UPSC NDA I 2015 Mathematics Solved Paper

Consider the vertices of the triangle
A=(−2,3)
B=(2,1)
C=(1,2)
Therefore, the equation of the line passing through B and C will be

y−1

x−2

=

y−2

x−1

xy−x−y+1=xy−2x−2y+4
x+y=3
y=−x+3
Therefore, equation of the line perpendicular to BC and passing through A will give the equation of the altitude from A. Now slope of the side BC is -1.
Hence, slope of the altitude from A to BC will be 1.
Thus,

y−3

x+2

=1
y−3=x+2
y−x=5
Solving the equations of the side BC and the altitude from A will give us the point of intersection of the altitude and the side BC.
Hence
y−x=5...(1)
x+y=3...(2)
2y=8
y=4
x=−1
Thus the foot of the perpendicular from A to BC is given by P.
y=4
x=−1
p=(−1,4)