Problem: find line perpendicular to 4x+7y+3=0 passing through (-2,1).
Solution: We will first find the general form of the perpendicular line. The perpendicular line has the form 7x-4y+k=0 .........................(1) by switching the coefficients of x and y, and switching the sign of one of the two coefficients. This way, the slope of (1) multiplied by that of the original equation is -1, a condition that the two lines are perpendicular. The value of k is to be determined from the given point (-2,1).
To find k, we substitute x=-2, y=1 into equation (1) and solve for k. 7(-2)-4(1)+k=0 => k=14+4=18 Therefore the required line is 7x-4y+18=0
If you know the slope of a line and a point the line passes through, you can find the equation of the line using the formula:
y - y_1 = m(x - x_1)
You are given a point the line passes through, (-2, 1). Now you need to slope of your line. You are given the equation of a line that your line is perpendicular to. The slopes of perpendicular lines are negative reciprocals. If you find the slope of the given line, you can find the slope of the perpendicular line.
Given line:
4x + 7y + 3 = 0
Solve for y:
7y = -4x - 3
y = (-4/7)x - 3/7
The slope of the given line is -4/7. The slope of the perpendicular is 7/4.
Now we can find the equation of the line using the given point and the slope we just found.
