`2x-3y-3=0; (2x)/(3)+4y+(1)/(2)=0`

Find the point of intersection of the straight lines : (i) 2x+3y-6=0 , 3x-2y-6=0 (ii) x=0 , 2x-y+3=0 (iii) (x)/(3)-(y)/(4)=0 , (x)/(2)+(y)/(3)=1

2x-3y+1=0 x+4y+3=0

2x+3y=0 3x+4y=5

If (2, -2),(-1,2), (3,5) are the vertices of a triangle then the equation of the side not passing through (2,-2) is (A) x + 2y +1 = 0 (B) 2x - y- 4= 0 (C) X-3y -3= 0 (D) 3x – 4y + 11 = 0

Solve:(1)/(2(2x+3y))+(12)/(7(3x-2y))=(1)/(2)(7)/(2x+3y)+(4)/(3x-2y)=2 where 2x+3y!=0 and 3x-2y!=0

Match the following. {:("Line, parabola"," "," ""Pole"),("I". 2x-3y+4=0,y^(2)=4x," "(a)(4,6)),("II". 2x+3y-4=0,y^(2)=4x," "(b)(2,3)),("III". x-2y+4=0,y^(2)=6x," "(c)(-2,-3)),("IV". 3x+4y-4=0,x^(2)=4y," "(d)(-3//2,-1)):}

Let (2,3) be the focus of a parabola and x + y = 0 and x-y= 0 be its two tangents. Then equation of its directrix will be (a) 2x - 3y = 0 (b) 3x +4y = 0 (c) x +y = 5 (d) 12x -5y +1 = 0

The locus of the image of the point (2,3) in the line (x-2y+3)+lambda(2x-3y+4)=0 is (lambda in R)( a) x^(2)+y^(2)-3x-4y-4=0 (b) 2x^(2)+3y^(2)+2x+4y-7=0(c)x^(2)+y^(2)-2x-4y+4=0(d) none of these

x- y + 3 = 0, 2x + 3y - 4 = 0 .

2(2x+y+5)+3(x-3y-1)=0 2x-3y+1=0