Math  /  Algebra

QuestionUse the given conditions to write an equation for the line in point-slope form and general form. Passing through (6,1)(-6,1) and parallel to the line whose equation is 5x2y3=05 x-2 y-3=0
The equation of the line in point-slope form is \square (Type an equation. Use integers or fractions for any numbers in the equation.) The equation of the line in general form is \square =0=0. (Type an expression using xx and yy as the variables. Simplify your answer. Use integers or fractions for

Studdy Solution
Convert the point-slope form equation to general form.
Start with the point-slope equation: y1=52(x+6) y - 1 = \frac{5}{2}(x + 6)
Distribute the slope: y1=52x+15 y - 1 = \frac{5}{2}x + 15
Add 1 to both sides: y=52x+16 y = \frac{5}{2}x + 16
Multiply through by 2 to eliminate the fraction: 2y=5x+32 2y = 5x + 32
Rearrange to get the general form Ax+By+C=0Ax + By + C = 0: 5x2y+32=0 5x - 2y + 32 = 0
The equation of the line in general form is: 5x2y+32=0 5x - 2y + 32 = 0

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