Coordinate Geometry
Mathematics Class Tenth
Solution NCERT Exercise 7.3
Area of Triangle whose coordinates of whose vertices are given
In Coordinate Geometry when coordinate of vertices are given, then Area of Triangle =1/2[x1(y2–y3) + x2(y3–y1) + x3(y3–y1)]
Question (1) Find the area of the triangle whose vertices are :
(i) (2, 3), (–1, 0), (2, –4)
Solution
Given coordinate of vertices of triangle are (2, 3), (–1, 0), (2, –4)
Then Area of given triangle = ?
We know that, In Coordinate Geometry when coordinate of vertices are given, then Area of Triangle =1/2[x1(y2–y3) + x2(y3–y1) + x3(y3–y1)]
Here we have, x1 =2, y1 = 3
And, x2 = –1, y2 = 0
And, x3 = 2, y3 = –4
Thus, Area of given triangle
=1/2[2(0–(–4)) + –1(–4–3) + 2(3–0)]
= 1/2 [2(0 + 4) + -1(- 7) + 2 × 3 ]
= 1/2 [2 × 4 + 7 + 6]
= 1/2 [8 + 7 + 6]
`=1/2xx21`
`=21/2` square unit
Thus, Area of given triangle = `21/2` square unit Answer
(ii) (–5, –1), (3, –5), (5, 2)
Solution
Given, coordinates of vertices of triangle are (–5, –1), (3, –5), (5, 2)
We know that, In Coordinate Geometry when coordinate of vertices are given, then Area of Triangle =1/2[x1(y2–y3) + x2(y3–y1) + x3(y3–y1)]
Here we have, x1 =–5, y1 = –1
And, x2 =3 , y2 = –5
And, x3 = 5, y3 = 2
Thus, Area of given triangle
= 1/2 [–5(–5 – 2) + 3(2 – (–1)) + 5(–1 – (–5))
= 1/2 [(- 5 (- 7)) + 3(2+1) + 5 (- 1 + 5)]
= 1/2 [35 + (3 × 3) + (5 × 4)]
= 1/2 [35 + 9 +20]
`=1/2xx64 = 32`
Thus, Area of given triangle 32 square unit Answer
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)
Solution
Given, coordinates of points are (7, - 2), (5, 1), (3, k)
And, points are collinear, this means area of triangle = 0
Thus, value of k = ?;
We know that, In Coordinate Geometry when coordinate of vertices are given, then Area of Triangle =1/2[x1(y2–y3) + x2(y3–y1) + x3(y3–y1)]
Here we have, x1 =7, y1 = - 2
And, x2 = 5, y2 = 1
And, x3 = 3, y3 = k
Thus, area of given triangle
= 1/2 [7{1 - k} + 5{k - (- 2)} + 3{- 2 - 1}]
⇒ 0 = 1/2 [7 - 7k + 5{k + 2} + 3 × {- 3}]
⇒ 0 = 1/2 [7 - 7k + 5k + 10 - 9]
⇒ 0 = 1/2 [7 - 2k + 1]
⇒ 0 = 1/2 [8 - 2k]
`=> 0 =1/2xx8 - 1/2xx2k`
⇒ 0 = 4 - k
Thus, k = 4
Thus, value of k = 0 Answer
(ii) (8, 1), (k, - 4), (2, - 5)
Solution
Given, coordinates of points are (8, 1), (k, - 4), (2, - 5)
And, points are collinear, this means area of triangle = 0
Thus, value of k = ?;
We know that, In Coordinate Geometry when coordinate of vertices are given, then Area of Triangle =1/2[x1(y2–y3) + x2(y3–y1) + x3(y3–y1)]
Here we have, x1 = 8, y1 = 1
And, x2 = k, y2 = - 4
And, x3 = 2, y3 = - 5
Thus, Area of triangle
= 1/2 [8{- 4 - (-5)} + k{– 5 – 1} + 2{1 – (– 4)}]
⇒ 0 = 1/2 [8 {– 4 + 5} + k {– 6} + 2 {1+4}]
⇒ 0 = 1/2 [8 × 1 – 6k + 2 × 5]
⇒ 0 = 1/2 [8 – 6k + 10]
⇒ 0 = 1/2 [18 – 6k]
`=> 0 = 1/2xx18 – 1/2xx6k`
⇒ 0 = 9 – 3k
⇒ 3k = 9
`:. k = 9/3 =3`
Thus, value of k = 3 Answer
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
Given, Vertices of triangle are (0, –1), (2, 1) and (0, 3)
Thus, Area of the triangle formed by joining given mid points = ?
Ratio of both of the triangle = ?
Let, ABC is a triangle coordinates of whose vertices are given in the question.
Let, D, E and F are the mid points of the sides of given triangle.
Here, vertices of triangle are A(0, –1), B(2, 1) and C(0, 3)
We know that, In Coordinate Geometry when coordinate of vertices are given, then Area of Triangle =1/2[x1(y2–y3) + x2(y3–y1) + x3(y3–y1)]
Here, we have
Here we have, x1 = 0, y1 = –1
And, x2 = 2, y2 = 1
And, x3 = 0, y3 = 3
Calculation of Area of triangle ABC coordinate of whose vertices are given
= 1/2 [{1(1 – 3)} + {2(3 – (–1))} + {0(–1 – 1)}]
= 1/2 [{1(– 2)} + {2(3+1)} + 0]
= 1/2 [– 2 + {2 × 4}]
= 1/2 [– 2 + 8]
`=1/2xx6 =3`
Thus, Area of triangle ABC = 3 square unit
Calculation of coordinate of mid-point D of AB
We know that, coordinate of mid-point joining the line of two points (x1, y1) and (x2, y2)
`=((x_1+x_2)/2, (y_1+y_2)/2)`
Coordinates of mid-point of AB
`=((0+(-1))/2,\ (-1 + 1)/2)`
`=(-1/2, 0/2)`
Thus, Coordinate of D `=(1/2, 0)`
Calculation of coordinate of mid-point E of BC
Coordinate of mid-point E of BC
`=((2+0)/2, (1+3)/2)`
`=(2/2, 4/2)`
⇒ E = (1, 2)
Calculation of coordinate of mid-point F of AC
Coordinate of mid-point F of AC
`=((0+0)/2, (-1+3)/2)`
`=(0/2, 2/2)`
⇒ F = (0, 1)
Calculation of Area of triangle DEF, coordinate of whose vertices are given
Thus, coordinate of DEF are `=(1/2, 0)`, (1, 2) and (0, 1)
Here we have, x1`=1/2`, y1=0
x2=1, y2=2
x3=0, y3=1
Area of triangle DEF
= 1/2 [{1/2 (2 – 1)} + {1(1 – 0)} + {0(0 – 2)}]
`= 1/2 [{1/2xx1} + {1xx1} + {0xx-2}]`
= 1/2 [1 + 1 + 0]
`= 1/2xx2 =1`
Thus, Area of triangle DEF = 1 square unit
Now, ratio of small triangle to the big triangle = 1:2
Thus, vertices of mid points are `=(1/2, 0)`, (1, 2) and (0, 1) and ratio of given triangle = 1:2 Answer
Question (4) Find the area of the quadrilateral whose vertices, taken in order, are (–4, –2), (v3, –5), (3, –2) and (2, 3)
Solution
Given, vertices of a quadrilateral are (–4, –2), (–3, –5), (3, –2) and (2, 3)
Thus, Area of quadrilateral =?;
Let, the given quadrilateral is A(–4, –2), B(–3, –5), C(3, –2) and (2, 3)
Now, let join the points A and C of the quadrilateral, which is the diagonal of given quadrilateral.
This diagonal divides the quadrilateral into two triangles.
We know that, In Coordinate Geometry when coordinate of vertices are given, then Area of Triangle =1/2[x1(y2–y3) + x2(y3–y1) + x3(y3–y1)]
Calculation of Area of triangle ABC whose coordinates are given
Here, we have
Here we have, x1 = –4, y1 = –2
And, x2 = –3, y2 = –5
And, x3 = 3, y3 = –2
Thus, Area of triangle ABC
= 1/2 [{–4(–5 –(–2))} + {–3(–2 – (–2))} + {3(–2 – (–5))}]
= 1/2 [{–4 (–5 + 2)} + {–3 (–2 + 2)} + {3(–2 + 5)}]
= 1/2 [{–4 × (–3)} + {–3 × 0} + {3 × 3]
= 1/2 [12 + 0 + 9]
`=1/2xx21`
Thus, Area of triangle ABC `=21/2`
Calculation of Area of triangle ACD whose coordinates are given
Here, we have x1 = –4, y1 = –2
And, x2 = –3, y2 = –2
And, x3 = 2, y3 = 3
Thus, Area of triangle ACD
= 1/2 [{–4(–2 –3)} + {3(3 – (–2))} + {2(–2 – (–2))}]
= 1/2 [{–4(–5)} + {3(3+2)} + {2(–2 + 2)}]
= 1/2 [20 + {3 × 5} + {2 × 0}]
= 1/2 [20 + 15 + 0]
`=1/2xx35`
Thus, Area of triangle ACD `=35/2`
Thus, Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle BCD
`=21/2+35/2`
`=(21+35)/2`
`=56/2 =28`
Thus, area of given quadrilateral = 28 square unit Answer
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 triangle ADC whose vertices are A(4, –6), B(3, –2) and C(5, 2).
Solution:
Given, vertices of triangle ABC are A(4, –6), B(3, –2) and C(5, 2)
Let median AD divides the given triangle into two parts.
Now, we know that, coordinate of mid-point joining the line of two points (x1, y1) and (x2, y2)
`=((x_1+x_2)/2, (y_1+y_2)/2)`
Thus, coordinates of mid-point D of BC
`=((3+5)/2, (-2+2)/2)`
`=(8/2, 0)`
⇒ Coordinates of D is (4, 0)
We know that, in Coordinate Geometry when coordinate of vertices are given, then Area of Triangle =1/2[x1(y2–y3) + x2(y3–y1) + x3(y3–y1)]
Calculation of Area of triangle ABD
Here, we have x1 = 4, y1 = –6
And, x2 = 3, y2 = –2
And, x3 = 4, y3 = 0
Thus, Area of triangle ABD
= 1/2 [{4(–2 –0)} +{3(0 – (–6)} + {4(–6 – (–2))}]
= 1/2 [{4(–2)} + 3 × 6 + {4(–6 + 2)}]
= 1/2 [–8 + 18 + {4(–4)}]
= 1/2 [–8 + 18 – 16]
`= 1/2xx(-6)`
Thus, Area of triangle ABD = –3 square unit.
Calculation of Area of triangle ADC
Here, we have x1 = 4, y1 = –6
And, x2 = 4, y2 = 0
And, x3 = 5, y3 = 2
Thus, Area of triangle ABD
= 1/2 [{4(0 – 2)} + {4(2 – (–6))} + {5(–6 – 0)}]
= 1/2 [{4(–2)} + {4 (2+6)} + {5 (–6)}]
= 1/2 [–8 + 4 × 8 – 30]
= 1/2 [–8 + 32 – 30]
`=1/2 xx(–6)`
Thus, area of triangle ADC = –3
Thus, Area of triangle ABD = Area of triangle ADC
Thus, a median of a triangle divide it into two triangles of equal areas. Proved
Reference: