# C1 coordinate geometry area of triangle questionWatch

Thread starter 3 years ago
#1
So I was doing this question....

"The line l1 passes through the points P(–1, 2) and Q(11, 8).

(a) Find an equation for l1 in the form y = mx + c, where m and c are constants.

The line l2 passes through the point R(10, 0) and is perpendicular to l1. The lines l1 and l2intersect at the point S.

(b) Calculate the coordinates of S.

(c) Show that the length of RS is 3√5.

(d) Hence, or otherwise, find the exact area of triangle PQR. "

Can someone explain why for part (d), you can't just use 1/2 x b x h, where the base is length QR and the height is length PR? I drew a diagram and this is what looked sensible, but it got me the wrong answer. The mark scheme says you have to use length RS and then work out length PQ which gives you another surd, but what does length RS have to do with triangle PQR?

Thanks
0
3 years ago
#2
(Original post by jessyjellytot14)
So I was doing this question....

"The line l1 passes through the points P(–1, 2) and Q(11, 8).

(a) Find an equation for l1 in the form y = mx + c, where m and c are constants.

The line l2 passes through the point R(10, 0) and is perpendicular to l1. The lines l1 and l2intersect at the point S.

(b) Calculate the coordinates of S.

(c) Show that the length of RS is 3√5.

(d) Hence, or otherwise, find the exact area of triangle PQR. "

Can someone explain why for part (d), you can't just use 1/2 x b x h, where the base is length QR and the height is length PR? I drew a diagram and this is what looked sensible, but it got me the wrong answer. The mark scheme says you have to use length RS and then work out length PQ which gives you another surd, but what does length RS have to do with triangle PQR?

Thanks
you're supposed to use 0.5xbxh. the height isn't PR the height is RS, If you want to draw a rough sketch, feel free to and you'll find that RS is the height ^-^
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