Q: The equation of the line joining through -1 3 amd parallel to the line joining 6 3 2 -3?

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solve the equation for y to get the slope.y=-2x-1/2substitute (3,3) into the equation/3=2(3)+band solve for b.-3=+by=2x-3 is the equation with the same slope(parallel) and goes through (3,3)

If you mean y = -65x-4 then the parallel equation is y = -65x-66

When a straight line equation is parallel to another equation the slope remains the same but the y intercept changes

If you mean: y = -x-3 passing through point of (-1, 3) Then the equation of the parallel line is: y =-x+2

The given expression is not an equation because it has no equality sign but for a line to be parallel to another line they both will have the same slope but with different y intercepts

Related questions

The parallel equation will have the same slope but with a different y intercept

Write an equation in slope-intercept form for the line that passes through the given point and is parallel to the given line (-7,3); x=4

For a straight line equation to be parallel to another straight line equation they both must have the same slopes but different y intercepts.

Without an equality sign it cannot be considered to be a straight line equation. But if you mean: y = 25x-3 then the parallel line is y = 25x-61

If you mean: y = 6x-4 then the parallel equation is y = 6x+10

If you mean: y = 2x-4 and (1, 5) then the parallel equation is y = 2x+3

Parallel straight line equations have the same slope but with different y intercepts

solve the equation for y to get the slope.y=-2x-1/2substitute (3,3) into the equation/3=2(3)+band solve for b.-3=+by=2x-3 is the equation with the same slope(parallel) and goes through (3,3)

If you mean: y = -23x+3 then the parallel equation is y = -23x+164

5

The locus of points where the electric fields are parallel to the line joining the charges Q and -Q forms an equipotential surface. This surface is perpendicular to the line joining the charges and passes through the midpoint of the line connecting them. The electric field lines are always perpendicular to the equipotential surfaces.

y -x + 1