NCERT Solutions for Exercise 7.3 Class 10 Maths Chapter 7 - Coordinate Geometry

Coordinate Geometry Class 10 Chapter 7 Exercise: 7.1

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

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

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

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

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

$A= 10.5\:unit^2$

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:

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

Substituting the values in the above equation, we obtain

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

$= \frac{1}{2}\left [ 35+9+20 \right ] = 32 \ square\ units.$

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:

$Area = \frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ] = 0$

Substituting the values in the above equation, we obtain

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

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

$\Rightarrow -2k+8 = 0$

$\Rightarrow k = 4$

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:

$Area = \frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ] = 0$

Substituting the values in the above equation, we obtain

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

$\Rightarrow 8-6k+10 = 0$

$\Rightarrow 6k = 18$

$\Rightarrow k = 3$

Hence, the points are collinear for k = 3 .

Q3 Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Answer:

From the figure:

The coordinates of the point P, Q, and R are:

Point P is the midpoint of side AB, hence the coordinates of P are :

$P(x_{1},y_{1}) = \left (\frac{0+2}{2}, \frac{3+1}{2} \right ) = \left (1, 2 \right )$

Point Q is the midpoint of side AC, hence the coordinates of Q are :

$Q(x_{2},y_{2}) = \left (\frac{2+0}{2}, \frac{1-1}{2} \right ) = \left (1, 0 \right )$

Point R is the midpoint of side BC, hence the coordinates of R are :

$R(x_{3},y_{3}) = \left (\frac{0+0}{2}, \frac{-1+3}{2} \right ) = \left (0, 1 \right )$

Hence, the area of the triangle formed by the midpoints PQR will be,

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

$= \frac{1}{2}\left [ (2-1)+1(1-0)+0(0-2) \right ]$

$=\frac{1}{2}(1+1) = 1\ square\ units.$

And the area formed by the triangle ABC will be:

$Area_{(ABC)} = \frac{1}{2}\left [ 0(1-3)+2(3-(-1))+0(-1-1) \right ]$

$= \frac{1}{2}\left [ 8 \right ] = 4\ square\ units.$

Thus, the ratio of Area of $\triangle PQR$ to the Area of $\triangle ABC$ will be $1:4$ .

Q4 Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

Answer:

From the figure:

The coordinates are $A(-4,-2),\ B(-3,-5),\ C(3,-2)\ and\ D(2,3)$

Divide the quadrilateral into 2 parts of triangles.

Then the area will be, $ABC + ADC$

Area of the triangle formed by ABC will be,

$Area_{(ABC)} = \frac{1}{2}\left [ (-4)((-5)-(-2))+(-3)((-2)-(-2))+3((-2)-(-2)) \right ]$ $= \frac{1}{2}\left [ 12+0+9 \right ] = \frac{21}{2}\ Square\ units.$

Area of the triangle formed by ADC will be,

$Area_{(ADC)} = \frac{1}{2}\left [ (-4)((-2)-(-3))+3(3-(-2))+2((-2)-(-2)) \right ]$ $= \frac{1}{2}\left [ 20+15+0 \right ] = \frac{35}{2}\ Square\ units.$

Therefore, the area of the quadrilateral will be:

$= \frac{21}{2}+\frac{35}{2} = 28\ square\ units.$

Alternatively,

The points A and C are in the same ordinates.

Hence, the length of base AC will be $(3-(-4)) = 7\ units.$

Therefore,

Area of triangle ABC:

$= \frac{1}{2} \times (Base) \times (Height) = \frac{1}{2}\times(7)(3)$

Area of triangle ADC:

$= \frac{1}{2} \times (Base) \times (Height) = \frac{1}{2}\times(7)(5)$

Therefore, the area will be, $\frac{1}{2}\times(7)\times(5+3) =28\ square\ units.$

Q5 You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for D ABC whose vertices are A(4, – 6), B(3, –2) and C(5, 2).

Answer:

From the figure:

The coordinates of midpoint M of side BC is:

$M = \left ( \frac{3+5}{2}, \frac{-2+2}{2} \right ) = \left ( 4,0 \right )$

Now, calculating the areas of the triangle ABM and ACM :

Area of triangle, ABM:

$Area_{(ABM)} = \frac{1}{2}\left [ 4((-2)-0)+3(0-(-6))+4((-6)-(-2)) \right ]$

$= \frac{1}{2}\left [ -8+18-16 \right ] = 3\ Square\ units.$

Area of triangle, ACM:

$Area_{(ACM)} = \frac{1}{2}\left [ 4(0-(-2))+4(2-(-6))+5((-6)-0) \right ]$

$= \frac{1}{2}\left [ -8+32-30 \right ] = -3\ Square\ units.$

However, the area cannot be negative, Therefore, area of $\triangle ACM$ is 3 square units.

Clearly, the median AM divided the $\triangle ABC$ in two equal areas.