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In real life, coordinate geometry can be used in navigation, GPS tracking, computer graphics, game development, and for designing various structures. It is an integral part of Mathematics. This chapter is the link between geometry and algebra. It deals with geometrical figures in coordinate systems and coordinates like the x-axis and y-axis to locate the exact position of the points in two-dimensional planes. This chapter also consists of an introductory phase into coordinate geometry, which gradually makes students aware of topics like distance formulas and section formulas.
This chapter can be challenging for students who don't have their basic concepts clear. This article allows every student to understand all the necessary concepts and strengthen their knowledge of coordinate geometry. The NCERT solutions for Class 10 Maths Chapter 7 coordinate geometry will offer a systematic and structured approach for the exercise problems in the NCERT textbook to prepare well for your board exams by providing detailed solutions to all the exercise questions. These solutions also provide a valuable resource to the students to enhance their performance in their board exams. Follow the Coordinate Geometry Class 10 Mathematics Notes page to learn more!
Distance Formulae:
For a line defined by two points A(x1, y1) and B(x2, y2), the distance between these points can be calculated using the formula:
Distance AB = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Section Formula:
When a point P divides a line AB, with A(x1, y1) and B(x2, y2) as endpoints, in a ratio of m:n, the coordinates of point P can be found using the formula:
Point P = $\{\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n} \}$
Midpoint Formula:
The midpoint of a line AB, defined by A(x1, y1) and B(x2, y2), can be determined using the following formula:
Midpoint P = $\{ \frac{x_1+x_2}2,\frac{(y_1+y_2}2\}$
As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters
Area of a Triangle:
Consider a triangle formed by points A(x1, y1), B(x2, y2), and C(x3, y3). The area of this triangle can be calculated using the formula:
Area ∆ABC = $\frac12$|x1(y2 − y3) + x2(y3 – y1) + x3(y1 – y2)|
Class 10 Maths chapter 7 solutions Exercise: 7.1 Page number: 105-106 Total questions: 10 |
Q1. (i) Find the distance between the following pairs of points : (2, 3), (4, 1)
Answer:
Given points: (2, 3), (4, 1)
Distance between the points will be: $(x_{1},y_{1})$ and $(x_{2},y_{2})$
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$⇒D= \sqrt{(4-2)^2+(1-3)^2} = \sqrt{4+4} = 2\sqrt{2}$
Q1. (ii) Find the distance between the following pairs of points : (– 5, 7), (– 1, 3)
Answer:
Given points: (– 5, 7), (– 1, 3)
Distance between the points will be: $(x_{1},y_{1})$ and $(x_{2},y_{2})$
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$⇒D= \sqrt{(-1+5)^2+(3-7)^2} = \sqrt{16+16} = 4\sqrt{2}$
Q1. (iii) Find the distance between the following pairs of points :(a, b), (– a, – b)
Answer:
Given points: (a, b), (– a, – b)
Distance between the points will be: $(x_{1},y_{1})$ and $(x_{2},y_{2})$
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$⇒D= \sqrt{(-a-a)^2+(-b-b)^2} = \sqrt{4(a^2+b^2)} = 2\sqrt{a^2+b^2}$
Answer:
Given points: (0, 0) and (36, 15)
Distance between the points will be: $(x_{1},y_{1})$ and $(x_{2},y_{2})$
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$⇒D= \sqrt{(36-0)^2+(15-0)^2} =\sqrt{1296+225} = \sqrt{1521} = 39$
The distance between the two towns A and B is, thus, 39 km for the given town location
$(0,0)$ and $(36,15)$.
Q3. Determine if the points (1, 5), (2, 3), and (– 2, – 11) are collinear.
Answer:
Let the points (1, 5), (2, 3), and (– 2, – 11) represent the vertices A, B, and C of the given triangle, respectively.
$A = (1,5),\ B = (2,3),\ C = (-2,-11)$
Therefore,
$AB = \sqrt{(1-2)^2+(5-3)^2} = \sqrt{5}$
$BC = \sqrt{(2-(-2))^2+(3-(-11))^2} = \sqrt{4^2+14^2} = \sqrt{16+196} = \sqrt{212}$ $CA = \sqrt{(1-(-2))^2+(5-(-11))^2} = \sqrt{3^2+16^2} = \sqrt{9+256} = \sqrt{265}$ Since these are not satisfied.
$AB+BC \neq CA$
$BA+AC \neq BC$
$BC+CA \neq BA$
As these cases are not satisfied.
Hence, the points are not collinear.
Q4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Answer:
The distance between two points $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$ is given by:
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
So, we have the following points: (5, – 2), (6, 4) and (7, – 2) assuming it to be the vertices of triangles A, B, and C, respectively.
$AB = \sqrt{(5-6)^2+(-2-4)^2} = \sqrt{1+36} = \sqrt{37}$
$BC = \sqrt{(6-7)^2+(4+2)^2} = \sqrt{1+36} = \sqrt{37}$
$CA = \sqrt{(5-7)^2+(-2+2)^2} = \sqrt{4+0} = 2$
Therefore, AB = BC
Here two sides are equal in length.
Therefore, ABC is an isosceles triangle.
Answer:
The coordinates of the points:
$A(3,4),\ B(6,7),\ C(9,4),$ and $D(6,1)$ are the positions of 4 friends.
The distance between two points $A(x_{1},y_{1}),$ and $B(x_{2},y_{2})$ is given by:
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
Hence,
$AB = \sqrt{(3-6)^2+(4-7)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$
$BC = \sqrt{(6-9)^2+(7-4)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$
$CD = \sqrt{(9-6)^2+(4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$
$AD= \sqrt{(3-6)^2+(4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$
And the lengths of diagonals:
$AC = \sqrt{(3-9)^2+(4-4)^2} =\sqrt{36+0} = 6$
$BD = \sqrt{(6-6)^2+(7-1)^2} =\sqrt{36+0} = 6$
So, here it can be seen that all sides of quadrilateral ABCD are of the same length and diagonals are also having the same length.
Therefore, quadrilateral ABCD is a square, and Champa is saying right.
Answer:
Let the given points $(-1,-2),\ (1,0),\ (-1,2),$ and $(-3,0)$ represent the vertices A, B, C, and D of the given quadrilateral, respectively.
The distance formula:
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$AB= \sqrt{(-1-1)^2+(-2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$
$BC= \sqrt{(1+1)^2+(0-2)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$
$CD= \sqrt{(-1+3)^2+(2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$
$AD= \sqrt{(-1+3)^2+(-2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$
Finding the length of the diagonals:
$AC= \sqrt{(-1+1)^2+(-2-2)^2} =\sqrt{0+16} = 4$
$BD= \sqrt{(1+3)^2+(0-0)^2} =\sqrt{16+0} = 4$
It is clear that all sides are of the same lengths, and also, the diagonals have the same lengths.
Hence, the given quadrilateral is a square.
Q6. (ii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (–3, 5), (3, 1), (0, 3), (–1, – 4)
Answer:
Let the given points $(-3,5),\ (3,1),\ (0,3),$ and $(-1,-4)$ represent the vertices A, B, C, and D of the given quadrilateral, respectively.
The distance formula:
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$AB= \sqrt{(-3-3)^2+(5-1)^2} =\sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$
$BC= \sqrt{(3-0)^2+(1-3)^2} =\sqrt{9+4} = \sqrt{13}$
$CD= \sqrt{(0+1)^2+(3+4)^2} =\sqrt{1+49} = \sqrt{50} = 5\sqrt2$
$AD= \sqrt{(-3+1)^2+(5+4)^2} =\sqrt{4+81} = \sqrt{85}$
All the sides of the given quadrilateral have different lengths.
Therefore, it is only a general quadrilateral and not a specific one like a square, rectangle, etc.
Q6. (iii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (4, 5), (7, 6), (4, 3), (1, 2)
Answer:
Let the given points $(4,5),\ (7,6),\ (4,3),\ (1,2)$ represent the vertices A, B, C, and D of the given quadrilateral, respectively.
The distance formula:
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$AB= \sqrt{(4-7)^2+(5-6)^2} =\sqrt{9+1} = \sqrt{10}$
$BC= \sqrt{(7-4)^2+(6-3)^2} =\sqrt{9+9} = \sqrt{18}$
$CD= \sqrt{(4-1)^2+(3-2)^2} =\sqrt{9+1} = \sqrt{10}$
$AD= \sqrt{(4-1)^2+(5-2)^2} =\sqrt{9+9} = \sqrt{18}$
And the diagonals:
$AC =\sqrt{(4-4)^2+(5-3)^2} = \sqrt{0+4} = 2$
$BD =\sqrt{(7-1)^2+(6-2)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$
Here, we can observe that the opposite sides of this quadrilateral are of the same length.
However, the diagonals are of different lengths.
Therefore, the given points are the vertices of a parallelogram.
Q7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Answer:
Let the point which is equidistant from $A(2,-5)\ and \ B(-2,9)$ be $X(x,0)$ as it lies on X-axis.
Then, we have,
Distance AX $= \sqrt{(x-2)^2+(0+5)^2}$
and Distance BX $= \sqrt{(x+2)^2+(0+9)^2}$
According to the question, these distances are equal in length.
Hence, we have,
$\sqrt{(x-2)^2+(0+5)^2}$ $= \sqrt{(x+2)^2+(0+9)^2}$
Squaring both sides, we get,
$(x-2)^2+25 = (x+2)^2+81$
$⇒x^2-4x+4+25=x^2+4x+4+81$
$⇒4x+4x=-81+25$
$⇒8x=-56$
$\therefore x=-7$
Hence, the point is $X(-7,0)$.
Q8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
Answer:
Given the distance between the points $P(2,-3)$ and $Q(10,y)$ is 10 units.
The distance formula :
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
So, given $PQ = 10$ units
$PQ= \sqrt{(10-2)^2+(y-(-3))^2} = 10$
After squaring both sides,
$\Rightarrow (10-2)^2+(y-(-3))^2 = 100$
$\Rightarrow (y+3)^2 = 100 - 64$
$\Rightarrow y+3 = \pm 6$
$\Rightarrow y = 6 - 3$ or $y = -6-3$
Therefore, the values are $y = 3$ or $y=-9$.
Answer:
Given $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$ .
Then, the distances $PQ = RQ$.
Distance $PQ = \sqrt{(5-0)^2+(-3-1)^2} = \sqrt{25+16} = \sqrt{41}$
Distance $RQ = \sqrt{(x-0)^2+(6-1)^2} = \sqrt{x^2+25}$
$\Rightarrow \sqrt{x^2+25} = \sqrt{41}$
Squaring both sides, we get,
$⇒x^2+25=41$
$\Rightarrow x^2 = 16$
$\Rightarrow x = \pm 4$
The points are: $R(4,6)\ or\ R(-4,6.)$
CASE I: when R is $(4,6)$
The distances QR and PR.
$QR = \sqrt{(0-4)^2+(1-6)^2} = \sqrt{16+25} = \sqrt{41}$
$PR = \sqrt{(5-4)^2+(-3-6)^2} = \sqrt{1^2+(-9)^2} = \sqrt{1+81} = \sqrt{82}$
CASE II: when R is $(-4,6)$
The distances QR and PR.
$QR = \sqrt{(0-(-4))^2+(1-6)^2} = \sqrt{16+25} = \sqrt{41}$
$PR = \sqrt{(5-(-4))^2+(-3-6)^2} = \sqrt{9^2+(-9)^2} = \sqrt{81+81} = 9\sqrt{2}$
Answer:
Let the point $P(x,y )$ is equidistant from $A(3,6)$ and $B(-3,4)$ .
Then, the distances $AP =BP$
$AP = \sqrt{(x-3)^2+(y-6)^2}$ and $BP = \sqrt{(x-(-3))^2+(y-4)^2}$
$\Rightarrow \sqrt{(x-3)^2+(y-6)^2} = \sqrt{(x-(-3))^2+(y-4)^2}$
Squaring both sides, we obtain
$\Rightarrow (x-3)^2+(y-6)^2= (x+3)^2+(y-4)^2$
$\Rightarrow (2x)(-6)+(2y-10)(-2)= 0$ $\left [\because a^2-b^2 = (a+b)(a-b) \right ]$
$\Rightarrow -12x-4y+20 = 0$
$\Rightarrow 3x+y-5 = 0$
Thus, the relation is $3x+y-5 = 0$ between x and y.
Class 10 Maths chapter 7 solutions Exercise: 7.2 Page number: 111 Total questions: 10 |
Q1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Answer:
Let the coordinates of point $P(x,y)$ which divides the line segment joining the points $A(-1,7)$ and $B(4,-3)$ , internally, in the ratio $m_{1}:m_{2}$ then,
Section formula: $\left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
Substituting the values in the formula:
Here, $m_{1}:m_{2} = 2:3$
$\Rightarrow \left (\frac{2(4)+3(-1)}{2+3} , \frac{2(-3)+3(7)}{2+3} \right )$
$\Rightarrow \left (\frac{5}{5} , \frac{15}{5} \right )$
$⇒(1,3)$
Hence, the coordinate is $P \left (1, 3 \right )$.
Q2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Answer:
Let the trisection of the line segment $A(4,-1)$ and $B(-2,-3)$ have the points $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$
Then,
Section formula: $\left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
By observation point, P divides AB internally in the ratio $1:2$.
Hence, $m:n = 1:2$
Substituting the values in the equation, we get;
$\Rightarrow P\left (\frac{1(-2)+2(4)}{1+2} , \frac{1(-3)+2(-1)}{1+2} \right )$
$\Rightarrow P \left (\frac{-2+8}{3} , \frac{-3-2}{3} \right )$
$\Rightarrow P \left (2 , \frac{-5}{3} \right )$
And by observation point Q, divides AB internally in the ratio $2:1$
Hence, $m:n = 2:1$
Substituting the values in the equation above, we get
$\Rightarrow Q\left (\frac{2(-2)+1(4)}{2+1} , \frac{2(-3)+1(-1)}{2+1} \right )$
$\Rightarrow Q \left (\frac{-4+4}{3} , \frac{-6-1}{3} \right )$
$\Rightarrow Q\left (0 , \frac{-7}{3} \right )$
Hence, the points of trisections are $P \left (2 , \frac{-5}{3} \right )$ and $Q\left (0 , \frac{-7}{3} \right )$
Answer:
Niharika posted the green flag at the distance P, i.e.,
$\frac{1}{4}\times100\ m = 25\ m$ from the starting point of $2^{nd}$ line.
Therefore, the coordinates of this point $P$ are $(2,25).$
Similarly, Preet posted red flag at $\frac{1}{5}$ of the distance Q i.e.,
$\frac{1}{5}\times100\ m = 20\ m$ from the starting point of $8^{th}$ line.
Therefore, the coordinates of this point Q are $(8,20)$.
The distance $PQ$ is given by,
$PQ = \sqrt{(8-2)^2+(25-20)^2} = \sqrt{36+25} = \sqrt{61} m$
and the point at which Rashmi should post her Blue Flag is the mid-point of the line joining these points. Let this point be $R(x,y)$.
Then, by Section Formula,
$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
$⇒x = \frac{2+8}{2},\ y = \frac{25+20}{2}$
$⇒x = 5,\ y = 22.5$
Therefore, Rashmi should post her Blue Flag at 22.5 m on the 5th line.
Answer:
Let the ratio be $k:1$
Then, by section formula:
$P(x,y) = \left (\frac{kx_{2}+x_{1}}{k+1} , \frac{ky_{2}+y_{1}}{k+1} \right )$
Given point $P(x,y) = (-1,6)$
$-1 = \frac{6k-3}{k+1}$
$\Rightarrow -k-1 = 6k-3$
$\Rightarrow k = \frac{2}{7}$
Hence, the point $P$ divides the line AB in the ratio $2:7$.
Answer:
Let the point on the x-axis be $P(x,0)$, and it divides it in the ratio $k:1$.
Then, we have
Section formula:
$P(x,y) = \left (\frac{kx_{2}+x_{1}}{k+1} , \frac{ky_{2}+y_{1}}{k+1} \right )$
$⇒ \frac{ky_{2}+y_{1}}{k+1} = 0$
$⇒k =-\frac{y_{1}}{y_{2}}$
Hence, the value of k will be: $k =-\frac{-5}{5}= 1$
Therefore, the x-axis divides the line in the ratio $1:1$, and the point will be,
Putting the value of $k=1$ in the section formula.
$P(x,0) = \left ( \frac{x_{2}+x_{1}}{2}, 0 \right )$
$P(x,0) = \left ( \frac{1-4}{2}, 0 \right ) = \left ( \frac{-3}{2}, 0 \right )$
Answer:
Let the given points $A(1,2),\ B(4,y),\ C(x,6),\ D(3,5)$
Since the diagonals of a parallelogram bisect each other. Intersection point O of diagonals AC and BD also divides these diagonals.
Therefore, O is the midpoint of AC and BD.
The coordinates of the point O when it is the mid-point of AC.
$\left ( \frac{1+x}{2}, \frac{2+6}{2} \right ) \Rightarrow \left ( \frac{x+1}{2}, 4 \right )$
The coordinates of the point O when it is the mid-point of BD.
$\left ( \frac{4+3}{2}, \frac{5+y}{2} \right ) \Rightarrow \left ( \frac{7}{2}, \frac{5+y}{2} \right )$
Since both coordinates are of the same point O.
Therefore,
$\frac{x+1}{2} =\frac{7}{2}$ and $4 = \frac{5+y}{2}$
Or,
$x = 6\ and\ y = 3$
Answer:
As the centre point $C(2,-3)$ will be the midpoint of the diameter AB.
Then, the coordinates of point A will be $A(x,y)$.
Given point $B(1,4)$ .
Therefore,
$(2,-3) = \left ( \frac{x+1}{2}, \frac{y+4}{2} \right )$
$\frac{x+1}{2} = 2\ and\ \frac{y+4}{2} = -3$
$\Rightarrow x = 3\ and\ y = -10$ .
Therefore, the coordinates of A are $(3,-10).$
Answer:
From the figure:
As $AP = \frac{3}{7}AB$
$\Rightarrow PB = \frac{4}{7}AB$ hence the ratio is 3:4,
Now, from the section formula, we can find the coordinates of Point P.
Section Formula:
$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
$⇒P(x,y)= \left (\frac{3(2)+4(-2)}{3+4} , \frac{3(-4)+4(-2)}{3+4} \right )$
$⇒P(x,y)= \left (\frac{6-8}{7} , \frac{-12-8}{7} \right )$
$⇒P(x,y)= \left (\frac{-2}{7} , \frac{-20}{7} \right )$
Answer:
From the figure:
Points C, D, and E divide the line segment AB into four equal parts.
Now, from the section formula, we can find the coordinates of Points C, D, and E.
Section Formula:
$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
Here, point D divides the line segment AB into two equal parts; hence
$D(x_{2},y_{2})= \left (\frac{-2+2}{2} , \frac{2+8}{2} \right )$
$⇒D(x_{2},y_{2})= \left (0 , 5 \right )$
Now, point C divides the line segment AD into two equal parts hence
$C(x_{1},y_{1})= \left (\frac{-2+0}{2} , \frac{2+5}{2} \right )$
$⇒C(x_{2},y_{2})= \left (-1 , \frac{7}{2} \right )$
Also, point E divides the line segment DB into two equal parts hence
$E(x_{1},y_{1})= \left (\frac{2+0}{2} , \frac{8+5}{2} \right )$
$⇒E(x_{2},y_{2})= \left (1 , \frac{13}{2} \right )$
Answer:
From the figure:
Let the vertices of the rhombus be:
$A(3,0),\ B(4,5),\ C(-1,4),\ D(-2,-1)$
Area of the rhombus ABCD is given by:
$= \frac{1}{2}\times$ Product of length of diagonals
Hence, we have to find the lengths of the diagonals AC and BD of the rhombus.
The distance formula:
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
Length of the diagonal AC:
$AC = \sqrt{(3-(-1))^2+(0-4)^2} = \sqrt{16+16} = 4\sqrt{2}$
Length of the diagonal BD:
$BD = \sqrt{(4-(-2))^2+(5-(-1))^2} = \sqrt{36+36} = 6\sqrt{2}$
Thus, the area will be,
$= \frac{1}{2}\times (AC)\times(BD)$
$= \frac{1}{2}\times (4\sqrt{2})\times(6\sqrt{2}) = 24$ square units
To ease the learning for students, the below links to the exercises can also be used. Two exercises have been segregated, and students can analyze each exercise at a time on their own will.
Here is the latest NCERT syllabus, which is very useful for students before strategizing their study plan.
Also, links to some reference books which are important for further studies.
Students can check the following links for more in-depth learning.
After solving all the exercises, if students want to practice more, then exemplar solutions can be practised. For that, the following links can be used.
In class 10 Maths Chapter 7 following topics of coordinate geometry are covered.
Let the two points in a Cartesian plane be $(x_1,y_1)$ and $(x_2,y_2)$.
So, their distance = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
If a point $P(x, y)$ divides the line segment joining $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$ in the ratio $m: n$, then the coordinates of $P$ are:
$x=\frac{m x_2+n x_1}{m+n}, \quad y=\frac{m y_2+n y_1}{m+n}$
The formula of the area of a triangle can be used to prove that three points $A(x_1,y_1), B(x_2,y_2)$ and $C(x_3,y_3)$ are collinear.
If the area of the triangle formed by these three points is zero(0), then the points are collinear.
Area of a triangle $=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$
Let the two points in a Cartesian plane be $(x_1,y_1)$ and $(x_2,y_2)$.
Then their midpoint $M(x,y)=M\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
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Income Certificate (for minority category benefits).
Marksheets (11th and 12th from the Karnataka State Board).
KCET Admit Card and Scorecard.
This process will allow you to secure a seat based on your KCET performance and your category .
check link for more details
https://medicine.careers360.com/neet-college-predictor
Hope this helps you .
Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.
Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.
hello Zaid,
Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.
best of luck!
According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.
You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.
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