Through the point P(4,1) a line is drawn to meet the line 3x-y=0 where PQ=11÷2√2 .
Determine equation of line.
Answers
Let the co-ordinates of point Q be (a, b).
Since the point Q lies on the line 3x + y = 0, we have: 3a + b = 0 .......(equation 1)
Now, given the distance between points P(4,1) and Q(a,b) is 11/2√2
Therefore,
√[(a-4)^2 + (b-1)^2] = 11/2√2
Squaring both sides,
(a-4)^2 + (b-1)^2 = (11/2√2)^2
Expanding, we get
a^2 -8a + 16 + (b^2 -2b + 1) = 121/8.......(equation 2)
From equation 1, b = -3a
Substituting this value in equation 2, we get
a^2 -8a + 16 + (-3a)^2 - 2(-3a) + 1 = 121/8
a^2 -8a + 16 + 9a^2 + 6a + 1 = 121/8
10a^2 -2a + 17 = 121/8
Subtracting 121/8 from both sides,
10a^2 -2a + 15/8 = 0
Multiplying by 8 throughout,
80a^2 - 16a + 15 = 0
Discriminant of this quadratic is
16^2 - 4 x 80 x 15 = 256-4800 = -4544 which is negative.
Therefore, the given quadratic has no real solutions. Therefore, there are no real values of a and b, which means a line with given conditions cannot be drawn in reality.
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