Pair of Straight Lines Questions :
Here we are going to see some practice questions on pair of straight lines.
Question 1 :
Prove that one of the straight lines given by ax2 + 2hxy + by2 = 0 will bisect the angle between the co-ordinate axes if (a + b)2 = 4h2
Given that :
(a + b)2 = 4h2
Taking square roots on both sides, we get
a + b = 2 h
By applying the value of 2h = (a + b) in the given equation, we get
ax2 + (a + b)xy + by2 = 0
ax2 + a xy + bxy + by2 = 0
ax(x + y) + by(x + y) = 0
(ax + by) (x + y) = 0
by = -ax and y = -x
Now let us consider the second equation y = -x.
Slope = -1
m = tan θ = -1
θ = 135
Question 2 :
If the pair of straight lines x2 − 2kxy − y2 = 0 bisect the angle between the pair of straight lines x2 − 2lxy − y2 = 0, Show that the later pair also bisects the angle between the former.
x2 − 2kxy − y2 = 0
Equation of angle bisector :
x2 - y2/a - b = xy/h
a = 1, 2h = -2k ==> h = -k, b = -1
x2 - y2/(1 + 1) = xy/(-k)
x2 - y2/2 = -xy/k
(x2 - y2) k = -2xy
kx2 + 2xy - ky2 = 0 -------(1)
x2 − 2lxy − y2 = 0 -------(2)
is the equation of the bisector of the angle between the same lines(given)
From (1) and (2) by comparing the coefficients we get
k/1 = 2/(-2l) = -k/(-1)
k/1 = -1/l = k
-1 = kl
Question 3 :
Prove that the straight lines joining the origin to the points of intersection of 3x2 + 5xy − 3y2 + 2x + 3y = 0 and 3x − 2y − 1 = 0 are at right angles.
3x − 2y = 1
3x2 + 5xy − 3y2 + (2x + 3y) (1) = 0
3x2 + 5xy − 3y2 + (2x + 3y) (3x − 2y) = 0
3x2 + 5xy − 3y2 + (6x2 - 4xy + 9xy - 6y2) = 0
9x2 + 10xy − 9y2 = 0
a = 9 and b = -9
a + b = 9 + (-9) = 0
Hence they are perpendicular.
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