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How to determine the equation pf a line?

Line L1 passes the center of the circle (x - 3) 2 + (y - 2) 2 = 4 and is perpendicular to the line L2, 2x - 3y = 0. Find the equation of the line L1.

how to find line equation if this line passes the center of a circle and perpendicular to another line?

Solution:

The line L1 has the equation y = k1x + b, in which k1is the slope of the Line L1 and b is the y-intercept of the line L1.

Because line L1 is perpendicular to line L2 and the line L2 has the equation 2x - 3y = 0. Now we find the slope of the line L2. Let k2 be the slope of the line L2.

2x - 3y = 0
2x = 3y
y = (2/3)x
so, k2 = 2/3

Because line L1 is perpendicular to line L2, let k1 be the slope of the line L1

k1 = - 1/k2
= -1 ÷ (2/3)
= -1 × 3/2
= -3/2

so, the equation of the line L1 is y = -(3/2) x + b. Now we determine the value of b.

Because the line L1 passes the center of the circle. Now we need to find the coordinate of the center of the circle. The circle equation is, (x - 3) 2 + (y - 2) 2 = 4. Compare the standard circle equation (x - x0) 2 + (y - y02) = r2, in which x0 is the x-coordinate of the center of the circle, y0 is the y-coordinate of the center of the circle. So, the x-coordinate of the center of the circle is 3 and y-coordinate of the center of the circle is 2. Because the center of the circle lies on the line L1 we substitute the center of the circle into the equation of line L1 to get the value of b.

2 = -(3/2) × 3 + b
b = 2 + 9/2
= 2 × (2/2) + 9/2
= 4/2 + 9/2
= 13/2

Therefore, the equation of the line L1 is y = - (3/2) x + 13/2. Watch the video for more details.