Lines 1 2 and 3 meet at the same point and line 3 is perpendicular to line 2 If the first two lines have equations 1x plus 2y equals 7 and line 2 3x plus 3y equals 12 Find an equation for line 3?

Answer 1

Write the equation of l1 and l2 in the slope-intercept form: l1 : x + 2y = 7 subtract 3x to both sides:

2y = -x + 7 divide by 2 to both sides:

y = -x/2 + 7/2

l2 :

3x + 3y = 12 subtract 3x to both sides:

3y = -3x + 12 divide by 3 to both sides:

y = -x + 4, where slope is -1. Since l3 is perpendicular to l2 then the slope of l3 is the negative reciprocal of the slope of l2 (m2 = -1). So that m3 = -(1/-1) = 1. Since the three lines intercept at the same point, let's find this point of intersection and use it to write the equation of l3 in the point-slope form. y = -x/2 + 7/2

y = -x + 4 subtract both equations: 0 = x/2 -1/2 multiply by 2 to both sides:

0 = x - 1 add 1 to both sides:

1 = x Substitute 1 for x at the second equation: y = -x + 4

y = -1 + 4

y = 3 So that the point of the intersection is (1, 3). Using this point and the slope m = 1, we write: (y - y1) = m(x - x1)

(y - 3) = 1(x -1)

y -3 = x - 1 add 3, and subtract x to both sides:

-x + y = 2

Thus the equation of l3 is -x + y = 2