1. standard form of a linear equation, like yours is ax2+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⟹−5y=7⟹y=−75
3. We know we find the equation with point-slope form, which is y−y1=m(x−x1) 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 y−3=m(x−3)Then we can find the slope of the graph from the two points. which is 2/5. Then we plug in the values and solvey−3=m(x−3)⟹y−3=25(x−3)⟹y−3=25x−65y=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. y−y1=m(x−x1)⟹y−y1=85(x−x1) We also have the point (10,10). So we can substitute the point in the equation. y−y1=85(x−x1)⟹y−10=85(x−10)⟹y−10=85x−16⟹y=85x−68/5 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. y−y1=m(x−x1)⟹y−2=m(x−1)We can find the slope from the points and see it is 1/3. So y−2=13(x−1)⟹y−2=13x−13⟹y=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!