Find the equation of the straight line joining (–2, 3 – 6p) to (–2 + 2p, 3).
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Find the equations to the straight lines which go through the origin and trisect the portion of the straight line 3x+y=12 which is intercepted between the axes of coordinates.
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3x+y=12
x=0⇒y=12
y=0⇒x=4
So the line is interecepted between A(4,0) ans B(0,12)
Let the points P and Q trisect the portion of the line AB
Now Q divides AB in 2:1
⇒Q(
2+1
2(0)+1(4)
,
2+1
2(12)+1(0)
)
Q(
3
4
,
3
24
)
So the equation of line joining origin and Q is
y−0=
⎝
⎜
⎜
⎛
3
4
−0
3
24
−0
⎠
⎟
⎟
⎞
(x−0)
y=
4
24
x
y=6x
Now P divides AQ in 1:1 , so P is the mid point of AQ
⇒P
⎝
⎜
⎜
⎛
2
4+
3
4
,
2
0+
3
24
⎠
⎟
⎟
⎞
P(
3
8
,
3
12
)
So, the equation of line joining origin to P is
y−0=
⎝
⎜
⎜
⎛
3
8
−0
3
12
−0
⎠
⎟
⎟
⎞
(x−0)
y=
8
12
x
2y=3x
So, the desired equations are y=6x and 2y=3x.
The answer
y=3x+9-6p