An algebraic expression ,using d as the variable four times a number
2 answers:
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1)We have to make a line perpendicular to y=2x+3, passing through the point (0,0).
y=2x+3
m=solpe=2
The slope of the line perpedicular to y=2x+3 is m´
m´=-1/m
m´=-1/2
Point slope form:
y-y₀=m(x-x₀)
(0,0)
m=-1/2
y-0=-1/2(x-0)
y=-x/2
Therefore, the line perpendicular to y=2x+3 is y=-x/2.
2)The point on the line y=2x+3 tha is closest to the origin is the point of intersection of the two lines.
y=2x+3
y=-x/2
We can solve this system of equations by equalization method.
2x+3=-x/2
least common multiple=2
4x+6=-x
4x+x=-6
5x=-6
x=-6/5 (=-1.2)
y=-x/2
y=-(-6/5)/2=6/10=3/5 (=0,6)
Therefore: the point of the line y=2x+3 that is closest to the origin is:
(-1.2 , 0.6)
y=2x+3
m=solpe=2
The slope of the line perpedicular to y=2x+3 is m´
m´=-1/m
m´=-1/2
Point slope form:
y-y₀=m(x-x₀)
(0,0)
m=-1/2
y-0=-1/2(x-0)
y=-x/2
Therefore, the line perpendicular to y=2x+3 is y=-x/2.
2)The point on the line y=2x+3 tha is closest to the origin is the point of intersection of the two lines.
y=2x+3
y=-x/2
We can solve this system of equations by equalization method.
2x+3=-x/2
least common multiple=2
4x+6=-x
4x+x=-6
5x=-6
x=-6/5 (=-1.2)
y=-x/2
y=-(-6/5)/2=6/10=3/5 (=0,6)
Therefore: the point of the line y=2x+3 that is closest to the origin is:
(-1.2 , 0.6)