Find the general form of the equation of the straight line passing through: the point (2,5) and parallel to the line 3x-y-7+0 the origin (which is (0,0)) and parallel to the line 4x-5y+6+0 the point (-2,3) and perpendicular to the line 2x+y=9 the point (3,-4) and perpendicular to the line x-y+5=0 show that the two lines x-2y+7+0 and 2x+y-16=0 are perpendicular
y = 3x -1 5y = 4x y = x/2 + 4 y =-x -1
@lochana -it is better to explain how you got these results'. Please read terms and conditions
i got a problem with another question (-2,7)m=1/2
i guess you mean you want an equation of the line passing throug (-2,7) with a slope m of 1/22
yea but the m=\[\frac{ 1}{ 2 }\]
the general form is y - y1 = m(x - x1) where m = slope and (x1,y1) is a point on the line so plugging in the values: y - 7 = (1/2)( x - (-2)) y - 7 = (1/2) ( x + 2) multiply through by 2 to get rid of the 1/2 2y - 14 = x + 2 2y = x + 16
its worth comitting the general form y-y1 = m(x - x1) to memory
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