Logo
Jeeify
MCQMediumJEE 2023Slope & Angle Between Lines

JEE Mathematics 2023 Question with Solution

If the orthocentre of a triangle, whose vertices are (1,2)(1,2), (2,3)(2,3), and (3,1)(3,1) is (α,β)(\alpha, \beta), then the quadratic equation whose roots are α+4β\alpha + 4\beta and 4α+β4\alpha + \beta is:

  • A

    x2−19x+90=0x^2 - 19x + 90 = 0

  • B

    x2−18x+80=0x^2 - 18x + 80 = 0

  • C

    x2−22x+120=0x^2 - 22x + 120 = 0

  • D

    x2−20x+99=0x^2 - 20x + 99 = 0

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The vertices of the triangle are (1,2)(1,2), (2,3)(2,3) and (3,1)(3,1). Its orthocentre is H(α,β)H(\alpha, \beta).

Find: The quadratic equation whose roots are α+4β\alpha + 4\beta and 4α+β4\alpha + \beta.

From the solution working, use the fact that the altitude is perpendicular to the opposite side.

Triangle with vertices A(1,2), B(2,3), C(3,1) and orthocentre H(alpha,beta) marked inside, with slopes of sides and altitude relation shown.

Using perpendicular slopes,

(β−3α−2)(1−2)=−1\left(\frac{\beta-3}{\alpha-2}\right)\left(\frac{1}{-2}\right)=-1

so,

β−3=2α−4\beta-3=2\alpha-4 β=2α−1\beta=2\alpha-1

Also,

mAH×mBC=−1m_{AH}\times m_{BC}=-1 (β−2α−1)(−2)=−1\left(\frac{\beta-2}{\alpha-1}\right)(-2)=-1 2β−4=α−12\beta-4=\alpha-1

Substituting β=2α−1\beta=2\alpha-1,

2(2α−1)=α+32(2\alpha-1)=\alpha+3 3α=53\alpha=5 α=53,β=73\alpha=\frac{5}{3}, \quad \beta=\frac{7}{3}

Hence,

H(53,73)H\left(\frac{5}{3},\frac{7}{3}\right)

Now compute the required roots:

α+4β=53+283=333=11\alpha+4\beta=\frac{5}{3}+\frac{28}{3}=\frac{33}{3}=11 β+4α=73+203=273=9\beta+4\alpha=\frac{7}{3}+\frac{20}{3}=\frac{27}{3}=9

Therefore the quadratic equation with roots 1111 and 99 is

x2−(11+9)x+(11)(9)=0x^2-(11+9)x+(11)(9)=0 x2−20x+99=0x^2-20x+99=0

Therefore, the correct option is D.

Note: The solution labels the option as B, but its worked equation is x2−20x+99=0x^2-20x+99=0, which matches option D in the given options.

Using sum and product of roots

Given: α=53\alpha=\frac{5}{3} and β=73\beta=\frac{7}{3} from the orthocentre calculation.

Find: The quadratic polynomial formed by roots α+4β\alpha+4\beta and 4α+β4\alpha+\beta.

First root:

α+4β=53+4⋅73=11\alpha+4\beta=\frac{5}{3}+4\cdot\frac{7}{3}=11

Second root:

4α+β=4⋅53+73=94\alpha+\beta=4\cdot\frac{5}{3}+\frac{7}{3}=9

So the sum of roots is

11+9=2011+9=20

and the product of roots is

11×9=9911\times 9=99

Hence the required quadratic equation is

x2−20x+99=0x^2-20x+99=0

So the correct option is D.

Common mistakes

  • Using the orthocentre as the midpoint or centroid is incorrect. The orthocentre is the intersection point of altitudes, so perpendicular-slope relations must be used instead.

  • Taking the slope of a side incorrectly leads to wrong altitude equations. For example, for the side through (2,3)(2,3) and (3,1)(3,1), the slope must be computed carefully before applying the negative reciprocal.

  • Forming the quadratic as x2+(sum of roots)x+product of roots=0x^2+(\text{sum of roots})x+\text{product of roots}=0 is wrong. For roots r1r_1 and r2r_2, the correct form is x2−(r1+r2)x+r1r2=0x^2-(r_1+r_2)x+r_1r_2=0.

Practice more Slope & Angle Between Lines questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Slope & Angle Between Lines

Related questions