Question

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

Answer 1

The Correct Option is B

The correct answer is option (B) : x²−20x+99=0

\((\frac{β−3}{α−2})(\frac{1}{−2})=−1\)
\(β−3=2α−4\)
\(β=2α−1\)
\(m_{AH}\times m_{BC}=−1\)
\(⇒(\frac{β−2}{α−1})(−2)=−1\)
\(⇒2β−4=α−1\)
\(⇒2(2α−1)=α+3\)
\(⇒3α=5\)
\(α=\frac{5}{3}, β=\frac{7}{3} ​\)
\(⇒H(\frac{5}{3}, \frac{7}{3})\)
\(α+4β= \frac{5}{3} + \frac{28}{3} ​= \frac{33}{3} ​=11\)
\(β+4α= \frac{7}{3} + \frac{20}{3} = \frac{27}{3} = 9\)
\(x^2  −20x+99=0\)

Answer 2

Step 1: Finding slopes of perpendicular altitudes: \[ m = \frac{-1}{2} = -\frac{1}{2} \] \[ \text{Here } m_{BH} \times m_{AC} = -1 \] \[ \beta - 3 = 2\alpha - 4 \] \[ \beta = 2\alpha - 1 \] Step 2: Using midpoint formula and relations, we find: \[ \alpha = \frac{5}{3}, \quad \beta = \frac{7}{3} \]

Step 3: Now, computing roots: \[ \alpha + 4\beta = 11, \quad 4\alpha + \beta = 9 \] \[ x^2 - (11 + 9)x + 11 \times 9 = 0 \] \[ x^2 - 20x + 99 = 0 \]