Math  /  Geometry

Question-- webassign.net/web/Student/Assignment-Responses/submit?pos=48dep=360168108tags=au Consider the following lines. Line 1: 3x4y=123 x-4 y=12 Line 2: a line perpendicular to 3x4y=123 x-4 y=12 that contains the point (3,4)(3,-4)

Studdy Solution
Use the point-slope form to find the equation of Line 2, which passes through the point (3,4)(3, -4).
The point-slope form is: yy1=m(xx1) y - y_1 = m(x - x_1)
Substitute m=43m = -\frac{4}{3}, x1=3x_1 = 3, and y1=4y_1 = -4: y+4=43(x3) y + 4 = -\frac{4}{3}(x - 3)
Simplify: y+4=43x+4 y + 4 = -\frac{4}{3}x + 4
Subtract 4 from both sides: y=43x y = -\frac{4}{3}x
The equation of Line 2 is: y=43x y = -\frac{4}{3}x
The equation of the line perpendicular to 3x4y=123x - 4y = 12 and passing through the point (3,4)(3, -4) is:
y=43x \boxed{y = -\frac{4}{3}x}

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