Question 170094
a) Start with the "slope-intercept" form of a linear equation:
y = mx+b where the slope, m = 2 so now you have:
y = 2x+b Now you need to find b, the y-intercept.  Substitute the coordinates of the given point through which the line passes (6, 2).
2 = 2(6)+b Solve for b.
2 = 12+b Subtract 12 from both sides.
-10 = b Now you can write the final equation:
y = 2x-10
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b) First, find the slope using the slope formula:
Remember that the slope is defined as "rise over run" or the difference in the y-coordinates over the difference in x-coordinates.
{{{m = (y[2]-y[1])/(x[2]-x[1])}}} Substitute: (2, 4) and (8, 16)
{{{m = (16-4)/(8-2)}}}
{{{m = 12/6}}}
{{{m = 2}}}
Now you can start the equation with:
y = mx+b and substitute m = 2.
y = 2x+b To find b, substitute the x- and y-coordinates of either one of the two given points.  Let's use the point (2, 4), so we get:
4 = 2(2)+b
4 = 4+b 
b = 0 
The final equation is:
y = 2x
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c) Use the work in a) as a guide for solving this one.
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d) Find the slope by recalling that perpendicular lines have slopes that are the negative reciprocal of each other.
In othe words, in perpendicular lines, the product of the slopes is always equal to -1.
So, the slope of the line in the given equation (3x+4y = 12) is found by converting the equation to the "slope-intercept" form (y = mx+b) and reading the slope directly from the resulting equation.
{{{3x+4y = 12}}}  Subtract 3x from both sides.
{{{4y = -3x+12}}} Divide both sides by 4.
{{{y = (-3/4)x+3}}} So, by inspection, you can see that the slope is {{{m = (-3/4)}}}
But the perpendicular line would have a slope that is the negative reciproacal of {{{(-3/4)}}} which would be {{{(-1)/(-3/4) = 4/3}}}, so...
{{{y = (4/3)x+b}}} Now you can find b by substituting the x- and y-coordintes of the given point (1, 1) through which the line passes.
{{{1 = (4/3)(1)+b}}} Subtract {{{4/3}}} from both sides.
{{{-1/3 = b}}} The final equation is:
{{{y = (4/3)x-1/3}}}
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Starting with the general form of an equation of a circle with its center at (h, k) and a radius of r:
{{{(x-h)^2+(y-k)^2 = r^2}}} Substitute (3, 3) for (h, k) and 9 for r
{{{x-3)^2+(y-3)^2 = 9^2}}}
{{{highlight((x-3)^2+(y-3)^2 = 81)}}}