Co-ordinate Geometry: equation of l1 is y-x=2, of l2 is 2y+3x=-24; find PQR's area

[sojeee]

The equation of l₁ y-x=2

The equation of l ₂ 2y+3x=-24

The point of intersection is (-5.6, -3.6) this is the point, P
l₁ intersects the x-axis at Q, and l₂ intersects the x-axis at R

Find the area of triangle PQR.

My working:
y=0
0 = x + 2
x = -2

y=0
2 x 0 + 3x = -24
x = -8

QP^2 = 3.6^2 + 3.6^2
QP = 18sqrt2/5

RP^2 = 2.4^2 + 3.6^2
RP = 6sqrt13/5

Area = 1/2 x 18sqrt2/5 x 6sqrt13/5
= 11.01 (2.d.p)

However, the answer in the book is 10.8 and I cannot figure out what I have done wrong unless the answer is wrong. Please help.

 

[HallsofIvy]

That is not the correct formula for the area of a triangle! You have found the lengths of the side from Q to P and the side from R to P. But the formula for the area of a triangle is 1/2 height times base. The "base" is a side of the triangle but (unless the triangle is a right triangle which is not the case here) the "height" is not.

Yes, P is (-5.6, -2.6), Q is (-2, 0), and R is (-8, 0). The length of side RQ is |-8-(-2)|= 6. A good reason to use that, on the x-axis, is that the height is the y coordinate of the third point, P.

 

[pka]

The equation of l₁ y-x=2
The equation of l ₂ 2y+3x=-24
The point of intersection is (-5.6, -3.6) this is the point, P
l₁ intersects the x-axis at Q, and l₂ intersects the x-axis at R
Find the area of triangle PQR.
My working:
y=0
0 = x + 2
x = -2

y=0
2 x 0 + 3x = -24
x = -8
However, the answer in the book is 10.8 and I cannot figure out what I have done wrong unless the answer is wrong. Please help.

Did you actually plot the two lines and mark off the triangle?

Had you done, you would see that the base is six & the height is 3.6.
Now draw it out.