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(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

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

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(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

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