Answer to Question #114124 in Analytic Geometry for Randal Rodriguez

Question #114124
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.
1
Expert's answer
2020-05-11T19:14:29-0400

a)3×x+8×y4=0;y=38×x+0.5;(k=1k);y=83×x+b;9=83×8+b;b=9643=27643=913;y=83×x913;3×y=8×x91;8×x+3×y+91=0.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×x+3×y+91=0-8\times x+3\times y+91=0\\

b)d1=A×x+B×y+CA2+B2=3×8+8×(9)432+82=52×7373b)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)b) d1=52×7373d_1=\frac{52\times\sqrt73}{73}

c)y=38×x+0.5;k=k;y=38×x+b;3=38×4+b;b=24+128=368=4.5;y=38×x+4.5;8×y+3×x4.5=0;3×x+8×y4.5=0;b)d2=A×x+B×y+CA2+B2=3×4+8×3432+82=12+24432+82d2=3273=32×7373c)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=38×x+4.5;  3×x+8×y4.5=0;d2=32×7373c)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}




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