What is the equation of the line that passes through (#5, -2)# and #(3, 4)#?

3 Answers
Aug 13, 2018

Answer:

#y=-3x+13#

Explanation:

#"the equation of a line in "color(blue)"slope-intercept form"# is.

#•color(white)(x)y=mx+c#

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

#"to calculate m use the "color(blue)"gradient formula"#

#•color(white)(x)m=(y_2-y_1)/(x_2-x_1)#

#"let "(x_1,y_1)=(5,-2)" and "(x_2,y_2)=(3,4)#

#m=(4-(-2))/(3-5)=6/(-2)=-3#

#y=-x+clarrcolor(blue)"is the partial equation"#

#"to find c substitute either of the 2 given points into"#
#"the partial equation"#

#"using "(3,4)" then"#

#4=-9+crArrc=4+9=13#

#y=-3x+13larrcolor(red)"is the equation of the line"#

Aug 13, 2018

Answer:

The eqn. of #"line passes through "(5,-2) and (3,4)# is

#3x+y-13=0#

Explanation:

The eqn. of line passes through #A(x_1,y_1) and B(x_2,y_2)# is

#|(x,y,1),(x_1,y_1,1),(x_2,y_2,1)| =0#

We have , two points :#A(5,-2) and B(3,4)#

So, the eqn. of #"line passes through "A and B# is

#|(x,y,1),(5,-2,1),(3,4,1)| =0#

Expanding we get

#x(-2-4)-y(5-3)+1(20+6)=0#

#:.x(-6)-y(2)+1(26)=0#

#:.-6x-2y+26=0#

Divding each term by #(-2)#

#:.3x+y-13=0#

#:.#The eqn. of #"line passes through "(5,-2) and (3,4)# is

#3x+y-13=0#

Aug 14, 2018

Answer:

#y = -3x+13#

Explanation:

There is a useful formula which can be used to find the equation of a line if two points are known. It is based on the slope formula.

#(y-y_1)/(x-x_1) = (y_2-y_1)/(x_2-x_1)#

The points are #(5,-2) and (3,4)#

#(y-(-2))/(x-5) = (4-(-2))/(3-5)#

#(y+2)/(x-5) = (4+2)/(3-5) =color(blue)( 6/-2 = -3/1)" "larr# (this is the slope)

#(y+2)/(x-5) = -3/1#

#y+2 = -3(x-5)#

#y = -3x+15-2#

#y = -3x+13#