Question

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.
​

Answer 1

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

$\sf \: \implies \: \tan( \theta) = | \dfrac{ m_{1} \: - \: m_{2} }{1 + m _{1}m_{2} } | \\ \\ \sf \: \implies \: \tan(45 \degree) = | \frac{ m_{1} - 3}{1 + m_{1} \times 3} | \\ \\ \sf \: \implies \: 1 = | \frac{ m_{1} - 3}{1 + 3 m_{1}} | \\ \\ \sf \: \implies \: 1 + 3m_{1} \: = m_{1} \: - 3 \\ \\ \sf \: \implies \: 3 m_{1} \: - \: m_{1} \: = - 3 - 1$

$\sf \: \implies \: 2 m_{1} \: = - 4 \\ \\ \sf \: \implies \: m_{1} \: = - 2 \\ \\ \sf \: \implies \: equation \: of \: straight \: line \\ \\ \sf \: \implies \: (y - y_{1}) = m(x - x_{1}) \\ \\ \sf \: \implies \: (y - 2) = - 2(x - ( - 3)) \\ \\ \sf \: \implies \: y - 2 = - 2(x + 3)$

$\sf \: \implies \: y - 2 = - 2x - 6 \\ \\ \sf \: \implies \: y + 2x = - 4$

CASE = 2.

$\sf \: \implies \: 1 = - | \dfrac{ m_{1} - 3 }{1 + 3 m_{1}} | \\ \\ \sf \: \implies \: 1 + 3 m_{1} = - ( m_{1} - 3) \\ \\ \sf \: \implies \: 1 + 3m_{1} = - m_{1} + 3 \\ \\ \sf \: \implies \: 4m_{1} = 2 \\ \\ \sf \: \implies \: m_{1} = \frac{1}{2}$

$\sf \: \implies \: (y - 2) = \dfrac{1}{2} (x - ( - 3)) \\ \\ \sf \: \implies \: 2(y - 2) = (x + 3) \\ \\ \sf \: \implies \: 2y - 4 = x + 3 \\ \\ \sf \: \implies \: 2y - x = 7$

Answer 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.