Question

find equation of Line passing through (1,2) and making an angle of 45°with the line 2x+3y = 10​

Answer 1

Step-by-step explanation:

"What is the equation of the straight line through the line (1,-4) with an angle of 45 with the line 2x+3y+7=0?

Let the slope of the line through (1,-4) is m.

Slope of line 2x+3y+7=0 is -2/3.

+/- tan45 = (m+2/3)/(1-2/3.m).

+/- 1 = (3m+2)/(3-2m).

taking + ve sign.

3m+2 =3-2m.

or, 5m =1 or, m=1/5.

The equation of the line is:-

(y+4)= (1/5).(x-1).

or, x- 1 = 5y +20.

or, x - 5y = 21.

taking - ve sign.

3m+2 = -3+2m.

or, m=-5.

The equation of the line is:-

y+4 = -5.(x-1).

or, y+4.= -5x+5.

or, 5x +y =1.

Thus, x-5y=21 and 5x+y=1 are required

equation of lines. , Answer "

Answer 2

Step-by-step explanation:

"What is the equation of the straight line through the line (1,-4) with an angle of 45 with the line 2x+3y+7=0?

Let the slope of the line through (1,-4) is m.

Slope of line 2x+3y+7=0 is -2/3.

+/- tan45 = (m+2/3)/(1-2/3.m).

+/- 1 = (3m+2)/(3-2m).

taking + ve sign.

3m+2 =3-2m.

or, 5m =1 or, m=1/5.

The equation of the line is:-

(y+4)= (1/5).(x-1).

or, x- 1 = 5y +20.

or, x - 5y = 21.

taking - ve sign.

3m+2 = -3+2m.

or, m=-5.

The equation of the line is:-

y+4 = -5.(x-1).

or, y+4.= -5x+5.

or, 5x +y =1.

Thus, x-5y=21 and 5x+y=1 are required

equation of lines. , Answer "