**KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3.

## Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3

Question 1.

Find the area of the triangle whose vertices are:

i) (2, 3), (-1,0), (2,-4)

ii) (-5, -1), (3, -5), (5, 2)

Solution:

i) Let A (2, 3) = (x_{1}, y_{1})

B (-1, 0) = (x_{2}, y_{2})

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

Area of the triangle from the given data:

∴ Area of ∆ ABC = 10.5 sq. units

ii) Let A (-5, -1) = (x_{1}, y_{1})

B (3, -5) = (x_{2}, y_{2})

C (5, 2) = (x_{3}, y_{3}).

Area of the triangle from the given data :

∴ Area of ∆ ABC =32 sq. units

Question 2.

In each of the following find the value of ‘k’, for which the points are collinear.

i) (7, -2), (5, 1), (3, k)

ii) (8, 1), (k, -4), (2, -5)

Solution:

i) Let A (7,-2)= (x_{1}, y_{1})

B (5, 1) = (x_{2}, y_{2})

C (3, k) = (x_{3}, y_{3}).

Points are collinear, therefore the area of the triangle formed by these is zero (0).

-k + 4 = 0

-k = -4

∴ k = 4

ii) Let A (8, 1) = (x_{1}, y_{1})

B (k, -4) = (x_{2}, y_{2})

C (2, -5) = (x_{3}, y_{3}).

Area of Triangle ABC = 0

∴ ABC is a straight line.

Question 3.

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.

Solution:

Let the mid-points of ∆ABC are P, Q, R and also the mid-points of AB, BC and AC.

As per Mid-point formula,

\(=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)

Area of ∆PQR : Area of ∆ABC

∴ 1 : 4

Question 4.

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

Solution:

Area of quadrilateral ABCD =?

In the quadrilateral ABCD, diagonal AC is drawn which divides ∆ABC, ∆ACD.

Sum of these triangles is equal to the Area of the quadrilateral.

i) Now, Area of ∆ABC :

= 10.5 sq. units

ii) Area of ∆ACD

∴ Area of quadrilateral:

= Area of ∆ABC + Area of ∆ACD = 10.5 + 17.5 = 28 sq. units.

∴ Area of quadrilateral ABCD = 28 sq.units.

Question 5.

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 ∆ABC whose vertices are A(4, -6), B(3, -2) and C(5, 2).

Solution:

AD Median is drawn to side BC which is the vertex of ∆ABC.

Median AD divides AABC into two triangles ∆ABD and ∆ADC which are equal in area.

Now Median AD bisects BC.

∴ BD = DC.

i) As per the Mid-Points formula,

Coordinates of D are

= \(\sqrt{(4)^{2},(0)^{2}}\)

= \(\sqrt{16,0}\)

= (4,0)

∴ Coordinates of D are (4, 0)

ii) Now, Area of ∆ABD:

iiii) Now, Area of ADC:

= \(\frac{1}{2}\) × -6

= – 3 sq. units.

∴ Area of ∆ADC = -3 sq. units.

∴ Area of ∆ABC :

= Area of ∆ABD + Area of ∆ADC

= (-3) +(-3)

= -6 sq. units.

iv) Now, Area of ∆ABC : (Direct Method)

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