A pair of straight lines x^2 - 8x + 12 = 0 and y^2 - 14y + 45 = 0 are forming a square. Co-ordinates

English

Gujarati

Hindi

9.Straight Line

normal

A pair of straight lines $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$ are forming a square. Co-ordinates of the centre of the circle inscribed in the square are

A

$(3, 6)$

B

$(4, 7)$

C

$(4, 8)$

D

none

Solution

$\quad x^{2}-8 x+12=0$

$\Rightarrow(x-6)(x-2)=0$

$y^{2}-14 y+45=0$

$(y-5)(y-9)=0$

Thes sides of square are

x=2, x=6, y=5, y=9

The centre of circle in square will be

$\left(\frac{2+6}{2}, \frac{5+9}{2}\right)=\left(\frac{8}{2}, \frac{14}{2}\right)$

$=(4,7)$

Standard 11

Mathematics

Share

Similar Questions

Without using distance formula, show that points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are vertices of a parallelogram.

medium

The area of triangle formed by the lines $x = 0,y = 0$ and $\frac{x}{a} + \frac{y}{b} = 1$, is

easy

In a triangle $ABC,$ side $AB$ has the equation $2 x + 3 y = 29$ and the side $AC$ has the equation , $x + 2 y = 16$ . If the mid – point of $BC$ is $(5, 6)$ then the equation of $BC$ is :

normal

A square of side a lies above the $x$ -axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha ,(0 < \alpha < \frac{\pi }{4})$ with the positive direction of $x$-axis. The equation of its diagonal not passing through the origin is

hard

(AIEEE-2003)

If the straight lines $x + 3y = 4,\,\,3x + y = 4$ and $x +y = 0$ form a triangle, then the triangle is

hard

(AIEEE-2012)