In △abc, point p∈ ab is so that ap: bp=1: 3 and point m is the midpoint of segment cp . find the area of △abc if the area of △bmp is equal to 21m2.

56 m²

A diagram can be helpful.

Triangles with the same altitude will have areas proportional to the length of their bases.

The altitude from B to PC is the same for triangles BMP and BMC, so they have areas that are in the same proportion as MP : MC. Since M is the midpoint of CP, MP = MC and ABMP = ABMC = 21 m². Then ...

... ACPB = 21 m² + 21 m² = 42 m²

The altitude from C to AB is the same for triangles CPA and CPB, so those triangles have areas in the sampe proportion as AP : BP = 1 : 3. Then ...

... ACPA : ACPB = PA : PB = 1 : 3

... ACPA : 42 m² = 1 : 3

So, the area of ∆CPA is 1/3 of 42 m², or 14 m². The area of ABC is the sum of the areas of CPA and CPB, so is ...

... AABC = ACPA + ACPB = 14 m² + 42 m²

... AABC = 56 m²

the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

  -3         -3  

y=-4x+5