Question

15. (a) Find the equation of the line which is inclined at 135° with the negative direction of the r-axis and passes through the point (1, 2). (b) Find the equation of the line which is inclined at 45° with the positive direction of the y-axis and passes through the point (3,2).
PLZ ANSWER ME
I Will mark you as a brain list​

Answer 1

The slope of the line, m = tan (135°) = tan (45°) = -1; as 135° lies in 2nd quadrant and as such tan is negative by ASTC rule.

The line passes through a point (2,-3) and an arbitrary point (x,y).

Slope, m = (y - y1)/(x - x1)

y - y1 = m(x - x1); slope-point form

y + 3 = -1(x - 2)

y + 3 = -x + 2

Hence, equation of the line is

y = -x - 1; slope-intercept form; y = -(x+1)

y + x + 1 = 0; general form

Answer 2

${\huge{\pink{↬}}} \: \: {\huge{\underline{\boxed{\bf{\pink{Answer}}}}}}$

Here, m = slope of the line = tan 135º = tan (90º + 45º) = – cot 45º = –1, x1 = 2, y1 = –3

So, the equation of the line is y – y1 = m (x – x1)

i.e. y – (–3) = –1 (x – 2) or y + 3 = –x + 2 or x + y + 1 = 0