# KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3

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) = (x1, y1)
B (-1, 0) = (x2, y2)
C (2, -4) = (x3, y3).
Area of the triangle from the given data:

∴ Area of ∆ ABC = 10.5 sq. units

ii) Let A (-5, -1) = (x1, y1)
B (3, -5) = (x2, y2)
C (5, 2) = (x3, y3).
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:
The given three points will be collinear if the triangle formed by them has zero area.
(i) Let A(7, -2), B(5, 1) and C(3, k) be the vertices of a triangle.
∴ The given points will be collinear, if ar (∆ABC) = 0
or $$\frac{1}{2}$$ [7(1 – k) + 5(k + 2) + 3(-2 – 1)] = 0
⇒ 7 – 7k + 5k + 10 + (-6) – 3 = 0
⇒ 17 – 9 + 5k – 7k = 0
⇒ 8 – 2k = 0 ⇒ 2k = 8 ⇒ k = $$\frac{8}{2}$$ = 4
The required value of k = 4.
(ii) $$\frac{1}{2}$$ [8(- 4 + 5) + k(- 5 -1) + 2(1 + 4)] = 0
⇒ 8 – 6k + 10 = 0 ⇒ 6k = 18 ⇒ k = 3.

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:
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 ∆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 that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A(A, -6), B(3, -2) and C(5, 2).
Solution:
Here, the vertices of the triangle are A(4, -6), B(3, -2) and C(5, 2).
Let D be the midpoint of BC.
∴ The coordinates of the point D are
$$\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)$$ or ( 4, 0)
AD divides the triangle ABC into two parts i.e., ∆ABD and ∆ACD.

ar(∆ADC) = $$\frac{1}{2}$$ [4(0 – 2) + 4(2 + 6) + 5(-6 – 0)]
= $$\frac{1}{2}$$ [-8 + 32 – 30] = $$\frac{1}{2}$$ [-6] = -3
= 3 sq. units (numerically) ………… (2)
From (1) and (2),