Given the line y = - 3x + 6 and y = X-1, find the area of the triangle formed by their intersection point and X axis

Given the line y = - 3x + 6 and y = X-1, find the area of the triangle formed by their intersection point and X axis

The key is drawing

The area of the triangle formed by the line y = x + 4 and the line y = - x + 4 and the X axis is ()
A. 14B. 15C. 16D. 8

In the straight line y = x + 4, let y = 0, then x = - 4; let x = 0, then y = 4; therefore, the intersection of the straight line y = x + 4 and the coordinate axis is (- 4,0), (0,4); similarly, the intersection of the straight line y = - x + 4 and the coordinate axis is (4,0), (0,4). Therefore, s = 12 × 8 × 4 = 16

As shown in the figure, we know that the analytic formula of the straight line L is y = 3x + 6 (full answer)

(1) From the meaning of the title, we can know B (0,6), C (8,0),
Let the analytic expression of line L2 be y = KX + B, then 8K + B = 0b = 6,
The solution is k = - 34, B = 6,
Then the analytic expression of L2 is y = - 34x + 6;
(2) Solution 1: as shown in the figure, PD ⊥ L2 is made through P in D,
∵∠PDC=∠BOC=90°,∠DCP=∠OCB
∴△PDC∽△BOC
∴PDBO=PCBC
From the meaning of the title, we know that OA = 2, OB = 6, OC = 8
∴BC=OB2+OC2=10,PC=10-t
∴PD6=10-t10,
∴PD=35（10-t）
∴S△PCQ=12CQ��PD=12t��35（10-t）=-310t2+3t；
Solution 2: as shown in the figure, make QD ⊥ X axis at d through Q,
∵∠QDC=∠BOC=90°,∠QCD=∠BCO
∴△CQD∽△CBO
∴QDBO=QCBC
From the meaning of the title, we know that OA = 2, OB = 6, OC = 8
∴BC=OB2+OC2=10
∴QD6=t10
∴QD=35t
∴S△PCQ=12PC��QD=12（10-t）��35t=-310t2+3t；
（3）∵PC=10-t,CQ=t,
In order to make △ PCQ an isosceles triangle, CP = CQ, QC = QP, or PC = PQ should be satisfied,
When CP = CQ, t = 5 (seconds) is obtained from 10-T = t;
When QC = QP, qcbc = 12pcoc, that is, T10 = 12 (10-T) 8, t = 5013 (seconds);
When PC = PQ, 12cqoc = PCBC, that is, 12t8 = 10-t10, t = 8013 (seconds);
T = 5 or 5013 or 8013