Show that the pair of lines 3x^2-2xy-y^2=0 are parallel to the lines 3x^2-2xy-y^2-5x+y+2=0
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Answer:
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Step-by-step explanation:
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MATHS
Show that the following equation represents a pair of straight lines. Find the points of intersection and the acute angle between them.
(i) 3x
2
+7xy+2y
2
+5x+5y+2=0
(ii) 2x
2
−13xy−7y
2
+x+23y−6=0.
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ANSWER
for a general eq ax
2
+2hxy+by
2
+2gx+2fy+c=0 will represents pair of straight line if:
a,b,h
=0 and h
2
−ab>or=0
for 3x
2
+7xy+2y
2
+5x+5y+2=0 a=3,h=7/2,b=2
h
2
−ab=49/4−6=25/4>0 (represents straight line)
for 2x
2
−13xy−7y
2
+x+23y−6=0
a=2,h=−13/2,b=−7
h
2
−ab=169/4+14=225/4>0 (represents straight line)
Points of intersection is α=
ab−b
2
hf−bg
and β=
ab−h
2
hg−af
after substituting values of a,b,h,g,f we will get the points of intersection for the given lines.
acute angle θ=cos
−1
(a−b)
2
+4h
2
∣a+b∣
again, after putting the values of a,b,h will get the acute angle.
Step-by-step explanation:
first process
given, the equation is
3x^2-2xy-y^2=0
3x^2-3xy + xy -y^2=0
3x(x-y) + y (x-2) =0
(3x-y)(x-2) =0
Either
3x-y =0....... (1)
x-2 =0........(2)
the parallel to the lines 3x^2-2xy-y^2-5x+y+2=0
3x^2-x(y + 5) -y^2+y+2=0
comparing with ax^2 + bx + c =0
a = 3 , b = (y + 5) and c = y - y^2 +2
