Find the point on the line y = 2x + 3 that is closest to the origin.

Mathematics

1 answer:

ch4aika [34]2 years ago

3 0

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)

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write a linear equation that intersects y=x^2 at two points. Then write a second linear equation that intersects y=x^2 at just o

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We know that $y = x^2$ is a parabola, concave up, with vertex in the origin $(0,0)$

So, we may use three horizontal lines for our purpose: any horizontal line above the x axis will intersect the parabola twice. The x axis itself intersects the parabola once on the vertex, while any horizontal line below the x axis won't intercept the parabola.

Here's the examples:

The horizontal line $y = 4$ intercepts the parabola twice: the system $y = x^2,\ y = 4$ is solved by $x^2=4 \implies x = \pm 2$
The horizontal line $y=0$ intercepts the parabola only once: the system is $y=x^2,\ y=0$ which yields $x^2=0\implies x=0$ which is a repeated solution
The horizontal line $y=-5$ intercepts the parabola only once: the system is $y=x^2,\ y=-5$ which yields $x^2=-5$ which is impossible, because a squared number can't be negative.