Q: What is the straight line equation that has a slope of 1 over 3 and passing through a point of 2 and minus 5 on the line?

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x+7y+4 = 0 => y = -1/7x-4/7 The slope of the second equation is the reciprocal of the first equation with the minus sign changing to a plus sign. y = mx+c where m is the slope and c is the intercept on the y axis So: 7*4+c = 0 28+c = 0 c = -28 Therefore the perpendicular equation is: y = 7x-28 which can be expressed in the form of 7x-y-28 = 0

Without an equality sign and not knowing the plus or minus values of the given terms of the line which then can't be considered to be a straight line equation. In general for lines to be parallel to each other they will both have the same slope but different y intercepts.

The quadric equation is: negative b plus or minus the square root of b squared minus 4ac all over(divided by) 2a

x equals negative b plus or minus the square root of b squared minus 4bc over 2a

negative b plus or minus the square root of b minus 4 times ac all over 2 a

Related questions

3x-4y-6 = 0

Without an equality sign and not knowing the plus or minus values of y and 7 it can't be considered to be a straight line equation therefore finding its perpendicular equation is impossible.

I'm guessing that you mean "passing through the point (7,5)"? Use this to find the line equation: m(x - x1) = y - y1 where x1 & y1 are the coordinates of the point, and m is the slope x1 = 7, y1 = 5, m = -0.75 -0.75 * ( x - 7) = y - 5 Solve for y: y = -0.75*x + 10.25 Note the slope is -0.75, and plug in x = 7, to check your work that y = 5

It means then that the slope of the straight line equation will be minus

y = mx+c So: 1/2*1/3+c = -1/2 1/6+c = -1/2 c = -1/2-1/6 c = -2/3 Therefore the straight line equation is y = 1/2x-2/3 which also can be expressed in the form of: 3x-6y-4 = 0

Equation of original line is 4x + 3y - 5 = 0 that is, 3y = -4x + 5 or y = -(4/3)x + 5 Slope of original line = -4/3 Slope of line perpendicular to it = 3/4 General equation of perpendicular line: y = (3/4)x + c for some constant c or 4y = 3x + c' The point (-2,-3) is on this line so 4*(-3) = 3*(-2) + c' -12 = - 6 + c' so that c' = -6 The equation of the perpendicular line is 4y = 3x - 6

Without an equality sign and not knowing the plus or minus value of 5y the given terms can't be considered to be a straight lie equation.

Without an equality sign and not knowing the plus or minus values of the given terms it can't be considered to be a straight line equation. But if you mean 4x+y = 10 then y = -4x+10 and the parallel equation is y = -4x+6

It's the equation of a straight line with a slope of -3, that crosses the y-axis at y= -1 . Every point on the line is a solution to the equation. There are an infinite number of them.

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Without an equality sign and not knowing the plus or minus values of 2y and 5 it can't be considered to be a straight line equation.

Without an equality sign and not knowing the plus or minus values of 2y and 5 it can't be considered to be a straight line equation.