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1)Write the equation y=4/5x-7/10 in standard form. (Standard form is  where  is positive, and   and  are integers with greatest common divisor 1.)

2)Find the y-intercept of the line 3x-5y=7

3)Find the equation of the line that passes through the points (-2,1) and (3,3). Write your answer in the form y=mx+b .

4)Let line  l1 be the graph of 5x+8y=-9.Line l2 is perpendicular to line l1 and passes through the point (10,10). If line l1 is the graph of the equation y=mx+b , then find m+b

5)The perpendicular bisector of the line segment AB  is the line that passes through the midpoint of  AB and is perpendicular to AB.

Find the equation of the perpendicular bisector of the line segment joining the points(1,2)  and (7,4)  Enter your answer in the form "y=mx+b"

Thanks!

Nov 6, 2018

#1
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1.   I’m not sure what you meant by “(standard form is where is positive, and and are integers with greatest common divisor 1)”, but 8x-10y=7 (in Ax+By=C)

2.   y-intercept: -7/5

Nov 6, 2018
edited by EmperioDaZe  Nov 6, 2018
#2
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1. standard form of a linear equation, like yours is  $$ax^2+by = c$$. So it would be written as $$8x-10y = 7$$.

2. The y-intercept of a graph is when the x-value is 0. So in this equation, we can set x to 0. $$3(0) -5y = 7 \implies -5y = 7 \implies y = -\frac{7}{5}$$

3. We know we find the equation with point-slope form, which is $$y-y_1=m\left(x-x_1\right)$$ where m is the slope. $$y_1$$ is the y-value of any of the points. and $$x_1$$ is the value of the x-coordinate corresponding with the y-coordinate. So we can simplify to $$y-3=m\left(x-3\right)$$Then we can find the slope of the graph from the two points. which is 2/5. Then we plug in the values and solve$$y-3=m\left(x-3\right) \implies y-3=\frac{2}{5}\left(x-3\right) \implies y - 3 = \frac{2}{5}x - \frac{6}{5}$$$$\boxed{y = \frac{2}{5}x + \frac{9}{5}}$$

4. We can begin by putting line 1 in slope intercept form. 8y = -5x -9. y = -5/8x - 9/8. If line 2 is going to be perpindicular to line 1, their slopes are going to be opposite reciprocals of each other. So the slope of line 2 is 8/5. Then like in problem 3, we can use the point slope form. $$y-y_1=m\left(x-x_1\right) \implies y-y_1=\frac{8}{5}\left(x-x_1\right)$$ We also have the point (10,10). So we can substitute the point in the equation. $$y-y_1=\frac{8}{5}\left(x-x_1\right) \implies y-10=\frac{8}{5}\left(x-10\right) \implies y -10 = \frac{8}{5}x - 16 \implies y = \frac{8}{5}x -6$$8/5 is m, and -6 is b. 8/5 + (-6) = 8/5 - 6  = $$\boxed{-\frac{22}{5}}$$

5. We can start by finding the equation of the line segment using the point-slope form. $$y-y_1=m\left(x-x_1\right) \implies y-2=m\left(x-1\right)$$We can find the slope from the points and see it is 1/3. So $$y-2=\frac{1}{3}\left(x-1\right) \implies y - 2 = \frac{1}{3}x - \frac{1}{3} \implies y = \frac{1}{3}x + \frac{5}{3}$$ Then we know that the perpindicular bisector has an opposite reciprocal slope. So $$\boxed{y = -3x + \frac{5}{3}}$$is the equation of the perpendicular bisector.

Hope that helps!

Nov 6, 2018
#3
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