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Find the equation of the line perpendicular to 5x+3y=4 at the point where this line crosses the x-axis

Find the equation of the line perpendicular to 5x+3y=4 at the point where this line crosses the x-axis

Asked by: Guest | Views: 44
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Guest [Entry]

"Find the slope of the first line when y = mx + b and m=slope

5x + 3y = 4

3y = -5x + 4

3y/3 = -5x/3 + 4/3

y = -5x/3 + 4/3

Slope of first line, m=-5/3

The product of slopes of two perpendicular line is -1.

If slope of first line is -5/3, then the  slope of the other line perpendicular to it is 3/5.

(-5/3) (3/5) = -1

Find the x-intercept:

5x + 3(0) = 4

5x/5 = 4/5

x-intercept = 4/5

The point on the x-axis: (4/5, 0)

The slope, m = 3/5

Find the equation of the other line:

y-y₁ = m(x-x₁)

y - 0 = 3/5 (x - 4/5)

y = 3x/5 - 12/25   slope-intercept form

LCD: 25

(25) [y = 3x/5 - 12/25](25)

25y = 15x - 12

ANSWER:  15x - 25y = 12   Equation of the line in standard form"