2 answers

Find the equation of a line parallel to #2x+5y=3# and passing through #(2,-3)#?

Question:

Find the equation of a line parallel to #2x+5y=3# and passing through #(2,-3)#?

Answers

#2x+5y+11=0#

Explanation:

A line parallel to #ax+by=c# is of the type #ax+by=k#, i.e. only constant term is changed, while coefficients remain same.

hence a line parallel to #2x+5y=3# is of the type #2x+5y=k#

as it passes through #(2,-3)#, we have

#2xx2+5xx(-3)=k# or #4-15=k# i.e. #k=-11#.

Hence, equation is #2x+5y=-11# i.e. #2x+5y+11=0#

graph{(2x+5y+11)(2x+5y-3)((x-2)^2+(y+3)^2-0.01)=0 [-4.98, 5.02, -3.52, 1.48]}

Note: A line perpendicular to #ax+by=c# is of the type #bx-ay=k# i.e. while coefficients of #x# and #y# gets exchanged, sign of one of them too changes.

.

#y=-2/5x-11/5#

Explanation:

#• " parallel lines have equal slopes"#

#"rearrange "2x+5y=3" into "color(blue)"slope-intercept form"#

#rArry=-2/5x+3/5#

#"compare to "y=mx+blarrcolor(blue)" slope-intercept form"#

#"where m is the slope and b the y-intercept"#

#rArrm=-2/5#

#rArry=-2/5x+blarr" partial equation"#

#"to find b substitute "(2,-3)" into the partial equation"#

#-3=-4/5+brArrb=-11/5#

#rArry=-2/5x-11/5#

.

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