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)?

2 Answers
Aug 11, 2018

2x-y+2=02xy+2=0
is the equation of the straight line expected

Explanation:

Slope of the straight line,
ax+by+c=0ax+by+c=0
is given by
m_1 = -a/bm1=ab
The given line is

3y-x+1 = 03yx+1=0
Rearranging in the standard form,

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

or

x-3y-1 = 0x3y1=0

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

Thus,
m_1 = -1/(-3)m1=13
m_1 = 1/3m1=13

The slope of the new line is
m_2 = tan phim2=tanϕ
Here, the angle of inclination is
phi - 45^0 .ϕ450.

m_2 = tan 45^0m2=tan450

m_2 = 1m2=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=(m_1 + m_2)/(1 - m_1 m_2)m=m1+m21m1m2

Substituting for
m_1 = 1/3, m_2 = 1, m1=13,m2=1,
we have

m = (1/3+1)/(1-1/3*1)m=13+11131

=(1/3+3/3)/(3/3-1/3)=13+333313

=((1+3)/3)/((3-1)/3)=1+33313

=(1+3)/(3-1)=1+331

m=4/2m=42

m=2m=2
The new line passes through the point,
(2,0)(2,0)

Equation of a straight line passing through the point (a,b)(a,b), and having slope mm, is given by:

(y-b)/(x-a) = mybxa=m

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

Thus,

(y-2)/(x-0) = 2y2x0=2
Simplifying

(y-2)/x=2y2x=2

Cross multiplyng

y-2=2xy2=2x

Rearranging in the standard form, ax+by+c=0ax+by+c=0
we have

2x-y+2=02xy+2=0
is the equation of the straight line expected

Aug 11, 2018

2x-y-4=02xy4=0.

Explanation:

Note that the given line has slope 1/313.

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

Given that, the angle between these two lines is 45^@45.

:. tan45^@=|(m-1/3)/(1+1/3*m)|=|(3m-1)/(3+m)|.

:. (3m-1)/(3+m)=+-1.

:. 3m-1=3+m, or, 3m-1=-(3+m).

:. m=2, or, m=-1/2.

m gt 0 rArr m!=-1/2.

Thus, the reqd. line has slope 2 and passes thro. (2,0).

:."The Reqd. Line : "y-0=2(x-2)=2x-4,

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