CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 16

Question. Find area of a triangle whose vertices are
(i) \( (2,7) \), \( (3,-1) \) and \( (-5,6) \)
(ii) \( (4,4) \), \( (3,-16) \) and \( (3,2) \)
Answer:
(i) \( 28.5 \text{ sq. units} \)
(ii) \( 7 \text{ sq. units} \)

Question. Find the area of quadrilateral whose vertices are
(i) \( A(1,1) \), \( B(7,-3) \), \( C(12,2) \), \( D(7,21) \)
(ii) \( A(0,0) \), \( B(1,3) \), \( C(2,5) \), \( D(-1,4) \)
Answer:
(i) \( 132 \text{ sq. units} \)
(ii) \( 7 \text{ sq. units} \)

Question. Prove that the area of triangle whose vertices are \( (p, p-2) \), \( (p+2, p+2) \) and \( (p+3, p) \) is independent of \( p \).
Answer: Proof:
Let the vertices of the triangle be \( A(x_1, y_1) = (p, p-2) \), \( B(x_2, y_2) = (p+2, p+2) \), and \( C(x_3, y_3) = (p+3, p) \).
The area of a triangle with these vertices is given by:
\[ \text{Area} = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \]
Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} | p((p+2) - p) + (p+2)(p - (p-2)) + (p+3)((p-2) - (p+2)) | \]
\[ = \frac{1}{2} | p(2) + (p+2)(2) + (p+3)(-4) | \]
\[ = \frac{1}{2} | 2p + 2p + 4 - 4p - 12 | \]
\[ = \frac{1}{2} | -8 | = 4 \text{ sq. units} \]
Since the area of the triangle is \( 4 \text{ sq. units} \), which is a constant and does not contain \( p \), it is independent of \( p \).

Question. Find \( K \) so that the points \( A(k,1) \), \( B(2,1) \) and \( C(5,-1) \) are collinear [means area = 0]
Answer: \( k = 2 \)

Question. If the points \( (a,0) \), \( (0,b) \) and \( (1,1) \) are collinear, Show that \( \frac{1}{a} + \frac{1}{b} = 1 \)
Answer: Proof:
Since the points \( A(a,0) \), \( B(0,b) \), and \( C(1,1) \) are collinear, the area of the triangle formed by them must be zero:
\[ \frac{1}{2} | a(b - 1) + 0(1 - 0) + 1(0 - b) | = 0 \]
\[ | ab - a - b | = 0 \]
\[ ab - a - b = 0 \]
\[ ab = a + b \]
Dividing both sides by \( ab \), we get:
\[ 1 = \frac{a}{ab} + \frac{b}{ab} \]
\[ \frac{1}{a} + \frac{1}{b} = 1 \]

Question. Find the condition that the point \( (x,y) \) may lie on the line joining the points \( (3,4) \) and \( (1,2) \)
Answer: \( x - y + 1 = 0 \)

Question. If the vertices of a triangle are \( (1,k) \), \( (4,-3) \) and \( (-9,7) \) and its area is 15 square units. Find the value of \( k \).
Answer: \( k = -3, \frac{21}{13} \)

Question. Find \( k \) if the points \( A(2,3) \), \( B(4,k) \) and \( C(6,-3) \) are collinear
Answer: \( k = 0 \)

Question. Find the area of the triangle formed by joining the mid points of the sides of a triangle whose vertices are \( A(0,4) \), \( B(-1,3) \) and \( C(2,8) \). Find the ratio of this area to the area of the given triangle
Answer: Area is \( \frac{5}{8} \text{ sq. units} \) and the ratio is \( 1:4 \).

Question. Find a relation between \( x \) and \( y \) if the points \( (x,y) \), \( (1,2) \) and \( (7,0) \) are collinear.
Answer: \( x + 3y = 7 \)