 2

Hint 2: The Arsenal. You have the centroid

G = (\frac{10}{3}, \frac{7}{3})

. The centroid formula is the average of the vertices:

G = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})

. You will have two equations, one for x and one for y, but you will find yourself with three variables (

a, b, c

) and only two equations. The missing link is the Arithmetic Progression condition:

2 b = a + c

Hint 3

Hint 3: The Execution. Start by equating the x-coordinates:

\frac{a + 2 + a}{3} = \frac{10}{3}

. This will immediately give you the value of

a

. Once you have

a

, use the A.P. condition

2 b = a + c

and the y-coordinate equation

\frac{c + 2 b}{3} = \frac{7}{3}

to solve for

b

and

c

simultaneously.

Visualized Solution

The Geometric Setup

Vertices of the triangle: $A (a, c)$ , $B (2, b)$ , and $C (a, b)$ .
Notice that $A$ and $C$ share the same x-coordinate ( $a$ ).
$B$ and $C$ share the same y-coordinate ( $b$ ).
This forms a right-angled triangle at $C$ .

The Centroid Formula

The centroid $G$ of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ is given by:
$G = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})$
We are given $G = (\frac{10}{3}, \frac{7}{3})$ .

Applying the Centroid Formula

Substitute the vertices $A (a, c)$ , $B (2, b)$ , $C (a, b)$ into the formula:
$x$ -coordinate: $\frac{a + 2 + a}{3} = \frac{2 a + 2}{3}$
$y$ -coordinate: $\frac{c + b + b}{3} = \frac{c + 2 b}{3}$
Equating to given centroid: $(\frac{2 a + 2}{3}, \frac{c + 2 b}{3}) = (\frac{10}{3}, \frac{7}{3})$

Solving for $a$

Equating the $x$ -coordinates:
$\frac{2 a + 2}{3} = \frac{10}{3}$
$2 a + 2 = 10$
$2 a = 8 ⟹ a = 4$

Setting up the $y$ -coordinate Equation

Equating the $y$ -coordinates:
$\frac{c + 2 b}{3} = \frac{7}{3}$
$c + 2 b = 7$
We need another equation to solve for $b$ and $c$ .

The Arithmetic Progression Condition

The problem states that $a, b, c$ are in Arithmetic Progression (A.P.).
The condition for three terms to be in A.P. is:
$2 b = a + c$
We already found $a = 4$ , so $2 b = 4 + c$ .

Solving for $b$ and $c$

From the A.P. condition: $c = 2 b - 4$
Substitute this into our $y$ -coordinate equation: $c + 2 b = 7$
$(2 b - 4) + 2 b = 7$
$4 b = 11 ⟹ b = \frac{11}{4}$

The Quadratic Equation

We are given the quadratic equation: $a x^{2} + b x + 1 = 0$
Substitute the values we found: $a = 4$ and $b = \frac{11}{4}$
The equation becomes: $4 x^{2} + \frac{11}{4} x + 1 = 0$
Let's multiply by $4$ to clear the fraction: $16 x^{2} + 11 x + 4 = 0$

Sum and Product of Roots

For a quadratic equation $A x^{2} + B x + C = 0$ with roots $α, β$ :
Sum of roots: $α + β = - \frac{B}{A}$
Product of roots: $α β = \frac{C}{A}$
For our equation $16 x^{2} + 11 x + 4 = 0$ :
$α + β = - \frac{11}{16}$
$α β = \frac{4}{16} = \frac{1}{4}$

Rewriting the Target Expression

We need to find the value of: $α^{2} + β^{2} - α β$
We know the algebraic identity: $α^{2} + β^{2} = (α + β)^{2} - 2 α β$
Substitute this into the target expression:
$[(α + β)^{2} - 2 α β] - α β$
$= (α + β)^{2} - 3 α β$

Substituting the Values

Target: $(α + β)^{2} - 3 α β$
Substitute $α + β = - \frac{11}{16}$ and $α β = \frac{1}{4}$ :
$= (- \frac{11}{16})^{2} - 3 (\frac{1}{4})$

Final Calculation

Calculate the square: $(- \frac{11}{16})^{2} = \frac{121}{256}$
The expression becomes: $\frac{121}{256} - \frac{3}{4}$
Find a common denominator ( $256$ ):
$\frac{3}{4} = \frac{3 \times 64}{4 \times 64} = \frac{192}{256}$
Subtract: $\frac{121}{256} - \frac{192}{256} = \frac{121 - 192}{256}$

Conclusion

Final result: $\frac{121 - 192}{256} = - \frac{71}{256}$
Key Takeaway: Break down multi-concept problems. Use geometry (centroid) to find variables, then apply algebra (A.P. and quadratics).

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The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Solution Diagram

Analyzing the Setup

Imagine standing on a coordinate plane, looking at three points:

A (a, c)

B (2, b)

, and

C (a, b)

Notice how

A

and

C

share the same

x

-coordinate, while

B

and

C

share the same

y

-coordinate. This confirms that our triangle is a right-angled triangle, anchored at

C

The Centroid

Finding the Center of Mass

The centroid

G

is the average of the coordinates of the vertices. Given

G = (\frac{10}{3}, \frac{7}{3})

, we equate the

x

-coordinates:

\frac{a + 2 + a}{3} = \frac{10}{3}

This simplifies to

2 a + 2 = 10

, which yields $a = 4$ .

For the

y

-coordinates, we have:

\frac{c + b + b}{3} = \frac{7}{3} \Rightarrow c + 2 b = 7

Since

a, b, c

are in Arithmetic Progression (A.P.), we know that

2 b = a + c

. Substituting

a = 4

, we get

c = 2 b - 4

Substituting this into our

y

-coordinate equation:

(2 b - 4) + 2 b = 7 \Rightarrow 4 b = 11 \Rightarrow b = \frac{11}{4}

Using

c = 2 b - 4

, we find

c = 2 (\frac{11}{4}) - 4 = \frac{11}{2} - 4 = \frac{3}{2}

The Quadratic Realm

Vieta's Power

We now consider the quadratic equation

a x^{2} + b x + 1 = 0

. Substituting our values, we get:

4 x^{2} + \frac{11}{4} x + 1 = 0

Multiplying the entire equation by

4

to clear the fraction, we obtain:

16 x^{2} + 11 x + 4 = 0

We need to evaluate the expression

α^{2} + β^{2} - α β

. We use the algebraic identity:

α^{2} + β^{2} - α β = (α + β)^{2} - 3 α β

Final Calculation

From Vieta's formulas for the equation

16 x^{2} + 11 x + 4 = 0

, we identify:

α + β = - \frac{11}{16}, α β = \frac{4}{16} = \frac{1}{4}

Plugging these values into our identity:

(- \frac{11}{16})^{2} - 3 (\frac{1}{4}) = \frac{121}{256} - \frac{3}{4}

Converting to a common denominator:

\frac{121}{256} - \frac{192}{256} = - \frac{71}{256}

The final result of the expression is $- \frac{71}{256}$ .

Common Traps & FAQs

Wait, why do we use the identity

α^{2} + β^{2} - α β = (α + β)^{2} - 3 α β

instead of just finding the roots

α

and

β

That is a fantastic question! While you *could* use the quadratic formula to find the roots, it would involve calculating the discriminant and dealing with potentially messy radicals. In competitive exams like JEE, time is your most precious resource. Using Vieta's formulas—where the sum of roots

α + β = - \frac{B}{A}

and the product

α β = \frac{C}{A}

—allows you to manipulate the expression directly. It is faster, more elegant, and significantly reduces the chance of making a calculation error.

Is the triangle always right-angled in these types of problems?

Not always! In this specific problem, the coordinates were chosen such that the vertices share values, creating a right-angled triangle. This is a common technique in problem design to provide a subtle hint about the geometry. Always look for these patterns; they can simplify your visualization and help you verify your results.

Why is the final answer negative? Did I make a mistake?

Don't panic! A negative result is perfectly valid. The expression

α^{2} + β^{2} - α β

is not a squared term (which would always be positive), but rather a combination of terms. In this case, the product term

3 α β

was larger than the squared sum term

(α + β)^{2}

, leading to a negative value. Trust your algebra!

Select Answer:

Visualized Solution

The Geometric Setup

The Centroid Formula

Applying the Centroid Formula

Solving for $a$

Setting up the $y$ -coordinate Equation

The Arithmetic Progression Condition

Solving for $b$ and $c$

The Quadratic Equation

Sum and Product of Roots

Rewriting the Target Expression

Substituting the Values

Final Calculation

Conclusion

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Setup

The Centroid

The Quadratic Realm

Final Calculation

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Visualized Solution

The Geometric Setup

The Centroid Formula

Applying the Centroid Formula

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Setting up the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate Equation

The Arithmetic Progression Condition

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

The Quadratic Equation

Sum and Product of Roots

Rewriting the Target Expression

Substituting the Values

Final Calculation

Conclusion

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Setup

The Centroid

The Quadratic Realm

Final Calculation

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