If the equations 2x - 3y + 3 = 0,2x + y + 1 = | vedclass

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

Question

(C) Let the lines be $L_1: 2x - 3y + 3 = 0$,$L_2: 2x + y + 1 = 0$,and $L_3: 6x + 4y + 1 = 0$. First,check the slopes: Slope of $L_1$ is $m_1 = 2/3$. S

Vedclass · Accepted Answer

$8x^2 + 8y^2 + 18x - 20y + 17 = 0$

Vedclass · Answer

(C) Let the lines be $L_1: 2x - 3y + 3 = 0$,$L_2: 2x + y + 1 = 0$,and $L_3: 6x + 4y + 1 = 0$. First,check the slopes: Slope of $L_1$ is $m_1 = 2/3$. Slope of $L_3$ is $m_3 = -6/4 = -3/2$. Since $m_1 	imes m_3 = (2/3) 	imes (-3/2) = -1$,the lines $L_1$ and $L_3$ are perpendicular. Thus,the triangle is a right-angled triangle with the hypotenuse being the line $L_2$. The circle passing through the vertices of a right-angled triangle has the hypotenuse as its diameter. Find the vertices of the triangle: Intersection of $L_1$ and $L_2$: $2x - 3y + 3 = 0$ and $2x + y + 1 = 0$. Subtracting gives $-4y + 2 = 0 \Rightarrow y = 1/2$. Then $2x + 1/2 + 1 = 0$ $\Rightarrow 2x = -3/2$ $\Rightarrow x = -3/4$. Vertex $V_1 = (-3/4, 1/2)$. Intersection of $L_2$ and $L_3$: $2x + y + 1 = 0$ and $6x + 4y + 1 = 0$. Multiply first by $4$: $8x + 4y + 4 = 0$. Subtracting gives $2x + 3 = 0 \Rightarrow x = -3/2$. Then $2(-3/2) + y + 1 = 0$ $\Rightarrow -3 + y + 1 = 0$ $\Rightarrow y = 2$. Vertex $V_2 = (-3/2, 2)$. The diameter endpoints are $(-3/4, 1/2)$ and $(-3/2, 2)$. The equation of the circle is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$. $(x + 3/4)(x + 3/2) + (y - 1/2)(y - 2) = 0$. $x^2 + (3/2 + 3/4)x + 9/8 + y^2 - (2 + 1/2)y + 1 = 0$. $x^2 + (9/4)x + y^2 - (5/2)y + 17/8 = 0$. Multiplying by $8$: $8x^2 + 8y^2 + 18x - 20y + 17 = 0$.

If the equations $2x - 3y + 3 = 0$,$2x + y + 1 = 0$,and $6x + 4y + 1 = 0$ represent the sides of a triangle,then the equation of the circle passing through the vertices of this triangle is

Similar Questions

If in two circles,arcs of the same length subtend angles $60^{\circ}$ and $75^{\circ}$ at the centre,find the ratio of their radii.

If one end of a diameter of the circle $x^2 + y^2 - 4x - 6y + 11 = 0$ is $(3, 4)$,then the other end is

Let $S$ be the circumcircle of the triangle formed by the line $x-2y-4=0$ with the coordinate axes. If $P(-2, -4)$ is a point in the plane of the circle $S$ and $Q$ is a point on $S$ such that the distance between $P$ and $Q$ is the least,then $PQ=$

Suppose a circle passes through $(0, a)$ and $(b, h)$ having its centre at $(c, 0)$. Then the value of $c$ is

Let $L_1$ be a straight line passing through the origin and $L_2$ be the straight line $x+y=1$. If the intercepts made by the circle $x^2+y^2-x+3y=0$ on $L_1$ and $L_2$ are equal,then which of the following equations represent $L_1$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module