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

$a)3\times x +8\times y=4; \\y=\frac{4-3\times x }{8}=-\frac{3}{8}\times x+0.5;\\y=-\frac{3}{8}\times x+0.5;\\k'=\frac{8}{3};y=\frac{8}{3}\times x+b=\frac{8}{3}\times 8+b;\\-9=\frac{64}{3}+b;b=-9-\frac{64}{3}=-\frac{27}{3}-\frac{64}{3}=-\frac{91}{3};\\y=\frac{8}{3}\times x-\frac{91}{3};3y=8\times x-91;\\-8\times x+3 \times y+91=0\\b)d=\frac{ |A\times M_x+B\times M_y+C|}{\sqrt{A^2+B^2}};d=\frac{ |3\times 8+8\times (-9)-4|}{\sqrt{3^2+8^2}}=\frac{52}{\sqrt73}\\c)3\times x+8\times y-4=0;k=k'';\\y=-\frac{3}{8}\times x+b;3=-\frac{3}{8}\times 4+b;b=4.5;\\y=-\frac{3}{8}\times x+4.5;\\3\times x +8\times y-36=0;\\3\times x +8\times y-36=0;\\3\times x +8\times y-4=0; |d|=\frac{|C_2-C_1}{\sqrt{A^2+B^2}}=\frac{32}{\sqrt73}$

Answer:

$-8\times x+3 \times y+91=0$ ;

$\frac{52}{\sqrt73}$ ;

$3\times x +8\times y-36=0$ ;

$\frac{32}{\sqrt73}$