How to write this equation on standard form?
2015-08-31 5:26 am
What would the equation be in standard form that passes through (4,5) and is perpendicular to the line that passes through (1,-4) and (7,3)?
回答 (3)
2015-08-31 6:29 am
The equation of the line through (x₀,y₀), perpendicular to the line through (x₁,y₁) and (x₂,y₂) is:
(x₂-x₁)x + (y₂-y₁)y - (x₂-x₁)x₀ - (y₂-y₁)y₀ = 0
Here this is:
(7-1)x + (3--4)y - (7-1)4 - (3--4)5 = 0
6x + 7y - 24 - 35 = 0
6x + 7y - 59 = 0
(x₂-x₁)x + (y₂-y₁)y - (x₂-x₁)x₀ - (y₂-y₁)y₀ = 0
Here this is:
(7-1)x + (3--4)y - (7-1)4 - (3--4)5 = 0
6x + 7y - 24 - 35 = 0
6x + 7y - 59 = 0
2015-08-31 5:32 am
Slope of the line that passes through (1, -4) and (7, 3)
= (3 + 4)/(7 - 1)
= 7/6
Slope of the required line
= -1/(7/6)
= -6/7
The required line passes (4, 5) and with a slope of -6/7.
The equation of the required line is :
(y - 5) = (-6/7) (x - 4)
7(y - 5) = -6(x - 4)
7y - 35 = -6x + 24
6x + 7y - 59 = 0
= (3 + 4)/(7 - 1)
= 7/6
Slope of the required line
= -1/(7/6)
= -6/7
The required line passes (4, 5) and with a slope of -6/7.
The equation of the required line is :
(y - 5) = (-6/7) (x - 4)
7(y - 5) = -6(x - 4)
7y - 35 = -6x + 24
6x + 7y - 59 = 0
2015-08-31 5:28 am
Determine the slope of the line through (1,-4) and (7,3).
rise = -4 - 3 = -7
run = 1 - 7 = -6
slope = rise/run = 7/6
slope of any perpendicular to the line is -6/7
point-slope equation of the perpendicular:
y-5 = (-6/7)(x-4)
You can finish by converting the equation to standard form.
rise = -4 - 3 = -7
run = 1 - 7 = -6
slope = rise/run = 7/6
slope of any perpendicular to the line is -6/7
point-slope equation of the perpendicular:
y-5 = (-6/7)(x-4)
You can finish by converting the equation to standard form.
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