PDF chapter test TRY NOW

1. Find the coordinates of the point which divides the line segment joining the points \((3, -5)\) and \((-3, 4)\) internally, in the ratio \(1 : 2\).
 
Solution:
 
Let \(A(x_1, y_1) = (3, -5)\) , \(B(x_2, y_2) = (-3, 4)\) and \(P(x, y)\) be the required point.
 
Ratio, \((m : n) = 1 : 2\).
 
Section formula:
 
P(x,y)=mx2+nx1m+n,my2+ny1m+n
 
=1×3+2×31+2,1×4+2×51+2
 
=3+63,4103
 
=33,63
 
\(=\) \((1, -2)\)
 
Therefore, the coordinates of the required point are \((1, -2)\).
 
 
2. Find the ratio in which the line segment joining the points \((– 3, 10)\) and \((6, – 8)\) is divided by \((– 1, 6)\).
 
Solution:
 
Let \(A(x_1, y_1) = (-3, 10)\) , \(B(x_2, y_2) = (6, -8)\) and \(P(x, y) = (-1, 6)\).
 
The ratio \(m : n\) can also be written as mn:1.
That is \(k : 1\), where k=mn.
 
If the point \(P(x, y)\) divides the line segment in the ratio \(k : 1\), then the coordinates are:
 
P(x,y)=kx2+x1k+1,ky2+y1k+1
 
1,6=k×6+3k+1,k×8+10k+1
 
1,6=6k3k+1,8k+10k+1
 
Equate the coordinate of \(x\) values.
 
1=6k3k+1
 
\(-1(k + 1) = 6k - 3\)
 
\(-k - 1 = 6k - 3\)
 
\(-1 + 3 = 6k + k\)
 
\(2 = 7k\)
 
k=72
 
So, k:1=72:1=7:2
 
Therefore, the ratio is \(7:2\).