Co-ordinate Geometry is a branch of mathematics that focuses on the study of the position and movement of points, lines, and shapes in two-dimensional (2D) and three-dimensional (3D) space. It is a fundamental tool for understanding the fundamentals of physics, engineering, and many other scientific and engineering disciplines. In CBSE Class 10, co-ordinate geometry is an important topic that students must master in order to be successful in their future studies.
In this blog post, we will discuss the Exemplar Solution Class 10 Co-ordinate Geometry and some of the key concepts that students should understand in order to be successful in their CBSE Class 10 exams. We will also provide some tips and strategies for mastering this topic and preparing for the exam.
To begin, let’s define some important terms related to coordinate geometry. The coordinate system is used to represent points in space using numerical values. A point is identified by its coordinates, which are the numbers that represent its position in the coordinate system. The coordinates of a point are written in the form (x,y). The x-coordinate tells us the point’s distance from the y-axis, and the y-coordinate tells us the point’s distance from the x-axis.
Once we have the coordinates of a point, we can draw lines and shapes in the coordinate system. Lines are represented by their equations, which are written in the form y=mx+c, where m is the slope of the line and c is the constant term. Shapes are represented by their equations as well, but these equations are usually more complex than those used to represent lines.
Now that we have the basics of coordinate geometry out of the way, let’s discuss some of the key concepts that students should understand in order to be successful in their CBSE Class 10 exams. First, students should be familiar with the basic equations used to represent lines and shapes in the coordinate system. They should also understand how to calculate the slope of a line and the area of a shape.
In addition, students should understand how to calculate the distance between two points and the midpoint of a line. They should also know how to calculate the equation of a line given two points and the equation of a circle given its center and radius.
Finally, students should be familiar with the concept of transformations, which are used to move shapes around in the coordinate system. This includes understanding how to translate, rotate, and reflect shapes.
These are just some of the key concepts that students should understand in order to be successful in their CBSE Class 10 exams. With the right approach and a thorough understanding of the material, students can master coordinate geometry and be well-prepared for their exams. Good luck!
‹Exercise 7.2 Exercise 7.4›Solution:
AB = `\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }`
= `\sqrt{(-4 – (-5))^2 + (-2 – 6)^2 }`
=`\sqrt{(1)^2 + (-8)^2 }`
=`\sqrt{1 + 64}`
=`\sqrt{65}`
BC = `\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }`
= `\sqrt{(7 –(-4))^2 + (5 –(-2))^2 }`
=`\sqrt{(11)^2 + (7)^2 }`
=`\sqrt{121 + 49}`
=`\sqrt{170}`
CA = `\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }`
= `\sqrt{(-5 – 7)^2 + (6- 5)^2 }`
=`\sqrt{(-12)^2 + (1)^2 }`
=`\sqrt{144 + 1}`
=`\sqrt{145}`
Since AB ≠ BC ≠ CA
ABC is a scalene triangle.
Solution:
Let point P(x,0) on x-axis is at a distance of 2 √5 units from A(7, -4)
AP = `\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }`
`\sqrt{(7 – x)^2 + (- 4- 0)^2 }= 2 √5`
⇒ `\sqrt{49 + x^2 – 14x + 16 }= 2 √5`
⇒ `x^2 -14x + 65 = 20`
⇒ `x^2 -14x + 45= 0`
⇒ `x^2 -9x – 5x + 45 = 0`
⇒ ` x(x – 9) – 5(x – 9)= 0`
⇒` (x – 9)(x – 5)= 0`
⇒ x -9 = 0 and x – 5 = 0
⇒ x = 9 and x = 5
There are two such points (9, 0) and (5, 0)
Solution:
AB= `\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }`
= `\sqrt{(7-2)^2 + [3-(-2)]^2 }`
= `\sqrt{(5)^2 + (5)^2 }`
= `\sqrt{25 + 25 }`
=`\sqrt{50 }`
=`5\sqrt{2}`
BC= `\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }`
= `\sqrt{(11-7)^2 + (-1-3)^2 }`
= `\sqrt{(4)^2 + (-4)^2 }`
= `\sqrt{16 + 16 }`
=`\sqrt{32 }`
=`4\sqrt{2}`
CD= `\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }`
= `\sqrt{(6-11)^2 + [-6-(-1)]^2 }`
= `\sqrt{(-5)^2 + (-5)^2 }`
= `\sqrt{25 + 25 }`
=`\sqrt{50 }`
=`5\sqrt{2}`
DA= `\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }`
= `\sqrt{(2-6)^2 + [-2-(-6)]^2 }`
= `\sqrt{(-4)^2 + (4)^2 }`
= `\sqrt{16 + 16 }`
=`\sqrt{32 }`
=`4\sqrt{2}`
AC= `\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }`
= `\sqrt{(11-2)^2 + [-1-(-2)]^2 }`
= `\sqrt{(9)^2 + (1)^2 }`
= `\sqrt{81 + 1 }`
=`\sqrt{82 }`
BD= `\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }`
= `\sqrt{(6-7)^2 + [-6-(3)]^2 }`
= `\sqrt{(-1)^2 + (-9)^2 }`
= `\sqrt{1 + 81 }`
=`\sqrt{82 }`
Since, AB= CD, BC = DA and AC = BD.
i.e., pair of opposite sides and Diagonals are equal.
So, A, B, C and D are vertices of Rectangle.
Solution:
PQ = `\sqrt{(a + 3)^2 + (-5 + 14)^2 }= 9`
⇒ `\sqrt{(a + 3)^2 + (9)^2}=9`
⇒ `\sqrt {a^2 + 9 + 6a + 81 }= 9`
⇒ `a^2 +6a + 9 +81= 81`
⇒ `a^2 + 6a + 9 = 0`
⇒ `a^2 + 3a + 3a + 9 = 0`
⇒`a(a + 3) + 3 (a + 3)= 0`
⇒`(a + 3) (a + 3) = 0`
⇒ `a + 3 = 0` and `a + 3 = 0`
⇒ `a = -3` and `a = – 3`
Solution 5
Let P (X, Y) is equidistance between the points A(-5,4) and B (-1,6). Then, by using mid point formula
`P = (\frac {x_1 + x_2}{2},\frac {y_1 + y_2}{2})`
= `(\frac {- 5 – 1}{2},\frac {4 + 6}{2})`
= `(\frac {- 6}{2},\frac {10 }{2})`
= `(- 3, 5)`
Solution 6
Let M be the foot of perpendicular bisector of AB, which is mid point of AB. i.e.,
` M = = (\frac {x_1 + x_2}{2},\frac {y_1 + y_2}{2})`
= `(\frac {- 5 + 4}{2},\frac { – 2 -2}{2})`
= `(\frac {- 1}{2},\frac { – 4}{2})`
= `(\frac {- 1}{2},- 2)`
Here, x co-ordinate `= \frac{-1}{2}` and y co-ordinate `= -2`
Given that, Point Q lies on x -axis y = 0. So, Q = `(\frac{-1}{2},0)`
Further Q, A and B will form a isosceles triangle.
Solution:
Given that, A(5, 1), B(–2, –3) and C(8, 2m ) are collinear. i.e.,
Area of △ABC = 0
`\frac{1}{2} |x_1 (y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2) | = 0`
`\frac{1}{2} |5 (-3 -2) + (-2) (2m – 1) + 8 [1 – (-3)] | = 0`
`|5 (-3 -2m) + (-2) (2m – 1) + 8 [1 – (-3)] | = 0`
`|- 15 – 10m + (-2) (2m – 1) + 8 (4) | = 0`
`|- 15 – 10m – 4m + 2 + 32 | = 0`
`|- 14m + 19 | = 0`
or `-14m = – 19` or `m = 19/14`
Solution:
Given the point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y).
`PA = QA`
⇒ `PA^2=QA^2`
⇒ `(3-2)^2 + [8-(-4)]^2 ` = `(-10-2)^2 + [y- (-4)]^2 `
⇒ `(1)^2 + (12)^2 ` = `(-12)^2 + (y + 4)^2 `
⇒ `1 + 144 ` = `144 + y^2 + 16 + 8y`
⇒ `1` = `y^2 + 16 + 8y `
⇒ `y^2 + 8y + 15` = 0
⇒ `y^ 2 + 5y + 3y + 15`
or `y (y + 5) + 3 ( y + 5 )`
or `y + 5 = 0` and `y + 3 = 0`
or `y = – 5` and `y = -3`
Further, when P (3, 8) and Q (-10, -5)
`PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }`
= `\sqrt{(-10 -3)^2 + (-5 -8)^2 }`
= `\sqrt{(-13)^2 + (-13)^2 }`
= `\sqrt{169 + 169 }`
= `13\sqrt{2 }`
and
When P (3, 8) and Q (-10, -3)
`PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }`
= `\sqrt{(-10 -3)^2 + (-3 -8)^2 }`
= `\sqrt{(-13)^2 + (-11)^2 }`
= `\sqrt{169 + 121 }`
= `\sqrt{290 }`
Solution:
Given vertices are A (–8, 4), B(–6, 6) and C(–3, 9).
Area `= 1/2 |x_1 (y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2) |`
`= 1/2 |-8 (6 -9) + (-6) (9 -4) + (-3) (4 -6) |`
`= 1/2 |-8 (-3) + (-6) (5) + (-3) (-2) |`
`= 1/2 |24 -30 + 6 |`
`= 1/2 |0 |`
`= 0`
Solution:
Let point `P(x,0)` be the point be on x- axis divide the line segment joining the points (– 4, – 6) and (–1, 7) in k : 1. Then,
`p (x, 0) = [\frac{k(-1) + 1 (-4)}{K +1}, \frac{k(7) + 1 (-6)}{K+ 1}]`
`p (x, 0) = [\frac{-k -4}{K +1}, \frac{7k – 6}{K+ 1}]`
Equating both sides. We get, `0 = \frac{7k – 6}{K+ 1}`
` 7k – 6 = 0 or k = 6/7`
Required ratio is 6 : 7
Further, `x = \frac{-k -4}{K +1}`
` x = \frac{- 6/7 – 4}{6/7 + 1} = – 34/ 13`.
Co-ordinate of the point `( – 34/ 13, 0)`
Solution:
Let point `P(3/4, 5/12)` divides the line segment joining the points `A(1/2, 3/2)` and B (2, –5) in k : 1. Then,
`P(3/4, 5/12) = [\frac{k(2) + 1 (1/2)}{K +1}, \frac{k(-5) + 1 (3/2)}{K+ 1}]`
Equating
`3/4 = \frac{k(2) + 1 (1/2)}{K +1},`
`3 (k + 1) = 4[k(2) + 1 (1/2)],`
`3 k + 3 = 4(2k + 1/2)`
`3 k + 3 = 8k + 2`
` 1= 5k `
or `k = 1/5`. So, ratio is `1 : 5`
Solution:
Since P (9a – 2, –b) divides line segment joining A (3a + 1, –3) and B (8a, 5) in the ratio 3 : 1.
` P (9a- 2, -b) = [\frac{3(8a) + 1 (3a + 1)}{4}, \frac{3(5) + 1 (-3)}{4}] `
` P (9a- 2, -b) = [\frac{24a + 3a + 1}{4}, \frac{15 – 3}{4}] `
` P (9a- 2, -b) = (\frac{27a + 1 }{4}, 3) `
Equating both sides. We get,
`9a – 2= {27 a + 1}{4} and b = -3
`36a – 8 = 27 a + 1 or 9a = 9 or a = 1`
Solution 13
Given that, P (a,b) is mid point of A (10, –6) and B (k, 4).So,
` P (a, b) = 9{10 + k}{2},{- 6 + 4}{2}0`
` P (a, b) = ({10 + k}{2},-1)`
`a = {10 + k}{2} —(Equation 1) and b = – 1`
Put value of `b = -1 in a – 2b = 18` we get `a = 16`
Put value of a in equation (1).
` 16 = {10 + k}{2} or k = 22`
Further, `AB = \sqrt{(x_2 – x_ 1)^ 2+(y_2 – y_ 1)^ 2}`
`= \sqrt{(22 – 10)^ 2+(4 + 6)^ 2}`
`= \sqrt{(12)^ 2+(10)^ 2}`
`= \sqrt{144 + 100}`
`= \sqrt{244 }`
`= 2\sqrt{61 }`
Solution 14
Given that, Centre of a circle is (2a, a -7) and circle passes through (11, -9). Also, diametre of the circle is 10 √2 units. so, radius = 5 √2 units. Further,
radius = `\sqrt{(x_2 – x_ 1)^ 2+(y_2 – y_ 1)^ 2}`
=`\sqrt{(11 – 2a)^ 2+[-9 – (a -7)]^ 2}`
= `\sqrt{121 + 4a ^2 – 44 a +(-2-a)^ 2}`
= `\sqrt{121 + 4a ^2 – 44 a +4 + a^2 + 4a}`
= `\sqrt{125 + 5a ^2 -40a}` = `5 √2`
0r `125 + 5a ^2 -40a` = `50`
0r `5a ^2 -40a + 75` = `0`
0r `15 + a ^2 -8a` = `0`
0r `a ^2 -5a – 3a + 15` = `0`
0r `a(a – 5) – 3(a -5)` = `0`
or `a-5 = 0` and `a -3 = 0`
or `a= 5` and `a =3 `
Solution 15
Since point P lies divide line joining AB in 1 : 2. So,
`P = {1(5) + 2 (3)}{3}, {1(1) + 2 (2)}{3}) = (11/3, 5/3)`
Since P lies on the line `3x – 18y + k = 0`.
`3(11/3) – 18(5/3) + k = 0` or
` 11 -30 + k = 0 or k = 19`
Solution 16
Let point `A (x_1,y_1), B (x_2,y_2) and C (x_2,y_2)` be the points of 𝛥ABC. Given that, D, E and F are mid points of AB, BC and CA respectively. So,
`D (-1/2, 5/2) = ({x_1 + x_2}{2},{y_1 +y_2}{2})`
or `{x_1 + x_2}{2} = -1/2 and {y_1 +y_2}{2} = 5/2`
`or x_1 + x_2 = -1 (Equation i)and y_1 +y_2 = 5 (Equation ii)`
`E (7, 3) = ({x_2 + x_3}{2},{y_2 +y_3}{2})`
`or {x_2 + x_3}{2} = 7 and {y_2 +y_3}{2} = 3`
`or x_2 + x_3 = 14 (Equation iii) and y_2 +y_3 = 6 (Equation iv)`
(Equation 2)
`F (7/2, 7/2) = ({x_3 + x_1}{2},{y_3 +y_1}{2})`
`or {x_3 + x_1}{2} = 7/2 and {y_3 +y_1}{2} = 7/2`
`or x_3 + x_1 = 7 (Equation v) and y_1 +y_2 = 7 (Equation vi)`
Adding (i)(iii)(v) and (ii) (iv) (vi). We get,
`2 (x_1 + x_ 2 + x_ 3) = 20 and 2(y_1 + y_ 2 + y_ 3) = 18 `
or `(x_1 + x_ 2 + x_ 3) = 10 (Equation vii) and (y_1 + y_ 2 + y_ 3) = 9 (Equation viii)`
Subtracting (vii) – (i) and (viii) – (ii). We get,
`x_3 = 11 and y_3 = 4. Similarly x_1 = -4 , x_2 = 3 and y_1 =3 , y_2 = 2
Hence, A(-4,3) , B (3,2) and C (11,4).
Now, area `= 1/2 |x_1 (y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2) |`
`= 1/2 |(-4) (2 – 4) + 3 (4-3) + 11 (3 – 2) |`
`= 1/2 |(-4) (-2) + 3 (1) + 11 (1) |`
`= 1/2 |8 + 3 + 11 |`
`= 1/2 |22 |`
`= 11 unit`
Solution 17
Let A (2, 9), B (a, 5) and C (5, 5) are the vertices of right angled triangle.
Now,
`AB = \sqrt{(x_2 – x_ 1)^ 2+(y_2 – y_ 1)^ 2}`
` = \sqrt{(a – 2)^ 2+(5 -9)^ 2}`
` = \sqrt{a^2 + 4 – 4a + (-4)^2}`
` = \sqrt{a^2 + 4 – 4a + 16}`
` = \sqrt{a^2 – 4a + 20}`
` BC = \sqrt{(x_2 – x_ 1)^ 2+(y_2 – y_ 1)^ 2}`
`= \sqrt{(5 – a)^ 2+(5 – 5)^ 2}`
`= \sqrt{25 + a^2 – 10a + 0}`
`= \sqrt{a^2 – 10a + 25}`
CA = `\sqrt{(x_2 – x_ 1)^ 2+(y_2 – y_ 1)^ 2}`
= `\sqrt{(5 – 2)^ 2+(5 – 9)^ 2}`
= `\sqrt{(3)^2 + (- 4)^2}`
= `\sqrt{9 + 16}`
= `\sqrt{25}`
= `5`
Since, angle B is right angle.
`AB^ 2 + BC ^ 2 = AC ^2`
`a^2 – 4a + 20 + a^2 – 10a + 25 = 5 ^ 2 `
or `2a^2 – 14a + 45 = 25 `
or `2a^2 – 14a + 20 = 0 `
or `a^2 – 7a + 10 = 0 `
or `a^2 – 5a – 2a + 10 = 0 `
or `a (a -5)- 2(a – 5 ) = 0 `
or ` (a -5)(a – 2 ) = 0 `
or ` a = 5 and a = 2`
when a = 5, BC = 0 which is not possible. Hence a will be 2.
Now, Area `= 1/2 |2 (5 -5 ) + a (5 – 9) + 5 (9 – 5) |`
`= 1/2 |2 (0) + a (- 4) + 5 (4) |`
`= 1/2 |0 – 4a + 20 |`
When a = 2 ,
Area = `= 1/2 |0 – 4(2) + 20 |`
`= 1/2 |0 – 8 + 20 |= 0`
`= 1/2 | – 12|= 6`
Solution 18
Given that, `PR = 3/5PQ`
⇒ `PR = 3/5PQ`
⇒ `5/3 = PQ/PR`
⇒ `5/3 = (PR + RQ)/PR`
⇒ `5/3 = PR/PR + RQ/PR`
⇒ `5/3 = 1 + RQ/PR`
⇒ `5/3 – 1 = RQ/PR`
⇒`(5- 3)/3 = RQ/PR`
⇒`2/3 = RQ/PR`
or `3/2 = PR/RQ`
Hence point R (X,Y) be the point divide the line joining points PQ in 3 : 2
Here,
`x_1 = – 1, y_1 = 3`
`x_2 = 3` , y_2 = 5`
`m_1 = 3 , m_2= 2`
X = `\frac{m_2 x_1 + m_1 x_2}{m_1 + m_2 }` = `\frac{3(2) + 2(-1)}{3 + 2 }`
`= \frac{6 -2}{5 } =4/5`
Y = `\frac{m_2 y_1 + m_1 y_2}{m_1 + m_2 }` `\frac{3(5) + 2(3)}{3 + 2 }`
`= \frac{15 + 6}{5 } =21/5`
Hence required point P is (4/5,21/5)
are collinear.
Solution:
Given vertices are A (k + 1, 2k), B (3k, 2k + 3) and C (5k – 1, 5k). Since A,B and C are collinear. So,
Area `= 1/2 |x_1 (y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2) |= 0`
`⇒ 1/2 |(k + 1) (2k + 3 – 5 k) + (3k) (5k – 2k) + (5k – 1) (2k – 2k -3) |= 0 `
`⇒ 1/2 |(k + 1) (3 – 3 k) + (3k) (3k) + (5k – 1) (-3) |= 0 `
`⇒ 1/2 |3k – 3k^ 2 + 3 – 3k + 9k^2 – 15k + 3 |= 0 `
`⇒ 1/2 |6k^2 – 15k + 6 |= 0 `
`⇒ 3/2 |2k^2 – 5k + 2 |= 0 `
`⇒ 2k^2 – 5k + 2= 0 `
`⇒ 2k^2 – 4k – k + 2= 0 `
`⇒ 2k(k – 2) – 1(k – 2 )= 0 `
`⇒ (2k- 1 )(k – 2) = 0 `
`⇒ 2k- 1 = 0 and k – 2 = 0 `
`⇒ 2k= 1 and k = 2`
`⇒ k= 1/2 and k = 2`
Solution:
Let P (X, Y) be the point on the line 2x + 3 y – 5 = 0 which divides the line joining A (8, –9) and B (2, 1) in k: 1.
First of all, we need to find point P (X, Y)
Here, `x_1 = 8` , `x_ 2 = 2`
`y_ 1 =-9` , `y_2 = 1`
`m_1= k` and `m_2= 1`
`X = \frac {m_1x_2 + m_2x_1}{m_1 + m_ 2}`
` =\frac {k(2) + 1 (8)}{k + 1}`
`=\frac {2k + 8 }{k + 1}`
Y = `\frac {m_1y_2 + m_2y_1}{m_1 + m_ 2}`
`=\frac {k(1) + 1 (-9)}{k + 1}`
`=\frac {k -9 }{k + 1}`
Since point P `(\frac {2k + 8 }{k + 1}),\frac {k -9 }{k + 1})` lies on the line 2x + 3 y – 5 = 0. So,
`2 (\frac {2k + 8 }{k + 1}) + 3(\frac {k -9 }{k + 1}) – 5 = o`
`2 (2k + 8) + 3 (k -9) – 5 (k + 1) = 0`
`4k + 16 + 3 k – 27 – 5k -5 = 0`
`2 k – 16 = 0`
`2k = 16`
`K = 8`
So, required ratio is 8 : 1
Also, P = `(\frac {2(8) + 8 }{8 + 1}, \frac {8 -9 }{8 + 1})`
`= (\frac {16 + 8 }{9}, \frac {-1 }{9})`
`= (\frac {24}{9}, \frac {-1 }{9})`