# Find the equation of the straight line with positive gradient and inclined at an angle of 45° to the line 3y-x+1=0 and passing through the point (2,0)?

Aug 11, 2018

$2 x - y + 2 = 0$
is the equation of the straight line expected

#### Explanation:

Slope of the straight line,
$a x + b y + c = 0$
is given by
${m}_{1} = - \frac{a}{b}$
The given line is

$3 y - x + 1 = 0$
Rearranging in the standard form,

$- x + 3 y + 1 = 0$

or

$x - 3 y - 1 = 0$

Comparing
a = 1; b = -3; c = 0

Thus,
${m}_{1} = - \frac{1}{- 3}$
${m}_{1} = \frac{1}{3}$

The slope of the new line is
${m}_{2} = \tan \phi$
Here, the angle of inclination is
$\phi - {45}^{0} .$

${m}_{2} = \tan {45}^{0}$

${m}_{2} = 1$

Since the new line is having the positive slope with the straight line, the effective slope becomes combination of the straight line and the new line in the form given by

$m = \frac{{m}_{1} + {m}_{2}}{1 - {m}_{1} {m}_{2}}$

Substituting for
${m}_{1} = \frac{1}{3} , {m}_{2} = 1 ,$
we have

$m = \frac{\frac{1}{3} + 1}{1 - \frac{1}{3} \cdot 1}$

$= \frac{\frac{1}{3} + \frac{3}{3}}{\frac{3}{3} - \frac{1}{3}}$

$= \frac{\frac{1 + 3}{3}}{\frac{3 - 1}{3}}$

$= \frac{1 + 3}{3 - 1}$

$m = \frac{4}{2}$

$m = 2$
The new line passes through the point,
$\left(2 , 0\right)$

Equation of a straight line passing through the point $\left(a , b\right)$, and having slope $m$, is given by:

$\frac{y - b}{x - a} = m$

Here;
a=0; b=2; m=2

Thus,

$\frac{y - 2}{x - 0} = 2$
Simplifying

$\frac{y - 2}{x} = 2$

Cross multiplyng

$y - 2 = 2 x$

Rearranging in the standard form, $a x + b y + c = 0$
we have

$2 x - y + 2 = 0$
is the equation of the straight line expected

Aug 11, 2018

$2 x - y - 4 = 0$.

#### Explanation:

Note that the given line has slope $\frac{1}{3}$.

Let the reqd. line have slope $m > 0$.

Given that, the angle between these two lines is ${45}^{\circ}$.

$\therefore \tan {45}^{\circ} = | \frac{m - \frac{1}{3}}{1 + \frac{1}{3} \cdot m} | = | \frac{3 m - 1}{3 + m} |$.

$\therefore \frac{3 m - 1}{3 + m} = \pm 1$.

$\therefore 3 m - 1 = 3 + m , \mathmr{and} , 3 m - 1 = - \left(3 + m\right)$.

$\therefore m = 2 , \mathmr{and} , m = - \frac{1}{2}$.

$m > 0 \Rightarrow m \ne - \frac{1}{2}$.

Thus, the reqd. line has slope $2$ and passes thro. $\left(2 , 0\right)$.

$\therefore \text{The Reqd. Line : } y - 0 = 2 \left(x - 2\right) = 2 x - 4 ,$

$i . e . , 2 x - y - 4 = 0$.