Show that the pair of lines joining the origin to the points of intersection of the line
3x - y = 2 with the pair of lines 7x^2 - 4xy + 8y^2 + 2x - 4y - 8 = 0 are equally inclined with
the co-ordinate axes.
Answers
Given : pair of lines joining the origin to the points of intersection of the line 3x - y = 2 with the pair of lines 7x^2 - 4xy + 8y^2 + 2x - 4y - 8 = 0
To Find : Show that they are equally inclined with the co-ordinate axes.
Solution:
7x² - 4xy + 8y² + 2x - 4y - 8 = 0
3x - y = 2
=> y = 3x - 2
7x² - 4x(3x - 2) + 8(3x - 2)² + 2x - 4(3x - 2) - 8 = 0
=> 7x² - 12x² + 8x + 8(9x² -12x +4) + 2x - 12x + 8 - 8 = 0
=> -5x² - 2x + 72x² - 96x + 32 = 0
=> 67x² - 98x + 32 = 0
x = (98 ± √1028 ) /2(67)
=> x = (49 ± √257 ) / 67
x = (49 + √257 ) / 67
y = 3x - 2 => y = (13 + 3 √257 ) / 67
Angle with x axis
= Tan⁻¹| (13 + 3 √257 )/ (49 + √257 ) |
x = (49 - √257 ) / 67
y = 3x - 2 => y = (13 - 3 √257 ) / 67
Angle with y axis
= Tan⁻¹ | (49 - √257 ) / (13 - 3 √257 ) |
(49 - √257 ) = 2144/ (49 + √257 )
(13 - 3 √257 ) = -2144/ (13 + 3 √257 )
= Tan⁻¹ | - (13 + 3 √257 )/ (49 + √257 ) |
= Tan⁻¹ | (13 + 3 √257 )/ (49 + √257 ) |
Tan⁻¹| (13 + 3 √257 )/ (49 + √257 ) | = Tan⁻¹ | (13 + 3 √257 )/ (49 + √257 ) | ≈ 43.21°
Hence pair of lines joining the origin to the points of intersection of the line
3x - y = 2 with the pair of lines 7x^2 - 4xy + 8y^2 + 2x - 4y - 8 = 0 are equally inclined with the co-ordinate axes.
other way :
lines will be equally inclined with the co-ordinate axes if they are perpendicular to each other
Slopes are (13 + 3 √257 )/ (49 + √257 ) and (13 -3 √257 )/ (49 - √257 )
Multiplying both slope = (13 + 3 √257 ) (13 -3 √257 ) /((49 + √257 ) (49 - √257 ))
= (169 - 2313) /(2401 - 257)
= -2144/2144
= - 1
Hence lines are perpendicular to each other
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