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Straight Lines and Pair of Straight Lines question

2023 · 1 Feb · Shift 1 · Q33

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Straight Lines and Pair of Straight Lines
2023 · 1 Feb · Shift 1 · Q33

Straight Lines and Pair of Straight Lines question

2023 · 1 Feb · Shift 1 · Q33

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1

If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is

(\alpha,\beta)

, then the quadratic equation whose roots are

\alpha+4\beta

and

4\alpha+\beta

, is :

A
$x^2-20x+99=0$
B
$x^2-22x+120=0$
C
$x^2-19x+90=0$
D
$x^2-18x+80=0$

View written solutionFree

Correct answer: A

Given triangle vertices

Let $A(1,2),\quad B(2,3),\quad C(3,1).$ We need the orthocentre $H(\alpha,\beta)$ .

Find two altitudes

Altitude from $A$

Slope of $BC$ is $m_{BC}=\frac{1-3}{3-2}=-2.$ So the altitude from $A$ has slope $m=\frac{1}{2}$ since perpendicular slopes satisfy $m_1m_2=-1$ .

Equation through $A(1,2)$ : $y-2=\frac12(x-1).$ So $2y-4=x-1$ $x-2y+3=0.$

Altitude from $B$

Slope of $AC$ is $m_{AC}=\frac{1-2}{3-1}=-\frac12.$ So the altitude from $B$ has slope $2.$

Equation through $B(2,3)$ : $y-3=2(x-2)$ $y=2x-1.$

Find orthocentre by intersection of altitudes

Solve $y=2x-1$ and $x-2y+3=0.$

Substitute $y=2x-1$ into the second equation: $x-2(2x-1)+3=0$ $x-4x+2+3=0$ $-3x+5=0$ $x=\frac53.$ Then $y=2\cdot \frac53-1=\frac{10}{3}-1=\frac73.$

Hence, $\alpha=\frac53,\quad \beta=\frac73.$

Compute the required roots

The roots are $\alpha+4\beta=\frac53+4\cdot \frac73=\frac53+\frac{28}{3}=\frac{33}{3}=11,$

and $4\alpha+\beta=4\cdot \frac53+\frac73=\frac{20}{3}+\frac73=\frac{27}{3}=9.$

So the quadratic equation with roots $11$ and $9$ is $x^2-(11+9)x+(11)(9)=0$ $x^2-20x+99=0.$

Check options

A: $x^2-20x+99=0$ ✅
B: $x^2-22x+120=0$ ❌
C: $x^2-19x+90=0$ ❌
D: $x^2-18x+80=0$ ❌

Therefore, the correct option is A.

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