Reduce the line 2x-3y+5=0 (a) In slope - intercept form and hence find slope Y-intercept - Maths - Straight Lines

Question

Reduce the line 2x-3y+5=0
​(a) In slope - intercept form and hence find slope &
Y-intercept
​(b) In intercept form and hence find intercept on axes
​(c) In normal form and hence find perpendicular distance
from the origin and angle made by the perpendicular with the
positive x-axis

Brijendra Pal · Accepted Answer

Dear Student, 2x-3y+5=0 3y = 2x+5 y = 23x+53So comparing, we get y = mx+cslope m = 23, y axis intercept c = 53b)2x-3y = -5 2x-5+-3y-5=1x-5/2+y5/3= 1Comaring it with xa+yb= 1x-axis intercept -52,0, and y-axis intercept 53,0c)Normal form is xcosαp+ysinαp= 1 so comparing it with 2x-5+-3y-5=1cosαp=-25 and sinαp=35Dividing tanα=-32 so α=tan-1-32 this is the angle made by x axis
 Regards

Reduce the line 2x-3y+5=0 ​(a) In slope - intercept form and hence find slope & Y-intercept ​(b) In intercept form and hence find intercept on axes. ​(c) In normal form and hence find perpendicular distance from the origin and angle made by the perpendicular with the positive x-axis.

Reduce the line 2x-3y+5=0
(a) In slope - intercept form and hence find slope & Y-intercept
(b) In intercept form and hence find intercept on axes.
(c) In normal form and hence find perpendicular distance from the origin and angle made by the perpendicular with the positive x-axis.