Question
Mathematics Question on Coordinate Geometry
Let A(-1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of △APB is 10. If the locus of P is ax+by=15, then 5a+2b is:
A
512
B
−56
C
4
D
6
Answer
512
Explanation
Solution
The points are:
A(−1,1),B(2,3),P(h,k).
The area of △PAB is given as 10. Using the determinant formula for the area of a triangle:
Area=21h −1 2k13111.
Equating this to 10:
21h −1 2k13111=10.
Expand the determinant:
21[h(1−3)−k(−1−2)+1(−3−2)]=10, 21[−2h+3k−5]=10.
Simplify:
−2h+3k−5=20, −2h+3k=25.
The locus of P is obtained by replacing h with x and k with y:
−2x+3y=25.
Rewriting in the form ax+by=15:
5−2x+53y=15, −52x+53y=15.
Here:
a=−56,b=59.
Finally, calculate 5a+2b:
5a+2b=5(−56)+2(59), 5a+2b=−6+518=5−30+18=5−12.
Thus, the value of 5a+2b is:
5−12.