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# kuta software infinite algebra 1 graphing lines

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Graphing Linear Equations; Part 1 · (Algebra) Number Lines Part 1

Question:Hey, I can't seem to get graphing perpendicular lines on the Cartesian Graph. I understand that the slope-intercept formula for the second line must be the reciprocal and the opposite positive or negative sign than the original line, but when I get past this point, I am not able to lengthen the line to pass through the correct Y-axis intercept... How can I get it to pass through the right Y-axis intercept?? Hope i explained it enough, if not, just ask, thanks! :) Catbek: Yes, I understand that completly, but even when I have that part 100% correct I STILL don't get it to pass through the correct Y-intercept... :/ Ummm... my formula is: y=mx -/+ b

Answers:So you have the gradient and two points I am assuming. Use y - y1 = m(x - x1) to find the equation of the line. Then substitute in 0 for x into the equation to find the y intercept as when the line passes through the y intercept, x = 0.

Question:I have to graph two equations, and write the solution. for coincident lines, my teacher said you write either "all points fall on x = #" or "all points equal x = #" but I forget which one it is. which one is it, or are they both right?

Answers:I don't understand "all points fall on x = #" or "all points equal x = #" but for coincident lines, the equations are equivalent -- can be shown to be the same equation. The solution points for both equations fall on the same line.

Question:Line 1: -x -7y +9 = 0 Line 2: 4x+3y+14 =0 Line 1, Line 2, both lines or neither? 2,1 -5,2 7,5 I have to make a choice, if Line one is at 2,1, -5,2, 7,5, same with Line 2, or both lines at the same place or neither lines at none of these choices