The abscissae of two points P and Q are the r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the coordinates of points $P$ and $Q$ be $(x_1, y_1)$ and $(x_2, y_2)$ respectively.Given that $x_1$ and $x_2$ are the roots of $2x^2 + 4x - 7

Vedclass · Accepted Answer

$(-1, 2)$

Vedclass · Answer

(A) Let the coordinates of points $P$ and $Q$ be $(x_1, y_1)$ and $(x_2, y_2)$ respectively.Given that $x_1$ and $x_2$ are the roots of $2x^2 + 4x - 7 = 0$,the sum of the roots is $x_1 + x_2 = -\frac{4}{2} = -2$.Given that $y_1$ and $y_2$ are the roots of $3x^2 - 12x - 1 = 0$,the sum of the roots is $y_1 + y_2 = -\frac{-12}{3} = 4$.The centre of the circle with $PQ$ as a diameter is the midpoint of $PQ$,which is given by $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$.Substituting the values,we get $(\frac{-2}{2}, \frac{4}{2}) = (-1, 2)$.Thus,the correct option is $A$.

The abscissae of two points $P$ and $Q$ are the roots of the equation $2x^2 + 4x - 7 = 0$ and their ordinates are the roots of the equation $3x^2 - 12x - 1 = 0$. Then the centre of the circle with $PQ$ as a diameter is

Similar Questions

The sides of a square are $x = 2, x = 3, y = 1$ and $y = 2$. Find the equation of the circle drawn with the diagonal of the square as its diameter.

The parametric equations of the circle $2x^2 + 2y^2 = 9$ are

Find the equation of the circle with centre $(-a, -b)$ and radius $\sqrt{a^{2}-b^{2}}$.

The equation of the circle passing through the points $(1, \sqrt{2})$,$(7, \sqrt{2})$,and $(1, 3)$ is:

If $z=x+iy$,then the centre of the circle $\left|\frac{z-3}{z-2i}\right|=2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module