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hashtagmath

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1. standard form of a linear equation, like yours is  ax2+by=c. So it would be written as 8x10y=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=75y=7y=75

 

3. We know we find the equation with point-slope form, which is yy1=m(xx1) where m is the slope. y1 is the y-value of any of the points. and x1 is the value of the x-coordinate corresponding with the y-coordinate. So we can simplify to y3=m(x3)Then we can find the slope of the graph from the two points. which is 2/5. Then we plug in the values and solvey3=m(x3)y3=25(x3)y3=25x65y=25x+95

 

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. yy1=m(xx1)yy1=85(xx1) We also have the point (10,10). So we can substitute the point in the equation. yy1=85(xx1)y10=85(x10)y10=85x16y=85x68/5 is m, and -6 is b. 8/5 + (-6) = 8/5 - 6  = 225

 

5. We can start by finding the equation of the line segment using the point-slope form. yy1=m(xx1)y2=m(x1)We can find the slope from the points and see it is 1/3. So y2=13(x1)y2=13x13y=13x+53 Then we know that the perpindicular bisector has an opposite reciprocal slope. So y=3x+53is the equation of the perpendicular bisector. 

 

Hope that helps!

Nov 6, 2018