1.find the equation of the line passing through (4,-3) parallel through 2x+54+y=0
2.if the area of triangle formed with the vertices (t,2t)(-2,6)(3,1) is 5 square units find if A =
(-1,5)

2
If the area of triangle formed with the vertices (t,2t)(-2,6)(3,1)is 5² units find t if A =(-5,5)
write , if A =(-1,5) then what will be the area? and now ur question is complete
i gave u the hinds...u can try 
sry you are write A=(-1,5)
Equation is 2x+5y+4=0

Answers

  • qais
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2015-11-29T22:31:20+05:30

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1.since line is parallel to 2x+54+y=0, so the equation will be 2x+y+c=0
Now point (4,-3) lie in this line, 
so,c= -5
hence equation of line is 2x+y-5=0
2. we can use distance formula to find the length of sides and then
area= 1/2(base×height)
1 3 1
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2015-11-30T17:43:51+05:30

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1. equation given
   2x-5y +4 =0
a=2 ,b=-5 , c=4 
we know that equation of any line perpendicular to ax+by+c =0 is bx-ay +k =0
-5x -2y +k =0
passes through 
(4 ,-3)
-5(4)-2(-7)+k=0
-20+6+k =0
k =14 
required perpendicular line is -5x-2y+14 =0

2........
1/2 Ι x₁(y₂ - y₃)+x₂(y₃y₁)+x₃ (y₁y₂) Ι
(t,2t)        (-2,6)       (3,1)
 x₁,y₁       x₂,y₂       x₃,y₃

1/2 l t (6-1) +(-2)(1-2)+3(2t-6) l =5
1/2 l 6t -1t -2 +4t +6t -18 l =5
1/2 l 15t -20 l =5
15t -20 =5 x 2
15t=10 +20 
15t =30 
t =30/15 =2 
t =2

2 5 2