Let the abscissae of the two points P and Q be the roots of 2x2 – rx + p = 0 and the ordinates of P and Q be the roots of x2 – sx – q = 0. If the equation of the circle described on PQ as diameter is 2(x2 + y2) – 11x – 14y – 22 = 0, then 2r + s – 2q + is equal to…..

P ( x 1 , y 1 ) Q ( x 2 , y 2 ) x 1 + x 2 = r 2 , x 1 x 2 = P 2 y 1 + y 2 = s , y 1 , y 2 = − 9 ( x − x 1 ) ( x − x 2 ) + ( y − y 1 ) ( y − y 2 ) = 0 2 ( x 2 + y 2 ) − r x − 2 s y + p − 2 q = 0 r = 11, s = 7, p – 2q = -22

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Maths NCERT Exemplar Solutions Class 11th Chapter Two Straight Lines Ncert Solutions Maths class 11th Maths Class 11th

Let the abscissae of the two points P and Q be the roots of 2x² – rx + p = 0 and the ordinates of P and Q be the roots of x² – sx – q = 0. If the equation of the circle described on PQ as diameter is 2(x² + y²) – 11x – 14y – 22 = 0, then 2r + s – 2q + is equal to…..

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Raj Pandey | Contributor-Level 9

4 weeks ago

$P (x_{1}, y_{1})$

$Q (x_{2}, y_{2})$

$x_{1} + x_{2} = \frac{r}{2}, x_{1} x_{2} = \frac{P}{2}$

$y_{1} + y_{2} = s, y_{1}, y_{2} = - 9$

$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$

$2 (x^{2} + y^{2}) - r x - 2 s y + p - 2 q = 0$

r = 11, s = 7, p – 2q = -22

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Let the abscissae of the two points P and Q be the roots of 2x² – rx + p = 0 and the ordinates of P and Q be the roots of x² – sx – q = 0. If the equation of the circle described on PQ as diameter is 2(x² + y²) – 11x – 14y – 22 = 0, then 2r + s – 2q + is equal to…..

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