Write an equation of a line that passes through (3, -1) and is parallel and perpendicular to y = 6x - 4

Mathematics

1 answer:

Serhud [2]1 year ago

4 0

Here,

y=mx+c

the slope of the line 6x+4y= -16 is -20.

if the two lines are perpendicular, the slope of 1st line * slope of second line=-1

the slope of the 2nd line is -1/2

y-y1=m(x-x1)

y-13=-1/2(x- -18)

when you solve, you get the equation of the line as 2y+x=50

In basic geometry, if two geometry objects intersect at right angles (90 degrees or π / 2 radians), they are vertical.

If two lines intersect at right angles, the line is perpendicular to another line. Explicitly (1) if two lines intersect, the first line is perpendicular to the second line. (2) At the intersection, the straight line angle on one side of the first line is cut by the second line into two congruent angles. Verticality can be shown as symmetric. That is, if the first line is perpendicular to the second line, then the second line is also perpendicular to the first line. Therefore, two straight lines can be said to be perpendicular (to each other) without specifying the order.

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Check the picture below.

$\bf 0=AB^2+6AB-121\implies \stackrel{\textit{using the quadratic formula}}{AB=\cfrac{-6 \pm \sqrt{6^2-4(1)(-121)}}{2(1)}}
\\\\\\
AB=\cfrac{-6 \pm \sqrt{520}}{2}\implies AB=\cfrac{-6 \pm \sqrt{4\cdot 130}}{2}
\\\\\\
AB=\cfrac{-6 \pm \sqrt{2^2\cdot 130}}{2}\implies AB=\cfrac{-6 \pm 2\sqrt{130}}{2}
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\begin{cases}
\boxed{-3+\sqrt{130}}\\\\
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\end{cases}$

since the distance AB cannot be a negative value, thus is not -3-√(130).