The line L has equation of 4x + 3y = 7. Find the equation of the line perpendicular to L which passes through the point of intersection of L and the y-axis. Give answer in ax + by + c = 0.
slope of L: -4/3
slope of ┴: 3/4
L y-intercept: 7/3
so, now you have a point and a slope:
y-7/3 = 3/4(x-0)
To find the equation of the line perpendicular to line L, we first need to determine the slope of line L.
The given equation of line L is 4x + 3y = 7. To find the slope, we need to rearrange the equation in slope-intercept form (y = mx + b), where m represents the slope.
Let's solve for y in terms of x:
4x + 3y = 7
3y = -4x + 7
y = (-4/3)x + 7/3
From this equation, we can see that the slope of line L is -4/3.
Since the line perpendicular to line L has a slope that is negative reciprocal to the slope of L, we can determine its slope to be 3/4.
To find the point of intersection between line L and the y-axis, we set x = 0 in the equation of line L:
4(0) + 3y = 7
3y = 7
y = 7/3
Therefore, the point of intersection is (0, 7/3).
Now we have the slope of the perpendicular line (3/4) and a point it passes through (0, 7/3), so we can use the point-slope form of a line to find the equation of this line.
The point-slope form is given as: y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope of the line.
Plugging in the values, we have:
y - (7/3) = (3/4)(x - 0)
Simplifying:
y - 7/3 = (3/4)x
To convert this equation to the standard form (ax + by + c = 0), we multiply every term by 12 to eliminate fractions:
12y - 28 = 9x
Rearranging the equation:
9x - 12y + 28 = 0
Therefore, the equation of the line perpendicular to line L, passing through the point of intersection of L and the y-axis, in the standard form ax + by + c = 0 is:
9x - 12y + 28 = 0.