a line is perpedicular to the line 4x-3y=7. It also has the same y-intercept as

Question

a line is perpedicular to the line 4x-3y=7. It also has the same y-intercept as the line 5x-4y=20. What is the equation of the line in standard form?

Explanation / Answer

First find the slope of the line it's supposed to be perpendicular to, which you do by solving for y. 4x-3y = 7 =============> subtract 4x from both sides -3y = 7 - 4x ================> divide both sides by -3 y = (4/3)x - (7/3) So you know that the slope of this line is 4/3. For a line to be perpendicular, its slope has to be the negative inverse of the slope for the given line, in other words, if you have m = 4/3, then the perpendicular slope would be -1/m = -3/4 Then you find the y-intercept of the second line since it's supposed to be the same, again, by solving for y. 5x-4y = 20 =================> subtract 5x from both sides -4y = 20 - 5x =================> divide both sides by -4 y = (5/4)x - 5 So the y-intercept for this line is b = -5 In y-intercept form your equation would be: y = (-3/4)x - 5 The standard equation formula for a line is: Ax + By = C So rearrange your equation to look like that. (Add (3/4)x to both sides) (3/4)x + y = -5 Now you cannot have fractions as your numbers, so to get rid of the fraction, you are going to multiply the whole equation by 4. 4[(3/4)x + y = - 5] Distributing the 4 you get: 3x + 4y = -20

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