 RS Aggarwal Class 10 Solutions Chapter 16 - Coordinate Geometry Ex 16E (16.5)

RS Aggarwal Class 10 Chapter 16 - Coordinate Geometry Ex 16E (16.5) Solutions Free PDF

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 $\Delta ABC$. Then

(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 $\Delta ABC$. Then

(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 $\Delta ABC$. Then

(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 $\Delta ABC$. Then

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

Co-ordinates of midpoint of BC = Q(x2, y2) = $\left(\frac{4+2}{2},\frac{3+5}{2}\right )$ = (3, 4)

Co-ordinates of midpoint of AC = P(x3, y3) = $\left(\frac{2+2}{2},\frac{1+5}{2}\right )$ = (2, 3)

Now,

Area of $\Delta PQR$ = $\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x{3}(y_{1}-y_{2})\right\}$

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

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

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

= $\frac{1}{2}\left \{2\right \}$

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

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)

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

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

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

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

= $\frac{1}{2}\left \{10\right \}$

= 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 $\Delta ABC$ into two triangles of equal areas

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

$\frac{1+x_{2}}{2}=2$

$\Rightarrow$ x2 = 3

$\frac{-4+y_{2}}{2}= -1$

$\Rightarrow$ y2 = 2

$\frac{1+x_{3}}{2}= 0$

$\Rightarrow$ x3 = -1

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

$\Rightarrow$ y3 = 2

Let A (x1, y1) = A (1, -4), B (x2, y2) = B (3, 2) and C (x3, y3) = C (-1, 2). Now

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

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

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

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

= $\frac{1}{2}\left \{24\right \}$

= 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

$\frac{x+8}{2}= \frac{6+9}{2}$

$\Rightarrow$ x = 7

$\frac{y+2}{2}= \frac{1+4}{2}$

$\Rightarrow$ y = 3

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 ($8,\frac{7}{2}$) and D (x3, y3) = D (7, 3). Now

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

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

= $\frac{1}{2}*\frac{3}{2}$

= $\frac{3}{4}$ sq. unit

Thus, Area of triangle ADE = $\frac{3}{4}$ sq.units

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 (x1, y1) = A (1, -3), B (x2, y2) = B (4, p) and C (x3, y3) = C (-9, 7). Now

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

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

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

$\Rightarrow |10p + 60| = 30$

Therefore

$\Rightarrow$ 10p + 60 = -30 or 30

$\Rightarrow$ 10p = -90 or -30

$\Rightarrow$ p = -9 or -3

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

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

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

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

$\Rightarrow k = 3$

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

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

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

$\Rightarrow 28 + 5k + 18k + 9 = 106$

$\Rightarrow 37 + 23k = 106$

$\Rightarrow k = \frac{69}{23}$

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,

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

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

$\Rightarrow$ (-3) (-3) + 7 (-3) + x (6)

$\Rightarrow$ 9 -21 + 6x = 0

$\Rightarrow$ 6x – 12 = 0

$\Rightarrow$ 6x = 12

$\Rightarrow$ x = $\frac{12}{6}$

$\Rightarrow x = 2$

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,

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

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

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

$\Rightarrow$ y – 16 + 36 -12 + 3y = 0

$\Rightarrow$ 8 + 4y = 0

$\Rightarrow$ 4y = -8

$\Rightarrow$$y=\frac{-8}{4}=-2$

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,

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

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

$\Rightarrow$ –3y -15 -28 + 36 – 4y = 0

$\Rightarrow$ 7y = 36 – 43

$\Rightarrow$ y = -1

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,

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

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

$\Rightarrow$ –16k + 40 -18 + k + 2k2 = 0

$\Rightarrow$ 2k2 – 15k + 22 = 0

$\Rightarrow$ 2k2 – 11k – 4k + 22 = 0

$\Rightarrow$ k(2k – 11) – 2(2k – 11) = 0

$\Rightarrow$ (k-2) (2k-11) = 0

$\Rightarrow k=2 or k = \frac{11}{2}$

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,

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

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

$\Rightarrow$ 2y -10 + 4x + 7 – 7y = 0

$\Rightarrow$ 4x – 5y – 3 = 0

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,

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

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

$\Rightarrow$ 7x – 5x -25 + 5y -4y + 28 = 0

$\Rightarrow$ 2x + y + 3 = 0

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,

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

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

$\Rightarrow$ ab – a – b = 0

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,

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

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

$\Rightarrow$ -3b -15 -14a +36 -4b = 0

$\Rightarrow$ -14a -7b +21 = 0

$\Rightarrow$ 2a + b = 3

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

Hence, a = 2 and b = -1′

Practise This Question

Which of these chemical equations is not balanced?