If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is , then the quadratic equation whose roots are and is [2023]

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Solution

Q.
If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is $(α, β)$ , then the quadratic equation whose roots are $α + 4 β$ and $4 α + β$ is [2023]

1 $x^{2} - 19 x + 90 = 0$

2 $x^{2} - 20 x + 99 = 0$

3 $x^{2} - 18 x + 80 = 0$

4 $x^{2} - 22 x + 120 = 0$

Ans.
(2)

Let A(2, 3), B(1, 2), and C(3, 1) be the vertices of a triangle.

$Now, B C = \sqrt{{(3 - 1)}^{2} + {(1 - 2)}^{2}} and A C = \sqrt{{(3 - 2)}^{2} + {(1 - 3)}^{2}}$

$∴ ∆ A B C is an isosceles triangle.$

$∴ The median and altitude passing through (α, β) are the same.$

$Now, slope of A B = 1$

$∴ slope of C D = - 1$

$∴ Equation of line passing through (a, b) and having slope - 1 is given by a + b = c, where c is some constant.$

$Now, line a + b = c will pass through the midpoint of A B as ∆ A B C is isosceles.$

$\Rightarrow \frac{3}{2} + \frac{5}{2} = c = 4 \Rightarrow α + β = 4 ...(i)$

$Now, we need to find the equation where roots are α + 4 β and 4 α + β .$

$Sum of roots = (α + 4 β) + (4 α + β) = 5 α + 5 β = 5 (α + β)$

$= 5 \times 4 [Using (i)]$

$= 20$

$From the options, we can see that the equation x^{2} - 20 x + 99 = 0 has sum of roots as 20 .$

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