NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Class 10 maths ch 7 solutions Exercise: 7.1

Q1 (i) Find the distance between the following pairs of points : (2, 3), (4, 1)

Answer:

Given points: (2, 3), (4, 1)

Distance between the points will be: $(x_1,y_1)and(x_2,y_2)$

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$D=\sqrt{(4-2{)}^2+(1-3{)}^2}=\sqrt{4+4}=2\sqrt{2}$

Q1 (ii) Find the distance between the following pairs of points : (– 5, 7), (– 1, 3)

Answer:

Given points: (– 5, 7), (– 1, 3)

Distance between the points will be: $(x_1,y_1)and(x_2,y_2)$

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$D=\sqrt{(-1+5{)}^2+(3-7{)}^2}=\sqrt{16+16}=4\sqrt{2}$

Q1 (iii) Find the distance between the following pairs of points :(a, b), (– a, – b)

Answer:

Given points: (a, b), (– a, – b)

Distance between the points will be: $(x_1,y_1)and(x_2,y_2)$

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$D=\sqrt{(-a-a{)}^2+(-b-b{)}^2}=\sqrt{4(a^2+b^2)}=2\sqrt{a^2+b^2}$

Q2 Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Answer:

Given points: (0, 0) and (36, 15)

Distance between the points will be: $(x_1,y_1)and(x_2,y_2)$

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$D=\sqrt{(36-0{)}^2+(15-0{)}^2}=\sqrt{1296+225}=\sqrt{1521}=39$

The distance between the two towns A and B is, thus 39 km for given towns location

$(0,0)$ and $(36,15)$ .

Q3 Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

Answer:

Let the points (1, 5), (2, 3) and (– 2, – 11) be representing the vertices A, B, and C of the given triangle respectively.

$A=(1,5),B=(2,3),C=(-2,-11)$

Therefore,

$AB=\sqrt{(1-2{)}^2+(5-3{)}^2}=\sqrt{5}$

$BC=\sqrt{(2-(-2){)}^2+(3-(-11){)}^2}=\sqrt{4^2+{14}^2}=\sqrt{16+196}=\sqrt{212}$ $CA=\sqrt{(1-(-2){)}^2+(5-(-11){)}^2}=\sqrt{3^2+{16}^2}=\sqrt{9+256}=\sqrt{265}$ Since these are not satisfied.

$AB+BC\ne CA$

$BA+AC\ne BC$

$BC+CA\ne BA$

As these cases are not satisfied.

Hence the points are not collinear.

Q4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Answer:

The distance between two points $A(x_1,y_1)andB(x_2,y_2)$ is given by:

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

So, we have the following points: (5, – 2), (6, 4) and (7, – 2) assuming it to be the vertices of triangle A, B, and C respectively.

$AB=\sqrt{(5-6{)}^2+(-2-4{)}^2}=\sqrt{1+36}=\sqrt{37}$

$BC=\sqrt{(6-7{)}^2+(4+2{)}^2}=\sqrt{1+36}=\sqrt{37}$

$CA=\sqrt{(5-7{)}^2+(-2+2{)}^2}=\sqrt{4+0}=2$

Therefore, AB = BC

Here two sides are equal in length.

Therefore, ABC is an isosceles triangle.

Q5 In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Answer:

The coordinates of the points:

$A(3,4),B(6,7),C(9,4),andD(6,1)$ are the positions of 4 friends.

The distance between two points $A(x_1,y_1),andB(x_2,y_2)$ is given by:

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Hence,

$AB=\sqrt{(3-6{)}^2+(4-7{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

$BC=\sqrt{(6-9{)}^2+(7-4{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

$CD=\sqrt{(9-6{)}^2+(4-1{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

$AD=\sqrt{(3-6{)}^2+(4-1{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

And the lengths of diagonals:

$AC=\sqrt{(3-9{)}^2+(4-4{)}^2}=\sqrt{36+0}=6$

$BD=\sqrt{(6-6{)}^2+(7-1{)}^2}=\sqrt{36+0}=6$

So, here it can be seen that all sides of quadrilateral ABCD are of the same lengths and diagonals are also having the same length.

Therefore, quadrilateral ABCD is a square and Champa is saying right.

Q6 (i) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)

Answer:

Let the given points $(-1,-2),(1,0),(-1,2),and(-3,0)$ be representing the vertices A, B, C, and D of the given quadrilateral respectively.

The distance formula:

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$AB=\sqrt{(-1-1{)}^2+(-2-0{)}^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$

$BC=\sqrt{(1+1{)}^2+(0-2{)}^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$

$CD=\sqrt{(-1+3{)}^2+(2-0{)}^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$

$AD=\sqrt{(-1+3{)}^2+(-2-0{)}^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$

Finding the length of the diagonals:

$AC=\sqrt{(-1+1{)}^2+(-2-2{)}^2}=\sqrt{0+16}=4$

$BD=\sqrt{(1+3{)}^2+(0-0{)}^2}=\sqrt{16+0}=4$

It is clear that all sides are of the same lengths and also the diagonals have the same lengths.

Hence, the given quadrilateral is a square.

Q6 (ii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (–3, 5), (3, 1), (0, 3), (–1, – 4)

Answer:

Let the given points $(-3,5),(3,1),(0,3),and(-1,-4)$ be representing the vertices A, B, C, and D of the given quadrilateral respectively.

The distance formula:

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$AB=\sqrt{(-3-3{)}^2+(5-1{)}^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$

$BC=\sqrt{(3-0{)}^2+(1-3{)}^2}=\sqrt{9+4}=\sqrt{13}$

$CD=\sqrt{(0+1{)}^2+(3+4{)}^2}=\sqrt{1+49}=\sqrt{50}=5\sqrt{2}$

$AD=\sqrt{(-3+1{)}^2+(5+4{)}^2}=\sqrt{4+81}=\sqrt{85}$

All the sides of the given quadrilateral have different lengths.

Therefore, it is only a general quadrilateral and not a specific one like square, rectangle, etc.

Q6 (iii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (4, 5), (7, 6), (4, 3), (1, 2)

Answer:

Let the given points $(4,5),(7,6),(4,3),(1,2)$ be representing the vertices A, B, C, and D of the given quadrilateral respectively.

The distance formula:

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$AB=\sqrt{(4-7{)}^2+(5-6{)}^2}=\sqrt{9+1}=\sqrt{10}$

$BC=\sqrt{(7-4{)}^2+(6-3{)}^2}=\sqrt{9+9}=\sqrt{18}$

$CD=\sqrt{(4-1{)}^2+(3-2{)}^2}=\sqrt{9+1}=\sqrt{10}$

$AD=\sqrt{(4-1{)}^2+(5-2{)}^2}=\sqrt{9+9}=\sqrt{18}$

And the diagonals:

$AC=\sqrt{(4-4{)}^2+(5-3{)}^2}=\sqrt{0+4}=2$

$BD=\sqrt{(7-1{)}^2+(6-2{)}^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$

Here we can observe that the opposite sides of this quadrilateral are of the same length.

However, the diagonals are of different lengths.

Therefore, the given points are the vertices of a parallelogram.

Q7 Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Answer:

Let the point which is equidistant from $A(2,-5)andB(-2,9)$ be $X(x,0)$ as it lies on X-axis.

Then, we have

Distance AX: $=\sqrt{(x-2{)}^2+(0+5{)}^2}$

and Distance BX $=\sqrt{(x+2{)}^2+(0+9{)}^2}$

According to the question, these distances are equal length.

Hence we have,

$\sqrt{(x-2{)}^2+(0+5{)}^2}$ $=\sqrt{(x+2{)}^2+(0+9{)}^2}$

Solving this to get the required coordinates.

Squaring both sides we get,

$(x-2{)}^2+25=(x+2{)}^2+81$

$\Rightarrow (x-2+x+2)(x-2-x-2)=81-25=56$

$\Rightarrow (-8x)=56$

Or, $\Rightarrow x=-7$

Hence the point is $X(-7,0)$ .

Q8 Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Answer:

Given the distance between the points $P(2,-3)$ and $Q(10,y)$ is 10 units.

The distance formula :

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

So, given $PQ=10units$

$PQ=\sqrt{(10-2{)}^2+(y-(-3){)}^2}=10$

After squaring both sides

$\Rightarrow (10-2{)}^2+(y-(-3){)}^2=100$

$\Rightarrow (y+3{)}^2=100-64$

$\Rightarrow y+3=\pm 6$

$\Rightarrow y=6-3ory=-6-3$

Therefore, the values are $y=3or-9$ .

Q9 If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also, find the distances QR and PR.

Answer:

Given $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$ .

Then, the distances $PQ=RQ$ .

Distance $PQ=\sqrt{(5-0{)}^2+(-3-1{)}^2}=\sqrt{25+16}=\sqrt{41}$

Distance $RQ=\sqrt{(x-0{)}^2+(6-1{)}^2}=\sqrt{x^2+25}$

$\Rightarrow \sqrt{x^2+25}=\sqrt{41}$

Squaring both sides, we get

$\Rightarrow x^2=16$

$\Rightarrow x=\pm 4$

The points are: $R(4,6)orR(-4,6.)$

CASE I: when R is $(4,6)$

The distances QR and PR.

$QR=\sqrt{(0-4{)}^2+(1-6{)}^2}=\sqrt{16+25}=\sqrt{41}$

$PR=\sqrt{(5-4{)}^2+(-3-6{)}^2}=\sqrt{1^2+(-9{)}^2}=\sqrt{1+81}=\sqrt{82}$

CASE II: when R is $(-4,6)$

The distances QR and PR.

$QR=\sqrt{(0-(-4){)}^2+(1-6{)}^2}=\sqrt{16+25}=\sqrt{41}$

$PR=\sqrt{(5-(-4){)}^2+(-3-6{)}^2}=\sqrt{9^2+(-9{)}^2}=\sqrt{81+81}=9\sqrt{2}$

Q10 Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).

Answer:

Let the point $P(x,y)$ is equidistant from $A(3,6)$ and $B(-3,4)$ .

Then, the distances $AP=BP$

$AP=\sqrt{(x-3{)}^2+(y-6{)}^2}$ and $BP=\sqrt{(x-(-3){)}^2+(y-4{)}^2}$

$\Rightarrow \sqrt{(x-3{)}^2+(y-6{)}^2}=\sqrt{(x-(-3){)}^2+(y-4{)}^2}$

Squaring both sides: we obtain

$\Rightarrow (x-3{)}^2+(y-6{)}^2=(x+3{)}^2+(y-4{)}^2$

$\Rightarrow (2x)(-6)+(2y-10)(-2)=0$ $[\because a^2-b^2=(a+b)(a-b)]$

$\Rightarrow -12x-4y+20=0$

$\Rightarrow 3x+y-5=0$

Thus, the relation is $3x+y-5=0$ between x and y.

Class 10 maths ch 7 solutions Exercise: 7.2

Q1 Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Answer:

Let the coordinates of point $P(x,y)$ which divides the line segment joining the points $A(-1,7)$ and $B(4,-3)$ , internally, in the ratio $m_1:m_2$ then,

Section formula: $(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

Substituting the values in the formula:

Here, $m_1:m_2=2:3$

$\Rightarrow (\frac{2(4)+3(-1)}{2+3},\frac{2(-3)+3(7)}{2+3})$

$\Rightarrow (\frac{5}{5},\frac{15}{5})$

Hence the coordinate is $P(1,3)$ .

Q2 Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Answer:

Let the trisection of the line segment $A(4,-1)$ and $B(-2,-3)$ have the points $P(x_1,y_1)$ and $Q(x_2,y_2)$

Then,

Section formula: $(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

By observation point, P divides AB internally in the ratio $1:2$ .

Hence, $m:n=1:2$

Substituting the values in the equation we get;

$\Rightarrow P(\frac{1(-2)+2(4)}{1+2},\frac{1(-3)+2(-1)}{1+2})$

$\Rightarrow P(\frac{-2+8}{3},\frac{-3-2}{3})$

$\Rightarrow P(2,\frac{-5}{3})$

And by observation point Q, divides AB internally in the ratio $2:1$

Hence, $m:n=2:1$

Substituting the values in the equation above, we get

$\Rightarrow Q(\frac{2(-2)+1(4)}{2+1},\frac{2(-3)+1(-1)}{2+1})$

$\Rightarrow Q(\frac{-4+4}{3},\frac{-6-1}{3})$

$\Rightarrow Q(0,\frac{-7}{3})$

Hence, the points of trisections are $P(2,\frac{-5}{3})$ and $Q(0,\frac{-7}{3})$

Q3 To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag

Answer:

Niharika posted the green flag at the distance P, i.e.,

$\frac{1}{4}\times 100m=25m$ from the starting point of $2^{nd}$ line.

Therefore, the coordinates of this point $P$ are $(2,25).$

Similarly, Preet posted red flag at $\frac{1}{5}$ of the distance Q i.e.,

$\frac{1}{5}\times 100m=20m$ from the starting point of $8^{th}$ line.

Therefore, the coordinates of this point Q are $(8,20)$ .

The distance $PQ$ is given by,

$PQ=\sqrt{(8-2{)}^2+(25-20{)}^2}=\sqrt{36+25}=\sqrt{61}m$

and the point at which Rashmi should post her Blue Flag is the mid-point of the line joining these points. Let this point be $R(x,y)$ .

Then, by Section Formula,

$P(x,y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

$x=\frac{2+8}{2},y=\frac{25+20}{2}$

$x=5,y=22.5$

Therefore, Rashmi should post her Blue Flag at 22.5 m on the 5th line.

Q4 Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

Answer:

Let the ratio be : $k:1$

Then, By section formula:

$P(x,y)=(\frac{k{x}_2+x_1}{k+1},\frac{k{y}_2+y_1}{k+1})$

Given point $P(x,y)=(-1,6)$

$-1=\frac{6k-3}{k+1}$

$\Rightarrow -k-1=6k-3$

$\Rightarrow k=\frac{7}{2}$

Hence, the point $P$ divides the line AB in the ratio $2:7$ .

Q5 Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Answer:

Let the point on the x-axis be $P(x,0)$ and it divides it in the ratio $k:1$ .

Then, we have

Section formula:

$P(x,y)=(\frac{k{x}_2+x_1}{k+1},\frac{k{y}_2+y_1}{k+1})$

$⟹\frac{k{y}_2+y_1}{k+1}=0$

$k=-\frac{y_1}{y_2}$

Hence, the value of k will be: $k=-\frac{-5}{5}=1$

Therefore, the x-axis divides the line in the ratio $1:1$ and the point will be,

Putting the value of $k=1$ in section formula.

$P(x,0)=(\frac{x_2+x_1}{2},0)$

$P(x,0)=(\frac{1-4}{2},0)=(\frac{-3}{2},0)$

Q6 If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Answer:

Let the given points $A(1,2),B(4,y),C(x,6),D(3,5)$ .

Since the diagonals of a parallelogram bisect each other. Intersection point O of diagonal AC and BD also divides these diagonals.

Therefore, O is the mid-point of AC and BD.

The coordinates of the point O when it is mid-point of AC.

$(\frac{1+x}{2},\frac{2+6}{2})\Rightarrow (\frac{x+1}{2},4)$

The coordinates of the point O when it is mid-point of BD.

$(\frac{4+3}{2},\frac{5+y}{2})\Rightarrow (\frac{7}{2},\frac{5+y}{2})$

Since both coordinates are of same point O.

Therefore,

$\frac{x+1}{2}=\frac{7}{2}$ and $4=\frac{5+y}{2}$

Or,

$x=6andy=3$

Q7 Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).

Answer:

As the centre point $C(2,-3)$ will be the mid-point of the diameter AB.

Then, the coordinates of point A will be $A(x,y)$ .

Given point $B(1,4)$ .

Therefore,

$(2,-3)=(\frac{x+1}{2},\frac{y+4}{2})$

$\frac{x+1}{2}=2and\frac{y+4}{2}=-3$

$\Rightarrow x=3andy=-10$ .

Therefore, the coordinates of A are $(3,-10).$

Q8 If A and B are (- 2, - 2) and (2, - 4), respectively, find the coordinates of P such that AP = 3 AB 7 and P lies on the line segment AB.

Answer:

From the figure:

As $AP=\frac{3}{7}AB$

$\Rightarrow PB=\frac{4}{7}AB$ hence the ratio is 3:4,

Now, from the section formula, we can find the coordinates of Point P.

Section Formula:

$P(x,y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

$P(x,y)=(\frac{3(2)+4(-2)}{3+4},\frac{3(-4)+4(-2)}{3+4})$

$P(x,y)=(\frac{6-8}{7},\frac{-12-8}{7})$

$P(x,y)=(\frac{-2}{7},\frac{-20}{7})$

Q9 Find the coordinates of the points which divide the line segment joining A (– 2, 2) and B(2, 8) into four equal parts.

Answer:

From the figure:

Points C, D, and E divide the line segment AB into four equal parts.

Now, from the section formula, we can find the coordinates of Point C, D, and E.

Section Formula:

$P(x,y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

Here point D divides the line segment AB into two equal parts hence

$D(x_2,y_2)=(\frac{-2+2}{2},\frac{2+8}{2})$

$D(x_2,y_2)=(0,5)$

Now, point C divides the line segment AD into two equal parts hence

$C(x_1,y_1)=(\frac{-2+0}{2},\frac{2+5}{2})$

$C(x_2,y_2)=(-1,\frac{7}{2})$

Also, point E divides the line segment DB into two equal parts hence

$E(x_1,y_1)=(\frac{2+0}{2},\frac{8+5}{2})$

$E(x_2,y_2)=(1,\frac{13}{2})$

Q10 Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.

Answer:

From the figure:

Let the vertices of the rhombus are:

$A(3,0),B(4,5),C(-1,4),D(-2,-1)$

Area of the rhombus ABCD is given by;

$=\frac{1}{2}\times (Productoflengthsofdiagonals)$

Hence we have to find the lengths of the diagonals AC and BD of the rhombus.

The distance formula:

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Length of the diagonal AC:

$AC=\sqrt{(3-(-1){)}^2+(0-4{)}^2}=\sqrt{16+16}=4\sqrt{2}$

Length of the diagonal BD:

$BD=\sqrt{(4-(-2){)}^2+(5-(-1){)}^2}=\sqrt{36+36}=6\sqrt{2}$

Thus, the area will be,

$=\frac{1}{2}\times (AC)\times (BD)$

$=\frac{1}{2}\times (4\sqrt{2})\times (6\sqrt{2})=24squareunits.$

Class 10 maths ch 7 solutions Exercise: 7.3

Q 1 (i) Find the area of the triangle whose vertices are : (2, 3), (–1, 0), (2, – 4)

Answer:

As we know, the area of a triangle with vertices (x1,x2) ,(y1, y2) and (z1 z2 ) is given by :

$A=\frac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))$

So Area of a triangle whose vertices are(2, 3), (–1, 0)and (2, – 4) is

$A=\frac{1}{2}[2(0-(-4))+(-1)(-4-3)+2(3-0)]$

$A=\frac{1}{2}[8+7+6]$

$A=\frac{1}{2}[21]$

$A=\frac{21}{2}$

$A=10.5uni{t}^2$

Hence, the area of the triangle is 10.5 per unit square.

Q1 (ii) Find the area of the triangle whose vertices are (–5, –1), (3, –5), (5, 2)

Answer:

From the figure:

Area of the triangle is given by:

$Area=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Substituting the values in the above equation, we obtain

$Area=\frac{1}{2}[(-5)((-5)-(-2))+3(2-(1))+5(-1-(-5))]$

$=\frac{1}{2}[35+9+20]=32squareunits.$

Q2 (i) In each of the following find the value of ‘k’, for which the points are collinear. (7, –2), (5, 1), (3, k)

Answer:

The points (7, –2), (5, 1), (3, k) are collinear if the area of the triangle formed by the points will be zero.

Area of the triangle is given by:

$Area=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0$

Substituting the values in the above equation, we obtain

$\frac{1}{2}[7(1-k)+5(k-(-2))+3(-2-1)]=0$

$[7-7k+5k+10-9]=0$

$\Rightarrow -2k+8=0$

$\Rightarrow k=4$

Hence, the points are collinear for k=4 .

Q2 (ii) In each of the following find the value of ‘k’, for which the points are collinear. (8, 1), (k, – 4), (2, –5)

Answer:

The points (8,1), (k, -4), (2,-5) are collinear if the area of the triangle formed by these points will be zero.

Area of the triangle is given by:

$Area=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0$

Substituting the values in the above equation, we obtain

$\frac{1}{2}[8(-4-(-5))+k((-5)-1)+2(1-(-4))]=0$