# The point p(2,9)and(a,5)and(5,5)or the verticies of a triangle PQRright angle at Q.find the value of A and hence the arc of triangle PQR

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P (2, 9) Q ( a, 5) R (5, 5)

PQR is a right angle triangle. Q= 90 deg. So PQ is perpendicular to QR.

slope of PQ * slope of QR = -1

slope of QR = 0 => slope of QP is infinity. Hence (9-5)/(2-a) = infinity.

a = 2.

you see that QR is parallel to x axis. PQ is perpendicular t o x axis.

base = QR =3

height = PQ = 4

area = 1/2 * 3 * 4 = 6

Area =

PQR is a right angle triangle. Q= 90 deg. So PQ is perpendicular to QR.

slope of PQ * slope of QR = -1

slope of QR = 0 => slope of QP is infinity. Hence (9-5)/(2-a) = infinity.

a = 2.

you see that QR is parallel to x axis. PQ is perpendicular t o x axis.

base = QR =3

height = PQ = 4

area = 1/2 * 3 * 4 = 6

Area =