Coordinate Geometry
Mathematics Class Tenth
Solution of NCERT Exercise 7.1 part2
Question (3) Determine if the points (1,5), (2,3) and (–2, –11) are collinear.
Solution:
Given, points are (1,5), (2,3) and (–2, –11)
And, here to prove that the given points are collinear or not.
Let given points are
Thus, if the sum of distance between A and B, and B and C will be equal to the distance between A and C, then the given points are collinear otherwise they are not collinear.
That means, if AB + BC = AC, then the given points are collinear otherwise not.
Let, A = (1, 5), B = (2, 3) and C = (-2, -11)
We know that, in a plane distance between two given points PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
(a) Distance between A(1, 5) and B(2, 3)
Here, x1 = 1, y1 = 5
And, x2 = 2, and y2 = 3
Thus, AB `= sqrt((2-1)^2+(3-5)^2)`
`=sqrt(1^2+(-2)^2)`
`=sqrt(1+4)`
⇒ AB `= sqrt5` units - - - - - (i)
(b) Distance between B(2, 3) and C(–2, –11)
Here, x1 = 2, y1 = 3
And, x2 = –2, y2 = –11
Thus, using distance formula,
BC `= sqrt((-2-2)^2+(-11-3)^2)`
`=sqrt((-4)^2+(-14)^2)`
`=sqrt(16+196)`
⇒ BC `=sqrt(212)` - - - - - - (ii)
(c) Distance between A(1, 5) and C(–2, –11)
Here, x1 = 1, y1 = 5
And, x2 = –2, y2 = –11
Thus, by using distance formula
AC `= sqrt((-2-1)^2+(-11-5)^2)`
`=sqrt((-3)^2+(-16)^2)`
`=sqrt(9+256)`
⇒ AC `=sqrt(265)` - - - - - - (iii)
Now, AB + BC = AC
After substituting the value of AB and BC and AC From equation (i) and (ii) and (iii) we have
`= sqrt5 + sqrt(212) = sqrt(265)`
Now, since AB + BC `!=` AC
Thus, the given points are not collinear.
Alternate Method
Graphical method
Given, points are (1,5), (2,3) and (–2, –11)
And, here to prove that the given points are collinear or not.
Let given points are A, B and C
And, A = (1.5), B = (2, 3) and C = (–2, –11)
A graph is taken and given points are located as follows
Clearly, given points are not collinear.
Thus, non–collinear Answer
Question (4) Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.
Solution:
Given, points are (5, –2), (6, 4) and (7, –2)
And to check whether these points are vertices of an isosceles triangle or not.
Let the points are A(5, –2), B(6, 4) and C(7, –2)
We know that, in an isosceles triangle, the length of two sides are equal and the length of third unequal side is less than other sides.
Now, we have to calculate the AB, BC and AC
Using distance formula, we know that, in a plane the distance between two points
PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
(a) Calculation of distance between A(5, –2) and B(6, 4)
Here, x1 = 5, y1 = –2
And, x2 = 6, y2 = 4
Thus, using distance formula
AB `= sqrt((6-5)^2+(4-(-2))^2)`
`=sqrt(1^2+(4+2)^2)`
`=sqrt(1+6^2)`
`=sqrt(1+36)`
⇒ AB `=sqrt(37)` units
(b) Calculation of distance between B(6, 7) and C(4, –2)
Here, x1 = 6, y1 = 4
And, x2 = 7, and y2 = –2
Thus, using distance formula,
BC `=sqrt((7-6)^2+(-2-4)^2)`
`=sqrt(1^2+(-6)^2)`
`=sqrt(1+36)`
⇒ BC `=sqrt(37)` units
(c) Distance between A(5, 1) and (7, –2)
Here, x1 = 5, y1 = -2
And, x2 = 7, y2 = -2
Thus, by using distance formula
AC `= sqrt((7-5)^2+(-2-(-2))^2)`
`=sqrt(2^2(-2+2)^2)`
`=sqrt(4+0^2)`
`=sqrt4`
⇒ AC =2 units
Now, Since, in the given ΔABC
AB=BC`!=`AC and, AB = BC > AC
Thus, given triangle is an isosceles triangle Answer
Question (5) In a classroom, 4 friends are seated at the points A, B, C and D as shown in Figure. Champa and Chameli walk into the class and after observing for few minutes Champa asks Chameli, "Don't you think ABCD is a square?" Chameli disagrees. Using distance formula, find which of them is correct.
Solution:
From the given, figure it is clear that coordinates of points
A = (3, 4), B = (6, 7), C = (9, 4) and D = (6, 1)
Thus, it is to prove that, ABCD is a square or not.
Now, we know that, if all sides and diagonals are equal, then the given quadrilateral is a square.
Now, we know that,
Using distance formula, in a plane the distance between two given points
PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
(a) Calculation of distance between A = (3, 4) and B = (6, 7) using distance formula
Here, we have, x1 = 3, y1 = 4
And, x2 = 6 and y2 = 7
Thus, Using distance formula,
AB `=sqrt((6-3)^2+(7-4)^2)`
`=sqrt(3^2+3^2)`
`=sqrt(9+9)`
⇒ `=sqrt(18)` unit - - - - - (i)
(b) Calculation of distance between B(6, 7) and C(9, 4) using distance formula
Here, we have, x1 = 6, y1 = 7
And, x2 = 9, y2 = 4
Thus, using distance formula
BC `=sqrt((9-6)^2+(4-7)^2)`
`=sqrt(3^2+(-3)^2)`
`=sqrt(9+9)`
⇒ BC `=sqrt(18)` unit - - - - - - (ii)
(c) Calculation of distance between C(9, 4) and D(6, 1) using distance formula
Here, we have, x1 = 9, y1 = 4
And, x2 = 6, y2 = 1
Thus, using distance formula,
CD `=sqrt((6-9)^2+(1-4)^2)`
`=sqrt(3^2+(-3)^2)`
`=sqrt(9+9)`
⇒ CD `=sqrt(18)` unit - - - - - (iii)
(d) Calculation of distance between D(6, 1) and A(3, 4) using distance formula
Here, we have, x1 = 6, y1 = 1
And, x2 = 3, y2 = 4
Thus, using distance formula,
DA `=sqrt((3-6)^2+(4-1)^2)`
`=sqrt((-3)^2+3^2)`
`=sqrt(9+9)`
⇒ DA `=sqrt(18)` unit - - - - - - (iv)
(e) Calculation of distance between A(3, 4) and C(9, 4) using distance formula
Here, we have, x1 = 3, y1 = 4
And, x2 = 9, y2 = 4
Thus, using distance formula,
AC `=sqrt((9-3)^2+(4-4)^2)`
`=sqrt(6^2+0^2)`
⇒ AC `=sqrt(36)`
⇒ AC = 6 unit - - - - - - (v)
(f) Calculation of distance between B(6, 7) and D(6, 1) using distance formula
Here, we have, x1 = 6, y1 = 7
And, x2 = 6, y2 = 1
Thus, using distance formula,
BD `=sqrt((6-6)^2+(1-7)^2)`
`=sqrt(0^2+(-6)^2)`
`=sqrt(36)`
⇒ BD = 6 unit - - - - - - (vi)
Now, from equation (i), (ii), (iii), (iv), (v) and (v), it is clear that,
AB = BC = CD = AD = `sqrt(18)` which are sides
And AC = BD = 6 unit which are diagonals.
Thus, since all sides and diagonals are equal, thus ABCD is a square and hence Champa was correct. Answer
Question (6) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (–1, –2), (1, 0), (–1, 2), (–3, 0)
Solution
Given, points in a plane are (–1, –2), (1, 0), (–1, 2), (–3, 0)
Then, it is to detect that which type of quadrilateral is formed by these points.
Let given points are A, B, C and D
Thus, A(-1, -2), B(1, 0), C (-1, 2) and D (-3,0)
Now, we know that, according to distance formula, in a plane the distance between two points
PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
(a) Thus, the distance between A(–1, –2) and B(1, 0)
Here we have, x1 = -1, y1 = -2
And, x2 = 1, y2 = 0
Thus, using distance formula
AB `=sqrt((1-(-1))^2+(0-(-2))^2)`
`=sqrt((1+1)^2+(0+2)^2)`
`=sqrt(2^2+2^2)`
`=sqrt(4+4)`
`=sqrt(8)`
⇒ AB `=2sqrt2` units - - - - - - - (i)
(b) The distance between B(1, 0) and C(–1, 2)
Here we have, x1 = 1, y1 = 0
And, x2 = -1, y2 = 2
Thus, using distance formula
BC`=sqrt((-1-1)^2+(2-0)^2)`
`=sqrt((-2)^2+2^2)`
`=sqrt(4+4)`
`=sqrt8`
⇒ BC `=2sqrt2` units - - - - - - (ii)
(c) The distance between C(–1, 2) and D(–3, 0)
Here we have, x1 = -1, y1 = 2
And, x2 = -3, y2 = 0
Thus, using distance formula
CD `=sqrt((-3-(-1))^2+(0-2)^2)`
`=sqrt((-3+1)^2+2^2)`
`=sqrt((-2)^2+4)`
`=sqrt(4+4)`
`=sqrt8`
⇒ CD `=2sqrt2` units - - - - - - (iii)
(d) Distance between A(–1, –2) and D(–3, 0)
Here we have, x1 = –1, y1 = –2
And, x2 = –3, y2 = 0
Thus, by using distance formula, we have
AD `=sqrt((-3(-1))^2+(0-(-2))^2)`
`=sqrt((-3+1)^2+(0+2)^2)`
`=sqrt((-2)^2+2^2)`
`=sqrt(4+4)`
`=sqrt8`
⇒ AD `=2sqrt2` units - - - - - (iv)
(e) Distance between A(–1, –2) and C(–1, 2), i.e. length of one of the diagonal
Here we have, x1 = –1, y1 = –2
And, x2 = –1, y2 = 2
Thus, by using distance formula we have
AC `=sqrt((-1(-1))^2+(2-(-2))^2)`
`=sqrt((-1+1)^2+(2+2)^2)`
`=sqrt(0^2+4^2)`
`=sqrt(16)`
⇒ AC = 4 units - - - - - - (v)
(f) Distance between B(1, 0) and D(–3, 0), i.e. length of second diagonal
Here we have, x1 = 1, y1 = 0
And, x2 = –3, and y2 = 0
Thus, by using distance formula, we have
BD `=sqrt((-3-1)^2+(0-0)^2)`
`=sqrt((-4)^2+0^2)`
`=sqrt16`
⇒ BD =4 units - - - - - (vi)
Now, here from equation (i), (ii), (iii), (iv), (v) and (vi), we have
AB=BC=CD=AD `=2sqrt2` unit (all sides of quadrilateral)
and AC=BD = 4 unit (diagonals)
Since all sides are equal and diagonals are also equal,
Thus given quadrilateral is a square Answer
(ii) (–3, 5), (3, 1), (0, 3), (–1, –4)
Solution:
The given points are (–3, 5), (3, 1), (0, 3), (–1, –4)
Thus, it is to be detect that which type of quadrilateral is formed by these points.
Let, the points are A (–3, 5), B (3, 1), C (0, 3) and D(–1, –4)
Now, we know that, according to distance formula, in a plane the distance between two points
PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
(a) Calculation of distance between A(–3, 5) and B(3, 1)
Here we have, x1 = –3, y1 = 5
And, x2 = 3, and y2 = 1
Thus, by using distance formula
AB `=sqrt((3-(-3))^2+(1-5)^2)`
`=sqrt((3+3)^2+(-4)^2)`
`=sqrt(6^2+(-4)^2)`
`=sqrt(36+16)`
⇒ AB `=sqrt52` unit - - - - - (i)
(b) Calculation of distance between B(3, 1) and C(0, 3)
Here we have, x1 = 3, y1 = 1
And, x2 = 0, and y2 = 3
Thus, by using distance formula we have
BC `=sqrt((0-3)^2+(3-1)^2)`
`=sqrt((-3)^2+2^2)`
`=sqrt(9+4)`
⇒ BC `=sqrt(13)` unit - - - - - (ii)
(c) Calculation of distance between C(0, 3) and D(–1, –4)
Here we have, x1 = 0, y1 = 3
And, x2 = –1, and y2 = –4
Now, by using distance formula we have
CD `=sqrt((-1-0)^2+(-4-3)^2)`
`=sqrt((-1)^2+(-7)^2)`
`=sqrt(1+49)`
`=sqrt(50)`
`=sqrt(25xx2)`
⇒ CD `=5sqrt2` unit - - - - - (iii)
(d) Calculation of distance between A(–3, 5) and D(–1, –4)
Here we have, x1 = –3, y1 = 5
And, x2 = –1, and y2 = –4
Thus, by using distance formula, we have
AD `=sqrt((1-(-3))^2+(-4-5)^2)`
`=sqrt((-1+3)^2+(-4-5)^2)`
`=sqrt(2^2+(-9)^2)`
`=sqrt(4+81)`
⇒ AD `=sqrt(85)` unit - - - - - (iv)
Now, from equations (i), (ii), (iii) and (iv), it is clear that
Here, `AC!=BC!=CD!=DA`
This means no sides are equal for the given points
Thus only a general quadrilateral is formed by the given points. Answer
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
Given, points are (4, 5), (7, 6), (4, 3), (1, 2)
Thus, to find out that which type of quadrilateral is formed using given points.
Let, the given points are A(4, 5), B(7, 6), C(4, 3) and D(1, 2)
Now, we know that, according to distance formula, in a plane the distance between two points
PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
(a) Calculation of distance between A(4, 5) and B(7, 6)
Here we have, x1 = 4, y1 = 5
And, x2 = 7, and y2 = 6
Thus, by using distance formula we have
AB `=sqrt((7-4)^2+(6-5)^2)`
`=sqrt(3^2+1^2)`
`=sqrt(9+1)`
`=sqrt(10)` unit - - - - - (i)
(b) Calculation of distance between B(7, 6) and C(4, 3)
Here we have, x1 = 7, y1 = 6
And, x2 = 4, and y2 = 3
Thus, by using distance formula we have
BC `=sqrt((4-7)^2+(3-6)^2)`
`=sqrt((-3)^2+(-3)^2)`
`=sqrt(9+9)`
`=sqrt(18)`
⇒ BC `=3sqrt2` unit - - - - (ii)
(c) Calculation of distance between C(4, 3) and D(1, 2)
Here we have, x1 = 4, y1 = 3
And, x2 = 1, and y2 = 2
Thus, by using distance formula we have
CD `=sqrt((1-4)^2+(2-3)^2)`
`=sqrt((-3)^2+(-1)^2)`
`=sqrt(9+1)`
`=sqrt(10)` unit - - - - (iii)
(d) Calculation of distance between A(4, 5) and D(1, 2)
Here we have, x1 = 4, y1 = 5
And, x2 = 1, and y2 = 2
Thus, by using distance formula we have
AD `=sqrt((1-4)^2+(2-5)^2)`
`=sqrt((-3)^2+(-3)^2)`
`=sqrt(9+9)`
`=sqrt(18)`
⇒ AD `=3sqrt2` unit - - - - - (iv)
(e) Distance between A(4, 5) and C(4, 3), i.e. length of one of the diagonal
Here we have, x1 = 4, y1 = 5
And, x2 = 4, and y2 = 3
Thus, by using distance formula we have
AC `=sqrt((4-4)^2+(3-5)^2)`
`=sqrt(0^2+(-2)^2)`
`=sqrt(0+4)`
⇒ AC `=sqrt4=2` unit - - - - - (v)
(f) Calculation of length of second diagonal, i.e. distance between B(7, 6) and D(1, 2)
Here we have, x1 = 7, y1 = 6
And, x2 = 1, and y2 = 2
Thus, by using distance formual we have
BD `=sqrt((1-7)^2+(2-6)^2)`
`=sqrt((-6)^2+(-4)^2)`
`=sqrt(36+16)`
`=sqrt(52)`
`=2sqrt(13)` unit - - - - - (vi)
From equations (i), (ii), (iii) and (iv) we have
AB=CD`=sqrt(10)` unit and BC=AD`=3sqrt2` unit.
And, from equation (v) and (vi) we have
Diagonal, AC=2 unit and diagonal BD`=2sqrt13` unit
Thus, it is clear that opposite sides are equal but diagonals are not equal.
Thus, using given points a parallelogram is formed Answer
Reference: