The line l1 has equation 2x+5y=10. The line l2 passes through the point A(-9,-6) and is perpendicular to the line l1.

Question

a) Find the equation of the line l2.

b) Given that the lines l1 and l2 intersect at the point B, find the area of triangle ABO, where O is the origin.

I'm not sure how to solve part b. How can I calculate the area of the triangle based on the given information?

Answer 1

2x+5y =10

y = -2x/5 +2

has slope =-2/5

a perpendicular line has a negative inverse slope = 5/2

y = 5x/2 + b

-6=5(-9)/2 + b

b = -6 +45/2 = (45-12)/2 = 33/2 = 16 1/2

y =5x/2 +33/2 is the perpendicular line through (-9,-6)

the perpendicular lines intersect when

-2x/5 + 2 = 5x/2 +33/2

5x/2 +2x/5 = 2 -33/2

25x/10 + 4x/10 = 4/2 -33/2

21x/10 = -29/2=-145/10

x = (-145/10)(10/21) = -145/21= about -6.9

y = (-5/2)(-145/21) + 33/2

y= 725/42 + 693/42 = 1418/42 = 709/21

= 33 16/21= about 33.76

point of intersection is (-145/21, 709/21)

you could calculate the length of each side, given the 3 vertices, using the distance formula, a version of the Pythagorean Theorem,

then use Heron's formula

Area = sqr((p)(p-a)(p-b)(p-c)) where p= half the perimeter, and a, b and c are the 3 sides

no guarantees there are no errors in the above calculations, but that's a basic method for a solution