Question

Find the equation of the line passing through the point (1, 1)which are inclined to the line y = 2x+1 at an angle 45 degree

thanks

Answer 1

$\text{Let slope of unknown equation of line is m}$
$\text{we know,}\\\text{slope of any line } = - \frac{\text{coefficient of x}}{\text{coefficient of y}}$
$\text{here,Line:} \: 2x - y + 1 = 0$
so, slope of these line {m'} = 2

$\text{now, use formula}$
$tan{\theta}=\frac{|m_1-m_2|}{|1+m_1.m_2|}$
here, ∅ = 45° so, tan45° = |2 - m|/|1 + 2m|
1 = |2 - m |/|1 + 2m|
| 1 + 2m| = |2 - m|
1 + 2m = ±( 2 - m)

1 + 2m = 2 - m
3m = 1 => m = 1/3

again, 1 + 2m = -2 + m
m = -3

hence, slope of unknown line are 1/3 and -3

now, equation of line
$(y - y_1) = m(x - x_1)$
so, (y - 1) = 1/3(x - 1)
3y - 3 = x - 1 => x - 3y + 2 = 0

and (y - 1) = -3(x - 1)
y - 1 +3x - 3 = 0
3x + y - 4 = 0

$\text{hence, answer is}$
$\boxed{\boxed{\bold{x-3y+2=0,3x+y - 4 = 0}}}$

Answer 2

Let slope of unknown equation of line is m

\begin{gathered}\text{we know,}\\\text{slope of any line } = - \frac{\text{coefficient of x}}{\text{coefficient of y}}\end{gathered}

we know,

slope of any line =−

coefficient of y

coefficient of x

\text{here,Line:} \: 2x - y + 1 = 0here,Line:2x−y+1=0

so, slope of these line {m'} = 2

\text{now, use formula}now, use formula

\begin{gathered}tan{\theta}=\frac{|m_1-m_2|}{|1+m_1.m_2|} \\\end{gathered}

tanθ=

∣1+m

1

.m

2

∣

∣m

1

−m

2

∣

here, ∅ = 45° so, tan45° = |2 - m|/|1 + 2m|

1 = |2 - m |/|1 + 2m|

| 1 + 2m| = |2 - m|

1 + 2m = ±( 2 - m)

1 + 2m = 2 - m

3m = 1 => m = 1/3

again, 1 + 2m = -2 + m

m = -3

hence, slope of unknown line are 1/3 and -3

now, equation of line

(y - y_1) = m(x - x_1)(y−y

1

)=m(x−x

1

)

so, (y - 1) = 1/3(x - 1)

3y - 3 = x - 1 => x - 3y + 2 = 0

and (y - 1) = -3(x - 1)

y - 1 +3x - 3 = 0

3x + y - 4 = 0