1) Find the area of \(\Delta ABC\)
(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 \(\Delta ABC\)
(x1 = 1, y1 = 2), (x2 = -2, y2 = 3), (x3 = -3, y3 = -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 \(\Delta ABC\)
(x1 = -5, y1 = 7), (x2 = -4, y2 = -5), (x3 = 4, y3 = 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 \(\Delta ABC\)
(x1 = 3, y1 = 8), (x2 = -4, y2 = 2), (x3 = 5, y3 = -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 \(\Delta ABC\)
(x1 = 10, y1 = -6), (x2 = 2, y2 = 5), (x3 = -1, y3 = 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)
(x1 = 3, y1 = -1), (x2 = 9, y2 = -5), (x3 = 14, y3 = 0), (x4 = 9, y4 = 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)
(x1 = -5, y1 = -3), (x2 = – 4, y2 = – 6), (x3 = 2, y3 = -3), (x4 = 1, y4 = 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)
(x1 = -3, y1 = -1), (x2 = -2, y2 = -4), (x3 = 4, y3 = -1), (x4 = 3, y4 = 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)
(x1 = -5, y1 = 7), (x2 = -4, y2 = -5), (x3 = -1, y3 = -6), (x4 = 4, y4 = 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(x1, y1) = \(\left(\frac{2+4}{2},\frac{1+3}{2}\right )\)
Co-ordinates of midpoint of BC = Q(x2, y2) = \(\left(\frac{4+2}{2},\frac{3+5}{2}\right )\)
Co-ordinates of midpoint of AC = P(x3, y3) = \(\left(\frac{2+2}{2},\frac{1+5}{2}\right )\)
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)
(x1 = 7, y1 = -3), (x2 = 4, y2 = 1), (x3 = 3, y3 = -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)
(x1 = 7, y1 = -3), (x2 = 5, y2 = 3), (x3 = 4, y3 = 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(x2, y2) and (x3, y3) be the co-ordinates of B and C respectively. Since, the co-ordinates of A are (1, -4), therefore
\(\frac{1+x_{2}}{2}=2\)
\(\Rightarrow\)
\(\frac{-4+y_{2}}{2}= -1\)
\(\Rightarrow\)
\(\frac{1+x_{3}}{2}= 0\)
\(\Rightarrow\)
\(\frac{-4+y_{3}}{2}= -1\)
\(\Rightarrow\)
Let A (x1, y1) = A (1, -4), B (x2, y2) = B (3, 2) and C (x3, y3) = 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 (x1, y1) = A (6, 1), E (x2, y2) = E (\(8,\frac{7}{2} \)
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 \(\Delta ABC\)
Let A (x1, y1) = A (1, -3), B (x2, y2) = B (4, p) and C (x3, y3) = 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 (x1, y1) = A (k + 1, 1), B (x2, y2) = B (4, -3) and C (x3, y3) = C (7, -k). Now
Area (\(\Delta ABC\)
\(\Rightarrow 6 = \frac{1}{2}* [(k+1) (-3 + k) + 4(- k – 1) + 7 (1 + 3)] \)
\(\Rightarrow 6 = \frac{1}{2}* [k^{2} – 2k – 3 – 4k – 4 + 28]\)
\(\Rightarrow (k-3)^{2} = 0\)
\(\Rightarrow k = 3 \)
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 (x1, y1) = A (-2, 5), B (x2, y2) = B (k, -4) and C (x3, y3) = C (2k +1, 10). Now
Area (\(\Delta ABC\)
\(\Rightarrow 53 = \frac{1}{2}* [(-2) (-4 – 10) + k (10 – 5) + (2k +1) (5 + 4)]\)
\(\Rightarrow 53 = \frac{1}{2}* [28 + 5k + 9(2k + 1)]\)
\(\Rightarrow 28 + 5k + 18k + 9 = 106\)
\(\Rightarrow 37 + 23k = 106 \)
\(\Rightarrow k = \frac{69}{23} \)
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 (x1, y1) = A (2, -2), B (x2, y2) = B (-3, 8) and C (x3, y3) = 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 (x1, y1) = A (-5, 1), B (x2, y2) = B (5, 5) and C (x3, y3) = 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 (x1, y1) = A (5, 1), B (x2, y2) = B (1, -1) and C (x3, y3) = 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 (x1, y1) = A (8, 1), B (x2, y2) = B (3, -4) and C (x3, y3) = 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 (x1, y1) = A (x, 2), B (x2, y2) = B (-3, -4) and C (x3, y3) = 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 (x1, y1) = A (-3, 12), B (x2, y2) = B (7, 6) and C (x3, y3) = 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 (x1, y1) = P (1, 4), Q (x2, y2) = Q (3, y) and R (x3, y3) = 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 (x1, y1) = A (-3, 9), B (x2, y2) = B (2, y) and C (x3, y3) = 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 (x1, y1) = A (8, 1), B (x2, y2) = B (3, -2k) and C (x3, y3) = C (k, -5) are the given points.
It is given that the points A, B and C are collinear. Therefore,
\(\Rightarrow\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)
\(\Rightarrow 8 (-2k + 5) + 3(-5-1) + (k) (1+2k)=0\)
\(\Rightarrow\)
\(\Rightarrow\)
\(\Rightarrow\)
\(\Rightarrow\)
\(\Rightarrow\)
\(\Rightarrow k=2 or k = \frac{11}{2}\)
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 (x1, y1) = A (2, 1), B (x2, y2) = B (x, y) and C (x3, y3) = 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 (x1, y1) = A (x, y), B (x2, y2) = B (-5, 7) and C (x3, y3) = 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 (x1, y1) = A (a, 0), B (x2, y2) = B (0, b) and C (x3, y3) = 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 (x1, y1) = A (-3, 9), B (x2, y2) = B (a, b) and C (x3, y3) = 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′