Question

find the equation of the straight line passing through the point 1/3 ,2/5 with slope -3/4

Answer 1

so given tanθ= -3/4
and a point = 1/3 ,2/5 
so using equation tanθ = (y₁ - y)/(x₁ - x)
so let there be any point (x,y)
-3/4 = (y - 2/5)/(x - 1/3)
 -5/4 = (5y - 2)/(3x - 1)
so we find the equation as 
20y +15x = 3

Answer 2

$The\ slope-intercept\ form:y=mx+b\\\\m=-\frac{3}{4}\Rightarrow y=-\frac{3}{4}x+b\\\\th\e point\ (\frac{1}{3};\ \frac{2}{5})\to substitute\ x=\frac{1}{3}\ and\ y=\frac{2}{5}\ to\ equation\ of\ line:\\\\-\frac{3}{4}(\frac{1}{3})+b=\frac{2}{5}\\\\-\frac{1}{4}+b=\frac{2}{5}\\\\b=\frac{2}{5}+\frac{1}{4}\\\\b=\frac{8}{20}+\frac{5}{20}\\\\b=\frac{13}{20}\\\\Answer:y=-\frac{3}{4}x+\frac{13}{20}$

$\frac{3}{4}x+y=\frac{13}{20}\ \ \ \ |\times20\\\\15x+20y=13$