Questions 1
Answers 5


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