**1) Find the area of \(\Delta ABC\) whose vertices are:**

**(i) A(1, 2), B(-2, 3) and C(-3, -4)**

**(ii) A(-5, 7), B(-4, -5) and C(4, 5)**

**(iii) A(3, 8), B(-4, 2) and C(5, -1)**

**(iv) A(10, -6), B(2, 5) and C(-1, 3)**

**i) **A (1, 2), B (-2, 3) and C (-3, -4) are the vertices of \(\Delta ABC\)

(x_{1} = 1, y_{1} = 2), (x_{2} = -2, y_{2} = 3), (x_{3} = -3, y_{3} = -4)

Area of triangle ABC

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ 1(3-(-4))+(-2)(-4-2)+(-3)(2-3) \right \}\)

= \(\frac{1}{2}*\left\{1(3+4)-2(-6)-3(-1)\right\}\)

= \(\frac{1}{2}\left \{7+12+3\right \}\)

= \(\frac{1}{2}\left \{22\right \}\)

= 11sq.units

**ii) A(-5, 7), B(-4, -5) and C(4, 5)**

A (-5, 7), B (-4, -5) and C (4, 5) are the vertices of \(\Delta ABC\)

(x_{1} = -5, y_{1} = 7), (x_{2} = -4, y_{2} = -5), (x_{3} = 4, y_{3} = 5)

Area of triangle ABC

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ -5 (-5–5) + (-4)(5-7) + 4 (7-(-5)) \right \}\)

= \(\frac{1}{2}*\left\{-5 (-10) -4 (-2) + 4 (12)\right\}\)

= \(\frac{1}{2}\left \{50 + 8 + 48\right \}\)

= \(\frac{1}{2}\left \{106\right \}\)

= 53 sq.units

**iii) A(3, 8), B(-4, 2) and C(5, -1)**

A (3, 8), B (-4, 2) and C (5, -1) are the vertices of \(\Delta ABC\)

(x_{1} = 3, y_{1} = 8), (x_{2} = -4, y_{2} = 2), (x_{3} = 5, y_{3} = -1)

Area of triangle ABC

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ 3 (2 – (-1)) + (-4)(-1 – 8) + 5 (8 – 2)) \right \}\)

= \(\frac{1}{2}*\left\{3 (2 + 1) – 4 (-9) + 5 (6)\right\}\)

= \(\frac{1}{2}\left \{9 + 36 + 30\right \}\)

= \(\frac{1}{2}\left \{75\right \}\)

= 37.5 sq.units

**iv) A(10, -6), B(2, 5) and C(-1, 3)**

A (10, -6), B (2, 5) and C (-1, -3) are the vertices of \(\Delta ABC\)

(x_{1} = 10, y_{1} = -6), (x_{2} = 2, y_{2} = 5), (x_{3} = -1, y_{3} = 3)

Area of triangle ABC

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ 10 (5 -3) + (2)(3 – (-6)) + (-1) (-6 – 5) \right \}\)

= \(\frac{1}{2}*\left\{10 (2 ) + 2 (9) – 1 (11)\right\}\)

= \(\frac{1}{2}\left \{20 + 18 + 11\right \}\)

= \(\frac{1}{2}\left \{49\right \}\)

= 24.5 sq.units

**2) Find the area of quadrilateral ABCD whose vertices are A(3, -1), B(9, -5), C(14, 0) and D(9, 19).**

By joining A and C, we get two triangles ABC and ACD

Let A (3, -1), B (9, -5) and C (14, 0) and D (9, 19)

(x_{1} = 3, y_{1} = -1), (x_{2} = 9, y_{2} = -5), (x_{3} = 14, y_{3} = 0), (x_{4} = 9, y_{4} = 19)

Then

Area of triangle ABC

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ 3 (-5 -0) + (9)(0 + 1) + (14) (-1 + 5) \right \}\)

= \(\frac{1}{2}*\left\{3 (-5 ) + 9 (1) + 14 (4)\right\}\)

= \(\frac{1}{2}\left \{-15 + 9 + 56\right \}\)

= \(\frac{1}{2}\left \{50\right \}\)

= 25 sq.units

Area of triangle ACD

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ 3 (0 -19) + (14)(19 + (1)) + (9) (-1 – 0)) \right \}\)

= \(\frac{1}{2}*\left\{3 (-19 ) + 14 (20) + 9 (-1)\right\}\)

= \(\frac{1}{2}\left \{-57 + 280 – 9\right \}\)

= \(\frac{1}{2}\left \{214\right \}\)

= 107 sq.units

So, the area of the quadrilateral is 25 + 107 = 132 sq. units

**3) Find the area of quadrilateral PQRS whose vertices are P(-5, -3), Q (- 4, – 6) and R (2, -3) and S (1, 2)**

By joining P and R, we get two triangles PQR and PRS

Let P (-5, -3), Q (- 4, – 6) and R (2, -3) and S (1, 2)

(x_{1} = -5, y_{1} = -3), (x_{2} = – 4, y_{2} = – 6), (x_{3} = 2, y_{3} = -3), (x_{4} = 1, y_{4} = 2)

Then

Area of triangle PQR

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ -5 (-6 + 3) -4(-3 + 3) + 2 (-3 + 6) \right \}\)

= \(\frac{1}{2}*\left\{-5 (-3 ) – 4 (0) + 2 (3)\right\}\)

= \(\frac{1}{2}\left \{15 – 0 + 6\right \}\)

= \(\frac{1}{2}\left \{21\right \}\)

= 10.5 sq.units

Area of triangle PRS

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ -5 (-3 -2) + 2(2 + 3) + 1 (-3 + 3) \right \}\)

= \(\frac{1}{2}*\left\{-5 (-5 ) + 2 (5) + 1 (0)\right\}\)

= \(\frac{1}{2}\left \{25 + 10 + 0\right \}\)

= \(\frac{1}{2}\left \{35\right \}\)

= 17.5 sq.units

So, the area of the quadrilateral is 10.5 + 17.5 = 28 sq. units

**4) Find the area of quadrilateral ABCD whose vertices are A(-3, -1), B(-2, -4), C(4, -1) and D(3, 4).**

By joining A and C, we get two triangles ABC and ACD

Let A (-3, -1), B (-2, -4) and C (4, -1) and D (3, 4)

(x_{1} = -3, y_{1} = -1), (x_{2} = -2, y_{2} = -4), (x_{3} = 4, y_{3} = -1), (x_{4} = 3, y_{4} = 4)

Then

Area of triangle ABC

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ -3 (-4 + 1) – 2(-1 + 1) + 4 (-1 + 4) \right \}\)

= \(\frac{1}{2}*\left\{-3 (-3 ) -2 (0) + 4 (3)\right\}\)

= \(\frac{1}{2}\left \{9 + 0 + 12\right \}\)

= \(\frac{1}{2}\left \{21\right \}\)

= 10.5 sq.units

Area of triangle ACD

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ -3 (-1 – 4) + 4(4 + 1) + (3) (-1 + 1) \right \}\)

= \(\frac{1}{2}*\left\{-3 (-5 ) + 4 (5) + 3 (0)\right\}\)

= \(\frac{1}{2}\left \{15 + 20 + 0\right \}\)

= \(\frac{1}{2}\left \{35\right \}\)

= 17.5 sq.units

So, the area of the quadrilateral is 10.5 + 17.5 = 28 sq. units

**5) Find the area of quadrilateral ABCD whose vertices are A (-5, 7), B (-4, -5) and C (-1, -6) and D (4, 5).**

By joining A and C, we get two triangles ABC and ACD

Let A (-5, 7), B (-4, -5) and C (-1, -6) and D (4, 5)

(x_{1} = -5, y_{1} = 7), (x_{2} = -4, y_{2} = -5), (x_{3} = -1, y_{3} = -6), (x_{4} = 4, y_{4} = 5)

Then

Area of triangle ABC

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ -5 (-5 + 6) – 4(-6 – 7) – 1 (7 + 5) \right \}\)

= \(\frac{1}{2}*\left\{-5 (1 ) -4 (-13) – 1 (12)\right\}\)

= \(\frac{1}{2}\left \{-5 + 52 – 12\right \}\)

= \(\frac{1}{2}\left \{35\right \}\)

= 17.5 sq.units

Area of triangle ACD

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ -5 (-6 – 5) – 1(5 – 7) + (4) (7 + 6) \right \}\)

= \(\frac{1}{2}*\left\{-5 (-11 ) – 1 (-2) + 4 (13)\right\}\)

= \(\frac{1}{2}\left \{55 + 2 + 52\right \}\)

= \(\frac{1}{2}\left \{109\right \}\)

= 54.5 sq.units

So, the area of the quadrilateral is 17.5 + 54.5 = 72 sq. units

**6) Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5).**

The vertices of the triangle are A (2,1), B ( 4,3) and C (2,5)

Co-ordinates of midpoint of AB = P(x_{1}, y_{1}) = \(\left(\frac{2+4}{2},\frac{1+3}{2}\right )\)

Co-ordinates of midpoint of BC = Q(x_{2}, y_{2}) = \(\left(\frac{4+2}{2},\frac{3+5}{2}\right )\)

Co-ordinates of midpoint of AC = P(x_{3}, y_{3}) = \(\left(\frac{2+2}{2},\frac{1+5}{2}\right )\)

Now,

Area of \(\Delta PQR\)

= \(\frac{1}{2}*\left \{ 3 (4 -3) + 3(3 – 2) + 2 (2 – 4) \right \}\)

= \(\frac{1}{2}*\left\{3 (1 ) + 3 (1) + 2 (-2)\right\}\)

= \(\frac{1}{2}\left \{3 + 3 – 4\right \}\)

= \(\frac{1}{2}\left \{2\right \}\)

= 1 sq.unit

Hence, the area of the required triangle is 1 sq.unit

**7) A(7, -3), B(5, 3) and C(3, -1) are the vertices of a triangle ABC and AD is its median. Prove that the median AD divides triangle ABC into two triangles of equal areas.**

The vertices of the triangle are A (7,-3), B ( 5,3) and C (3,-1)

Co-ordinates of D = \(\left(\frac{5+3}{2},\frac{3-1}{2}\right )\)

For the area of the triangle ADC, let

Let A (7, -3), D (4, 1) and C (3, -1)

(x_{1} = 7, y_{1} = -3), (x_{2} = 4, y_{2} = 1), (x_{3} = 3, y_{3} = -1)

Area of triangle ADC

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ 7 (1 + 1) + 4(-1 + 3) + 3 (-3 – 1) \right \}\)

= \(\frac{1}{2}*\left\{7 (2 ) + 4 (2) + 3 (-4)\right\}\)

= \(\frac{1}{2}\left \{14 + 8 – 12\right \}\)

= \(\frac{1}{2}\left \{10\right \}\)

= 5 sq.unit

For the area of the triangle ABD, let

Let A (7, -3), B (5, 3) and D (4, 1)

(x_{1} = 7, y_{1} = -3), (x_{2} = 5, y_{2} = 3), (x_{3} = 4, y_{3} = 1)

Area of triangle ABD

= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\frac{1}{2}*\left \{ 7 (3 – 1) + 5(1 + 3) + 4 (-3 – 3) \right \}\)

= \(\frac{1}{2}*\left\{7 (2 ) + 5 (4) + 4 (-6)\right\}\)

= \(\frac{1}{2}\left \{14 + 20 – 24\right \}\)

= \(\frac{1}{2}\left \{10\right \}\)

= 5 sq.unit

Thus, Area of triangle ADC = Area of triangle ABD = 5 sq.units

Hence, AD divides \(\Delta ABC\)

**8) Find the area of triangle ABC with A(1, -4) and midpoints of sides through A being (2, -1) and (0, -1).**

Let(x_{2}, y_{2}) and (x_{3}, y_{3}) be the co-ordinates of B and C respectively. Since, the co-ordinates of A are (1, -4), therefore

\(\frac{1+x_{2}}{2}=2\)

\(\Rightarrow\)_{2} = 3

\(\frac{-4+y_{2}}{2}= -1\)

\(\Rightarrow\)_{2} = 2

\(\frac{1+x_{3}}{2}= 0\)

\(\Rightarrow\)_{3} = -1

\(\frac{-4+y_{3}}{2}= -1\)

\(\Rightarrow\)_{3} = 2

Let A (x_{1}, y_{1}) = A (1, -4), B (x_{2}, y_{2}) = B (3, 2) and C (x_{3}, y_{3}) = C (-1, 2). Now

Area (\(\Delta ABC\)

= \(\frac{1}{2}*\left \{ 1 (2 – 2) + 3(2 + 4) -1 (-4 – 2) \right \}\)

= \(\frac{1}{2}*\left\{1 (0 ) + 3 (6) -1 (-6)\right\}\)

= \(\frac{1}{2}\left \{0 + 18 + 6\right \}\)

= \(\frac{1}{2}\left \{24\right \}\)

= 12 sq.unit

Thus, Area of triangle ABC = 12 sq.units

**9) A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of triangle ADE.**

Let(x, y) be the co-ordinates of D and (x_{’}, y_{’}) be the co-ordinates of E. Since, the diagonals of a parallelogram bisect each other at the same point, therefore

\(\frac{x+8}{2}= \frac{6+9}{2} \)

\(\Rightarrow\)

\(\frac{y+2}{2}= \frac{1+4}{2}\)

\(\Rightarrow\)

Thus, the co-ordinates of D are (7,3)

E is the midpoint of DC, therefore

x’ = \(\frac{7+9}{2}\)

x’ = 8

y’= \(\frac{3+4}{2} \)

y’ = \(\frac{7}{2} \)

Thus, the co-ordinates of E are (\(8,\frac{7}{2}\)

Let A (x_{1}, y_{1}) = A (6, 1), E (x_{2}, y_{2}) = E (\(8,\frac{7}{2} \)_{3}, y_{3}) = D (7, 3). Now

Area (\(\Delta ABC\)

= \(\frac{1}{2}*\left \{ 6 (\frac{7}{2}-3) + 8(3 – 1) + 7 (1-\frac{7}{2}) \right \}\)

= \(\frac{1}{2}*\frac{3}{2}\)

= \(\frac{3}{4}\)

Thus, Area of triangle ADE = \(\frac{3}{4}\)

**10) If the vertices of \(\Delta ABC\) be A(1, -3), B(4, p) and C(-9, 7) and its area is 15 square units, find the values of p.**

Let A (x_{1}, y_{1}) = A (1, -3), B (x_{2}, y_{2}) = B (4, p) and C (x_{3}, y_{3}) = C (-9, 7). Now

Area (\(\Delta ABC\)

\(\Rightarrow 15 = \frac{1}{2}* [1 (p – 7) + 4(7 + 3) – 9 (-3-p)] \)

\(\Rightarrow 15 = \frac{1}{2}* [10p + 60] \)

\(\Rightarrow |10p + 60| = 30 \)

Therefore

\(\Rightarrow \)

\(\Rightarrow \)

\(\Rightarrow \)

Hence, p = -9 or p = -3

** 11) Find the value of K so that the area of the triangle with vertices A(k+1, 1), B(4, -3) and C(7, -k) is 6 square units.**

Let A (x_{1}, y_{1}) = A (k + 1, 1), B (x_{2}, y_{2}) = B (4, -3) and C (x_{3}, y_{3}) = C (7, -k). Now

\(\Rightarrow 6 = \frac{1}{2}* [(k+1) (-3 + k) + 4(- k – 1) + 7 (1 + 3)] \)

\(\Rightarrow 6 = \frac{1}{2}* [k^{2} – 2k – 3 – 4k – 4 + 28]\)

\(\Rightarrow (k-3)^{2} = 0\)

\(\Rightarrow k = 3 \)

Hence, k = 3

**12) For what value of k(k>0) is the area of the triangle with vertices (-2, 5), (k, -4) and (2k+1, 10) equal to 53 square units ?**

Let A (x_{1}, y_{1}) = A (-2, 5), B (x_{2}, y_{2}) = B (k, -4) and C (x_{3}, y_{3}) = C (2k +1, 10). Now

\(\Rightarrow 53 = \frac{1}{2}* [(-2) (-4 – 10) + k (10 – 5) + (2k +1) (5 + 4)]\)

\(\Rightarrow 53 = \frac{1}{2}* [28 + 5k + 9(2k + 1)]\)

\(\Rightarrow 28 + 5k + 18k + 9 = 106\)

\(\Rightarrow 37 + 23k = 106 \)

\(\Rightarrow k = \frac{69}{23} \)

Hence, k = 3

**13) Show that the following points are collinear:**

**(i) A(2, -2), B(-3, 8) and C(-1, 4)**

**(ii) A(-5, 1), B(5, 5) and C(10, 7)**

**(iii) A(5, 1), B(1, -1) and C(11, 4)**

**(iv) A(8, 1), B(3, -4) and C(2, -5)**

**i) **Let A (x_{1}, y_{1}) = A (2, -2), B (x_{2}, y_{2}) = B (-3, 8) and C (x_{3}, y_{3}) = C (-1, 4) be the given points. Now

= \(\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\left \{ 2 (8 – 4) + (-3)(4 + 2) +(-1) (-2 – 8) \right \}\)

= 8 – 18 +10

= 0

Hence, the given points are collinear.

**ii)**

Let A (x_{1}, y_{1}) = A (-5, 1), B (x_{2}, y_{2}) = B (5, 5) and C (x_{3}, y_{3}) = C (10, 7) be the given points. Now

= \(\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\left \{ -5 (5 – 7) + (5)(7 – 1) +(10) (1 – 5) \right \}\)

= 10 + 30 – 40

= 0

Hence, the given points are collinear.

**iii)**

Let A (x_{1}, y_{1}) = A (5, 1), B (x_{2}, y_{2}) = B (1, -1) and C (x_{3}, y_{3}) = C (11, 4) be the given points. Now

= \(\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\left \{ 5 (-1 – 4) + (1)(4 – 1) + (11) (1 + 1) \right \}\)

= -25 + 3 + 22

= 0

Hence, the given points are collinear.

**iv)**

Let A (x_{1}, y_{1}) = A (8, 1), B (x_{2}, y_{2}) = B (3, -4) and C (x_{3}, y_{3}) = C (2, -5). Be the given points. Now

= \(\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\left \{ 8 (- 4 + 5) + (3)(-5 – 1) +(2) (1 + 4) \right \}\)

= 8 – 18 +10

= 0

Hence, the given points are collinear.

**14) Find the value of x for which the points A(x,2), B(-3, -4) and C(7, -5) are collinear.**

Let A (x_{1}, y_{1}) = A (x, 2), B (x_{2}, y_{2}) = B (-3, -4) and C (x_{3}, y_{3}) = C (7, -5). So the condition for three collinear points is

= \(\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

= \(\left \{ x (- 4 + 5) -3(-5 – 2) + 7 (2 + 4) \right \}\)

= x + 21 + 42

x = – 63

Hence x = – 63

**15) For what value of x are the points A(-3, 12), B(7, 6) and C(x, 9) collinear ?**

A (x_{1}, y_{1}) = A (-3, 12), B (x_{2}, y_{2}) = B (7, 6) and C (x_{3}, y_{3}) = C (x, 9) are the given points.

It is given that the points A, B and C are collinear. Therefore,

\(\Rightarrow\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

\(\Rightarrow -3 (6-9) + 7 (9-12) + (x) (12-6)\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow x = 2\)

Therefore, when x = 2, the given points are collinear

**16) For what value of y are the points P(1, 4), Q(3, y) and R(-3, 16) are collinear ?**

P (x_{1}, y_{1}) = P (1, 4), Q (x_{2}, y_{2}) = Q (3, y) and R (x_{3}, y_{3}) = R (-3, 16) are the given points.

It is given that the points P, Q and R are collinear. Therefore,

\(\Rightarrow\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

\(\Rightarrow 1(y-16)+3(16-4)+(-3)(4-y)=0\)

\(\Rightarrow 1(y-16) + 3(12) -3(4-y)=0\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Therefore, when y = -2, the given points are collinear

**17) Find the value of y for which the points A(-3, 9), B(2, y) and C(4, -5) are collinear ? **

A (x_{1}, y_{1}) = A (-3, 9), B (x_{2}, y_{2}) = B (2, y) and C (x_{3}, y_{3}) = C (4, -5) are the given points.

It is given that the points A, B and C are collinear. Therefore,

\(\Rightarrow\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

\(\Rightarrow -3 (y+15) + 2(-5-9) + (4) (9-y)=0\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

**18) For what values of k are the points A(8, 1), B(3, -2k) and C(k, -5) are collinear.**

A (x_{1}, y_{1}) = A (8, 1), B (x_{2}, y_{2}) = B (3, -2k) and C (x_{3}, y_{3}) = C (k, -5) are the given points.

It is given that the points A, B and C are collinear. Therefore,

\(\Rightarrow\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

\(\Rightarrow 8 (-2k + 5) + 3(-5-1) + (k) (1+2k)=0\)

\(\Rightarrow\)^{2} = 0

\(\Rightarrow\)^{2} – 15k + 22 = 0

\(\Rightarrow\)^{2} – 11k – 4k + 22 = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow k=2 or k = \frac{11}{2}\)

Hence, k = 2 or k = \(\frac{11}{2}\)

**19) Find a relation between x and y, if the points A(2, 1), B(x, y) and C(7, 5) are collinear.**

A (x_{1}, y_{1}) = A (2, 1), B (x_{2}, y_{2}) = B (x, y) and C (x_{3}, y_{3}) = C (7, 5) are the given points.

It is given that the points A, B and C are collinear. Therefore,

\(\Rightarrow\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

\(\Rightarrow 2 (y – 5) + x (5-1) + (7) (1-y)=0\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, the required relation is 4x -5y -3 = 0

**20) Find a relation between x and y, if the points A(x, y), B(-5, 7) and C(-4, 5) are collinear.**

A (x_{1}, y_{1}) = A (x, y), B (x_{2}, y_{2}) = B (-5, 7) and C (x_{3}, y_{3}) = C (-4, 5) are the given points.

It is given that the points A, B and C are collinear. Therefore,

\(\Rightarrow\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

\(\Rightarrow x (7 – 5) + (-5) (5-y) + (-4) (y-7)=0\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, the required relation is 2x + y + 3 = 0

**21) Prove that the points A(a,0), B(0,b) and C(1, 1) are collinear, if \(\frac{1}{a}+\frac{1}{b}=1\).**

Consider the points A (x_{1}, y_{1}) = A (a, 0), B (x_{2}, y_{2}) = B (0, b) and C (x_{3}, y_{3}) = C (1, 1) are the given points.

It is given that the points A, B and C are collinear. Therefore,

\(\Rightarrow\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

\(\Rightarrow a (b – 1) + (0) (1-0) + (1) (0-b)=0\)

\(\Rightarrow\)

Dividing the equation by ab:

\(\Rightarrow1-\frac{1}{a}-\frac{1}{b}=0\)

\(\Rightarrow1-\left(\frac{1}{a}+\frac{1}{b}\right )=0\)

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)=1\)

Therefore, the given points are collinear if \(\left(\frac{1}{a}+\frac{1}{b}\right)=1\)

**22) If the points P(-3, 9), Q(a, b) and R(4, -5) are collinear and a+b = 1, find the values of a and b.**

Let A (x_{1}, y_{1}) = A (-3, 9), B (x_{2}, y_{2}) = B (a, b) and C (x_{3}, y_{3}) = C (4, -5) are the given points.

The given points are collinear if,

\(\Rightarrow\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)

\(\Rightarrow -3 (b + 5) + (a) (-5-9) + (4) (9-b)=0\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Now, solving a + b = 1 and 2a + b = 3, we get a = 2 and b = -1

Hence, a = 2 and b = -1′