If p=(2,5),and Q=(x,-7)then the possible values of x so that PQ=13 are
Answers
Answer:
-3 or 7
Step-by-step explanation:
√(x2-x1)²+(x2-x1)²
√(x2-x1)²+(x2-x1)²(√(x-2)²+(5+7)²)²=13²
√(x2-x1)²+(x2-x1)²(√(x-2)²+(5+7)²)²=13²(x-2)²+12²=13²
√(x2-x1)²+(x2-x1)²(√(x-2)²+(5+7)²)²=13²(x-2)²+12²=13²x²+4-4x+144=169
√(x2-x1)²+(x2-x1)²(√(x-2)²+(5+7)²)²=13²(x-2)²+12²=13²x²+4-4x+144=169x²-4x-21=0
√(x2-x1)²+(x2-x1)²(√(x-2)²+(5+7)²)²=13²(x-2)²+12²=13²x²+4-4x+144=169x²-4x-21=0x²+3x-7x-21=0
√(x2-x1)²+(x2-x1)²(√(x-2)²+(5+7)²)²=13²(x-2)²+12²=13²x²+4-4x+144=169x²-4x-21=0x²+3x-7x-21=0(x+3)-7(x+3)=0
√(x2-x1)²+(x2-x1)²(√(x-2)²+(5+7)²)²=13²(x-2)²+12²=13²x²+4-4x+144=169x²-4x-21=0x²+3x-7x-21=0(x+3)-7(x+3)=0(x+3)(x-7)=0
√(x2-x1)²+(x2-x1)²(√(x-2)²+(5+7)²)²=13²(x-2)²+12²=13²x²+4-4x+144=169x²-4x-21=0x²+3x-7x-21=0(x+3)-7(x+3)=0(x+3)(x-7)=0x=-3
√(x2-x1)²+(x2-x1)²(√(x-2)²+(5+7)²)²=13²(x-2)²+12²=13²x²+4-4x+144=169x²-4x-21=0x²+3x-7x-21=0(x+3)-7(x+3)=0(x+3)(x-7)=0x=-3 or
√(x2-x1)²+(x2-x1)²(√(x-2)²+(5+7)²)²=13²(x-2)²+12²=13²x²+4-4x+144=169x²-4x-21=0x²+3x-7x-21=0(x+3)-7(x+3)=0(x+3)(x-7)=0x=-3 orx=7