Question

3x + 8y = 4
(8, -9)
a. Find the equation of a line passing through the point given perpendicular to the line given.
b. Find the distance between the point and the line.
c. Find the equation of a line passing through the point (4, 3) parallel to the line given and its distance.

Answer 1

$a)3\times x+8\times y-4=0;\\ y=-\frac{3}{8}\times x+0.5;(k'=-\frac{1}{k});\\y=\frac{8}{3}\times x+b;\\-9=\frac{8}{3}\times 8+b;\\b=-9-\frac{64}{3}=\frac{-27-64}{3}=\frac{-91}{3};\\y=\frac{8}{3}\times x-\frac{91}{3};\\3\times y=8\times x-91;\\-8\times x+3\times y+91=0.$

Answer: a)$-8\times x+3\times y+91=0\\$

$b)d_1=\frac{|A\times x+B\times y+C|}{\sqrt{A^2+B^2}}=\frac{|3\times 8+8\times (-9)-4|}{\sqrt{3^2+8^2}}=\frac{52\times\sqrt73}{73}\\$

Answer:$b)$ $d_1=\frac{52\times\sqrt73}{73}$

$c)y=-\frac{3}{8}\times x+0.5;k=k'';\\y=-\frac{3}{8}\times x+b;\\3=-\frac{3}{8}\times 4+b;b=\frac{24+12}{8}=\frac{36}{8}=4.5;\\y=-\frac{3}{8}\times x+4.5;8\times y+3\times x-4.5=0;\\3\times x+8\times y-4.5=0;\\b)d_2=\frac{|A\times x+B\times y+C|}{\sqrt{A^2+B^2}}=\frac{|3\times 4+8\times3-4|}{\sqrt{3^2+8^2}}=\frac{|12+24-4|}{\sqrt{3^2+8^2}}\\d_2=\frac{32}{\sqrt{73}}=\frac{32\times\sqrt{73} }{73}$

Answer:

$c)y=-\frac{3}{8}\times x+4.5;\;\\3\times x+8\times y-4.5=0;\\d_2=\frac{32\times\sqrt{73} }{73}$