Find the equation of the straight lines
passing through the point (-3, 2) and
making an angle of 45° with the straight line
3x-y+4=0.
Answers
EXPLANATION.
Equation of straight line passing through
the point (-3, 2).
make angle of = 45°.
Straight lines equation = 3x - y + 4 = 0.
Slope of the equation = -a/b = -3/-1 = 3.
angle between lines
CASE = 2.
Pair of Straight Lines
Straight lines passing through the point ( - 3, 2 ) and making an angle of 45° with the straight line 3x - y + 4 = 0
Let the slope of the required line be m1
Slope of the given straight line 4x - y + 4 = 0 m2 = - a / b = - 3 / - 1 = 3
Angle between the pair of straight lines θ = 45°
tan θ = | ( m1 - m2) / ( 1 + m1m2) |
where m1 and m2 are slopes
tan 45° = | ( m1 - 3 ) / ( 1 + 3m1 ) |
1 = | ( m1 - 3 ) / ( 1 + 3m1 ) |
Case 1
1 = ( m1 - 3 ) / ( 1 + 3m1 )
1 + 3m1 = m1 - 3
3m1 - m1 = - 3 - 1
2m1 = - 4
m1 = - 2
Equation of straight line with slope - 2 with ( - 3, 2 ) point on it
y - y1 = m ( x - x1)
y - 2 = - 2( x + 3 )
y - 2 = - 2x - 6
2x + y + 4 = 0
Case 2
- 1 = ( m1 - 3 ) / ( 1 + 3m1 )
- 1 - 3m1 = m1 - 3
- 1 + 3 = m1 + 3m1
2 = 4m1
m1 = 1/2
Equation of straight line with slope 1/2 with ( - 3, 2 ) point on it
y - y1 = m ( x - x1)
y - 2 = 1/2 ( x + 3 )
2(y - 2) = x + 3
2y - 4 = x + 3
0 = x - 2y + 7
x - 2y + 7 = 0
Therefore 2x + y + 4 = 0 and x - 2y + 7 = 0 are the equations of the straight lines passing through the point (-3, 2) and making an angle of 45⁰ with the straight line 3x -y + 4 = 0.
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