NCERT Class 10 Maths Solutions: Coordinate Geometry
Question:
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (−3, 4).
Concept in a Minute:
The distance formula between two points (x1, y1) and (x2, y2) is given by sqrt((x2-x1)^2 + (y2-y1)^2). The concept of equidistance means that the distance from a point to two other points is equal.
Explanation:
Let the given point be P(x, y).
Let the two fixed points be A(3, 6) and B(−3, 4).
According to the question, the point P(x, y) is equidistant from A and B. This means the distance PA is equal to the distance PB.
PA = PB
We can use the distance formula to express PA and PB.
Distance formula: d = sqrt((x2 – x1)^2 + (y2 – y1)^2)
For distance PA:
x1 = x, y1 = y
x2 = 3, y2 = 6
PA = sqrt((3 – x)^2 + (6 – y)^2)
For distance PB:
x1 = x, y1 = y
x2 = -3, y2 = 4
PB = sqrt((-3 – x)^2 + (4 – y)^2)
Since PA = PB, we can equate the two expressions:
sqrt((3 – x)^2 + (6 – y)^2) = sqrt((-3 – x)^2 + (4 – y)^2)
To eliminate the square roots, we square both sides of the equation:
(3 – x)^2 + (6 – y)^2 = (-3 – x)^2 + (4 – y)^2
Now, expand the squared terms:
(3^2 – 2*3*x + x^2) + (6^2 – 2*6*y + y^2) = ((-3)^2 – 2*(-3)*x + x^2) + (4^2 – 2*4*y + y^2)
(9 – 6x + x^2) + (36 – 12y + y^2) = (9 + 6x + x^2) + (16 – 8y + y^2)
Combine the terms on each side:
9 – 6x + x^2 + 36 – 12y + y^2 = 9 + 6x + x^2 + 16 – 8y + y^2
x^2 – 6x + y^2 – 12y + 45 = x^2 + 6x + y^2 – 8y + 25
Now, we cancel out the terms that appear on both sides of the equation (x^2, y^2):
-6x – 12y + 45 = 6x – 8y + 25
Rearrange the terms to group x terms, y terms, and constant terms on one side. Let’s move all terms to the right side to make the x coefficient positive:
0 = 6x – (-6x) – 8y – (-12y) + 25 – 45
0 = 6x + 6x – 8y + 12y + 25 – 45
0 = 12x + 4y – 20
We can simplify this equation by dividing all terms by 4:
0 / 4 = (12x / 4) + (4y / 4) – (20 / 4)
0 = 3x + y – 5
So, the relation between x and y is:
3x + y – 5 = 0
or
3x + y = 5
Question:
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (− 1, 4) and (− 2, −1) taken in order.
[Hint: Area of a rhombus = $12$ (product of its diagonals)]
Concept in a Minute:
The key concept required to solve this problem is the formula for the area of a rhombus using its diagonals. A rhombus is a quadrilateral with all four sides equal in length. The diagonals of a rhombus bisect each other at right angles. The area of a rhombus can be calculated as half the product of the lengths of its diagonals. Distance formula is also needed to calculate the lengths of the diagonals. The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Explanation:
Let the vertices of the rhombus be A = (3, 0), B = (4, 5), C = (-1, 4), and D = (-2, -1) in order.
The diagonals of the rhombus are AC and BD.
First, we calculate the length of the diagonal AC using the distance formula:
Length of AC = $\sqrt{((-1) – 3)^2 + (4 – 0)^2}$
Length of AC = $\sqrt{(-4)^2 + (4)^2}$
Length of AC = $\sqrt{16 + 16}$
Length of AC = $\sqrt{32}$
Length of AC = $4\sqrt{2}$ units.
Next, we calculate the length of the diagonal BD using the distance formula:
Length of BD = $\sqrt{((-2) – 4)^2 + ((-1) – 5)^2}$
Length of BD = $\sqrt{(-6)^2 + (-6)^2}$
Length of BD = $\sqrt{36 + 36}$
Length of BD = $\sqrt{72}$
Length of BD = $6\sqrt{2}$ units.
Now, we can find the area of the rhombus using the formula:
Area of rhombus = $\frac{1}{2} \times (\text{length of diagonal 1}) \times (\text{length of diagonal 2})$
Area of rhombus = $\frac{1}{2} \times (\text{length of AC}) \times (\text{length of BD})$
Area of rhombus = $\frac{1}{2} \times (4\sqrt{2}) \times (6\sqrt{2})$
Area of rhombus = $\frac{1}{2} \times 24 \times (\sqrt{2} \times \sqrt{2})$
Area of rhombus = $\frac{1}{2} \times 24 \times 2$
Area of rhombus = 24 square units.
The final answer is $\boxed{24}$.
Question:
Name the type of quadrilateral formed, if any, by the following point, and give reasons for your answer:
(4, 5), (7, 6), (4, 3), (1, 2)
Concept in a Minute:
To determine the type of quadrilateral formed by given points, we need to calculate the lengths of all four sides and the lengths of the diagonals. Based on these lengths, we can identify the quadrilateral using properties of squares, rectangles, rhombuses, parallelograms, and trapeziums.
Explanation:
Let the given points be A(4, 5), B(7, 6), C(4, 3), and D(1, 2).
Step 1: Calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$.
AB = $\sqrt{(7-4)^2 + (6-5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}$
BC = $\sqrt{(4-7)^2 + (3-6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18}$
CD = $\sqrt{(1-4)^2 + (2-3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10}$
DA = $\sqrt{(4-1)^2 + (5-2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9+9} = \sqrt{18}$
Step 2: Observe the side lengths. We have AB = CD = $\sqrt{10}$ and BC = DA = $\sqrt{18}$. Since opposite sides are equal, the quadrilateral is at least a parallelogram.
Step 3: Calculate the lengths of the diagonals.
AC = $\sqrt{(4-4)^2 + (3-5)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{0+4} = \sqrt{4} = 2$
BD = $\sqrt{(1-7)^2 + (2-6)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36+16} = \sqrt{52}$
Step 4: Compare the lengths of the diagonals.
The lengths of the diagonals AC and BD are 2 and $\sqrt{52}$ respectively, which are not equal.
Step 5: Conclude the type of quadrilateral.
Since opposite sides are equal and the diagonals are not equal, the quadrilateral is a parallelogram.
The quadrilateral formed is a parallelogram because opposite sides are equal in length (AB = CD and BC = DA), but the diagonals are not equal in length (AC $\neq$ BD).
Question:
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Concept in a Minute:
The section formula is used to find the coordinates of a point that divides a line segment joining two points in a given ratio. If a point P(x, y) divides the line segment joining A(x1, y1) and B(x2, y2) in the ratio m:n, then the coordinates of P are given by:
x = (m*x2 + n*x1) / (m + n)
y = (m*y2 + n*y1) / (m + n)
Points of trisection divide a line segment into three equal parts. This means that the two points of trisection divide the line segment in the ratios 1:2 and 2:1.
Explanation:
Let the two given points be A = (4, -1) and B = (-2, -3).
We need to find the coordinates of the points of trisection of the line segment AB. Let these points be P and Q.
Points of trisection divide the line segment into three equal parts. This means that P divides AB in the ratio 1:2, and Q divides AB in the ratio 2:1.
To find the coordinates of point P, which divides AB in the ratio 1:2:
Here, (x1, y1) = (4, -1), (x2, y2) = (-2, -3), m = 1, and n = 2.
Using the section formula:
x_P = (1 * (-2) + 2 * 4) / (1 + 2) = (-2 + 8) / 3 = 6 / 3 = 2
y_P = (1 * (-3) + 2 * (-1)) / (1 + 2) = (-3 – 2) / 3 = -5 / 3
So, the coordinates of point P are (2, -5/3).
To find the coordinates of point Q, which divides AB in the ratio 2:1:
Here, (x1, y1) = (4, -1), (x2, y2) = (-2, -3), m = 2, and n = 1.
Using the section formula:
x_Q = (2 * (-2) + 1 * 4) / (2 + 1) = (-4 + 4) / 3 = 0 / 3 = 0
y_Q = (2 * (-3) + 1 * (-1)) / (2 + 1) = (-6 – 1) / 3 = -7 / 3
So, the coordinates of point Q are (0, -7/3).
Therefore, the coordinates of the points of trisection are (2, -5/3) and (0, -7/3).
Question:
Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear.
Concept in a Minute:
Collinearity of three points. Three points are collinear if they lie on the same straight line. This can be checked by calculating the slopes between pairs of points. If the slopes are equal, the points are collinear. Alternatively, the area of the triangle formed by these points will be zero if they are collinear.
Explanation:
To determine if the points (1, 5), (2, 3), and (−2, −11) are collinear, we can use the slope method.
Let the points be A = (1, 5), B = (2, 3), and C = (−2, −11).
Calculate the slope of the line segment AB:
Slope (m_AB) = (y2 – y1) / (x2 – x1)
m_AB = (3 – 5) / (2 – 1)
m_AB = -2 / 1
m_AB = -2
Calculate the slope of the line segment BC:
Slope (m_BC) = (y2 – y1) / (x2 – x1)
m_BC = (-11 – 3) / (-2 – 2)
m_BC = -14 / -4
m_BC = 7/2
Since the slope of AB (m_AB = -2) is not equal to the slope of BC (m_BC = 7/2), the points (1, 5), (2, 3), and (−2, −11) are not collinear.
Alternatively, using the area of a triangle method:
The area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by:
Area = 1/2 |x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)|
Substitute the given points:
x1 = 1, y1 = 5
x2 = 2, y2 = 3
x3 = -2, y3 = -11
Area = 1/2 |1(3 – (-11)) + 2(-11 – 5) + (-2)(5 – 3)|
Area = 1/2 |1(3 + 11) + 2(-16) + (-2)(2)|
Area = 1/2 |1(14) – 32 – 4|
Area = 1/2 |14 – 32 – 4|
Area = 1/2 |-22|
Area = 1/2 * 22
Area = 11
Since the area of the triangle formed by these points is 11, which is not zero, the points are not collinear.
Final Answer: The points are not collinear.
Question:
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
Concept in a Minute:
The distance between two points in a Cartesian coordinate system can be found using the distance formula, which is derived from the Pythagorean theorem. The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$.
Explanation:
First, we will find the distance between the points (0, 0) and (36, 15) using the distance formula.
Let $(x_1, y_1) = (0, 0)$ and $(x_2, y_2) = (36, 15)$.
Distance = $\sqrt{(36 – 0)^2 + (15 – 0)^2}$
Distance = $\sqrt{36^2 + 15^2}$
Distance = $\sqrt{1296 + 225}$
Distance = $\sqrt{1521}$
To find the square root of 1521, we can try to find its prime factors or estimate. We know that $30^2 = 900$ and $40^2 = 1600$. Since 1521 ends in 1, its square root must end in 1 or 9. Let’s try 39: $39 \times 39 = (40-1)(40-1) = 1600 – 40 – 40 + 1 = 1521$.
So, the distance between (0, 0) and (36, 15) is 39 units.
The second part of the question asks if we can now find the distance between the two towns A and B discussed in Section 7.2. To answer this, we need to know the coordinates of towns A and B as provided in Section 7.2 of the NCERT textbook. Without that information, we cannot calculate the distance between towns A and B.
Therefore, the distance between the points (0, 0) and (36, 15) is 39 units. However, the distance between the two towns A and B cannot be determined from the information given in this question alone. We need the specific coordinates of towns A and B from Section 7.2.
Question:
If A and B are (−2, −2) and (2, −4), respectively, find the coordinates of P such that $AP=37AB$ and P lies on the line segment AB.
Concept in a Minute:
This question involves the section formula in coordinate geometry. The section formula is used to find the coordinates of a point that divides a line segment joining two given points in a specific ratio. Since P lies on the line segment AB and AP = 3/7 AB, it means that P divides AB internally in the ratio 3:4.
Explanation:
Let the coordinates of point A be $(x_1, y_1) = (-2, -2)$ and the coordinates of point B be $(x_2, y_2) = (2, -4)$.
We are given that P is a point such that $AP = \frac{3}{7} AB$ and P lies on the line segment AB.
This means that the ratio of the length AP to the length PB is $3:4$.
So, P divides the line segment AB internally in the ratio $m:n = 3:4$.
The section formula for internal division states that if a point P(x, y) divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m:n$, then the coordinates of P are given by:
$x = \frac{mx_2 + nx_1}{m+n}$
$y = \frac{my_2 + ny_1}{m+n}$
In this case, $x_1 = -2$, $y_1 = -2$, $x_2 = 2$, $y_2 = -4$, $m = 3$, and $n = 4$.
Substitute these values into the section formula:
For the x-coordinate of P:
$x = \frac{(3)(2) + (4)(-2)}{3+4}$
$x = \frac{6 – 8}{7}$
$x = \frac{-2}{7}$
For the y-coordinate of P:
$y = \frac{(3)(-4) + (4)(-2)}{3+4}$
$y = \frac{-12 – 8}{7}$
$y = \frac{-20}{7}$
Therefore, the coordinates of point P are $(-\frac{2}{7}, -\frac{20}{7})$.
Question:
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Concept in a Minute:
The core concept needed is the distance formula, which is used to calculate the length of a line segment between two points in a coordinate plane. For a triangle to be isosceles, at least two of its sides must have equal lengths.
Explanation:
To check if the given points (5, -2), (6, 4), and (7, -2) form an isosceles triangle, we need to calculate the lengths of the three sides of the triangle and see if any two sides are equal.
Let the points be A = (5, -2), B = (6, 4), and C = (7, -2).
We use the distance formula: $d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$
1. Calculate the length of side AB:
$AB = \sqrt{(6 – 5)^2 + (4 – (-2))^2}$
$AB = \sqrt{(1)^2 + (4 + 2)^2}$
$AB = \sqrt{1^2 + 6^2}$
$AB = \sqrt{1 + 36}$
$AB = \sqrt{37}$
2. Calculate the length of side BC:
$BC = \sqrt{(7 – 6)^2 + (-2 – 4)^2}$
$BC = \sqrt{(1)^2 + (-6)^2}$
$BC = \sqrt{1 + 36}$
$BC = \sqrt{37}$
3. Calculate the length of side AC:
$AC = \sqrt{(7 – 5)^2 + (-2 – (-2))^2}$
$AC = \sqrt{(2)^2 + (-2 + 2)^2}$
$AC = \sqrt{2^2 + 0^2}$
$AC = \sqrt{4 + 0}$
$AC = \sqrt{4}$
$AC = 2$
Now, we compare the lengths of the sides:
AB = $\sqrt{37}$
BC = $\sqrt{37}$
AC = 2
Since AB = BC, two sides of the triangle have equal lengths. Therefore, the points (5, -2), (6, 4), and (7, -2) are the vertices of an isosceles triangle.
Question:
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Concept in a Minute:
The diagonals of a parallelogram bisect each other. This means the midpoint of one diagonal is the same as the midpoint of the other diagonal. The midpoint formula for a line segment with endpoints (x1, y1) and (x2, y2) is ((x1+x2)/2, (y1+y2)/2).
Explanation:
Let the vertices of the parallelogram be A = (1, 2), B = (4, y), C = (x, 6), and D = (3, 5) taken in order.
Since the vertices are taken in order, the diagonals of the parallelogram are AC and BD.
According to the property that diagonals bisect each other, the midpoint of diagonal AC must be the same as the midpoint of diagonal BD.
Midpoint of AC:
Using the midpoint formula, the midpoint of AC is ((1 + x)/2, (2 + 6)/2) = ((1 + x)/2, 8/2) = ((1 + x)/2, 4).
Midpoint of BD:
Using the midpoint formula, the midpoint of BD is ((4 + 3)/2, (y + 5)/2) = (7/2, (y + 5)/2).
Equating the midpoints:
Since the midpoints are the same, we can equate the corresponding coordinates:
For the x-coordinates: (1 + x)/2 = 7/2
For the y-coordinates: 4 = (y + 5)/2
Solving for x:
Multiply both sides of the x-coordinate equation by 2:
1 + x = 7
Subtract 1 from both sides:
x = 7 – 1
x = 6
Solving for y:
Multiply both sides of the y-coordinate equation by 2:
4 * 2 = y + 5
8 = y + 5
Subtract 5 from both sides:
y = 8 – 5
y = 3
Therefore, x = 6 and y = 3.
Question:
Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Concept in a Minute:
Section Formula: The section formula is used to find the coordinates of a point that divides a line segment joining two given points in a specific ratio. If a point P(x, y) divides the line segment joining A(x1, y1) and B(x2, y2) in the ratio m : n, then the coordinates of P are given by:
x = (m*x2 + n*x1) / (m + n)
y = (m*y2 + n*y1) / (m + n)
Explanation:
The problem asks to find the coordinates of a point that divides the line segment joining two given points, (–1, 7) and (4, –3), in the ratio 2 : 3. This is a direct application of the section formula.
Let the two given points be A = (x1, y1) = (–1, 7) and B = (x2, y2) = (4, –3).
Let the ratio in which the point divides the join be m : n = 2 : 3.
Let the coordinates of the point that divides the join be P(x, y).
Using the section formula for the x-coordinate:
x = (m*x2 + n*x1) / (m + n)
Substitute the given values: m = 2, n = 3, x1 = –1, x2 = 4.
x = (2 * 4 + 3 * –1) / (2 + 3)
x = (8 – 3) / 5
x = 5 / 5
x = 1
Using the section formula for the y-coordinate:
y = (m*y2 + n*y1) / (m + n)
Substitute the given values: m = 2, n = 3, y1 = 7, y2 = –3.
y = (2 * –3 + 3 * 7) / (2 + 3)
y = (–6 + 21) / 5
y = 15 / 5
y = 3
Therefore, the coordinates of the point are (1, 3).
Question:
Find the distance between the following pairs of points:
(−5, 7), (−1, 3)
Concept in a Minute:
The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a Cartesian coordinate system is given by the distance formula: $d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$. This formula is derived from the Pythagorean theorem.
Explanation:
We are asked to find the distance between the points (−5, 7) and (−1, 3).
Let the first point be $(x_1, y_1) = (-5, 7)$ and the second point be $(x_2, y_2) = (-1, 3)$.
Using the distance formula:
$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$
Substitute the coordinates of the given points into the formula:
$d = \sqrt{(-1 – (-5))^2 + (3 – 7)^2}$
First, calculate the differences in the x and y coordinates:
$x_2 – x_1 = -1 – (-5) = -1 + 5 = 4$
$y_2 – y_1 = 3 – 7 = -4$
Now, square these differences:
$(x_2 – x_1)^2 = (4)^2 = 16$
$(y_2 – y_1)^2 = (-4)^2 = 16$
Add the squared differences:
$(x_2 – x_1)^2 + (y_2 – y_1)^2 = 16 + 16 = 32$
Finally, take the square root of the sum:
$d = \sqrt{32}$
To simplify the square root of 32, we can find its prime factorization: $32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5$.
We can rewrite $\sqrt{32}$ as $\sqrt{16 \times 2}$.
Since $\sqrt{16} = 4$, we have:
$d = 4\sqrt{2}$
The distance between the points (−5, 7) and (−1, 3) is $4\sqrt{2}$ units.
Question:
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Concept in a Minute:
The distance formula between two points (x1, y1) and (x2, y2) is given by sqrt((x2-x1)^2 + (y2-y1)^2).
A point on the x-axis has coordinates of the form (x, 0).
Equidistant means the distance from the point to both given points is equal.
Explanation:
Let the point on the x-axis be P(x, 0).
We are given two points A(2, -5) and B(-2, 9).
Since P is equidistant from A and B, the distance PA is equal to the distance PB.
Using the distance formula:
PA = sqrt((x – 2)^2 + (0 – (-5))^2) = sqrt((x – 2)^2 + 5^2) = sqrt((x – 2)^2 + 25)
PB = sqrt((x – (-2))^2 + (0 – 9)^2) = sqrt((x + 2)^2 + (-9)^2) = sqrt((x + 2)^2 + 81)
Since PA = PB, we can equate the squares of the distances to eliminate the square root:
PA^2 = PB^2
(x – 2)^2 + 25 = (x + 2)^2 + 81
Now, expand the squared terms:
x^2 – 4x + 4 + 25 = x^2 + 4x + 4 + 81
Simplify both sides:
x^2 – 4x + 29 = x^2 + 4x + 85
Subtract x^2 from both sides:
-4x + 29 = 4x + 85
Now, isolate the x terms on one side and the constant terms on the other.
Add 4x to both sides:
29 = 8x + 85
Subtract 85 from both sides:
29 – 85 = 8x
-56 = 8x
Divide by 8 to find x:
x = -56 / 8
x = -7
Therefore, the point on the x-axis which is equidistant from (2, -5) and (-2, 9) is (-7, 0).
Question:
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Concept in a Minute:
The section formula is used to find the coordinates of a point that divides a line segment joining two given points in a specific ratio. The formula is:
If a point (x, y) divides the line segment joining points (x1, y1) and (x2, y2) in the ratio m:n, then
x = (mx2 + nx1) / (m + n)
y = (my2 + ny1) / (m + n)
Explanation:
Let the given points be A = (-3, 10) and B = (6, -8).
Let the point that divides the line segment AB be P = (-1, 6).
We need to find the ratio in which P divides AB. Let this ratio be k:1.
Using the section formula for the x-coordinate:
x = (kx2 + 1×1) / (k + 1)
-1 = (k * 6 + 1 * -3) / (k + 1)
-1 * (k + 1) = 6k – 3
-k – 1 = 6k – 3
-1 + 3 = 6k + k
2 = 7k
k = 2/7
So, the ratio is 2:7.
Let’s verify this with the y-coordinate:
y = (ky2 + 1y1) / (k + 1)
6 = (k * -8 + 1 * 10) / (k + 1)
6 * (k + 1) = -8k + 10
6k + 6 = -8k + 10
6k + 8k = 10 – 6
14k = 4
k = 4/14
k = 2/7
The ratio obtained from both x and y coordinates is the same, which is 2:7.
Therefore, the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6) in the ratio 2:7.
Question:
Find the ratio in which the line segment joining A (1, −5) and B (−4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.
Concept in a Minute:
Section Formula: If a point P(x, y) divides the line segment joining A(x1, y1) and B(x2, y2) in the ratio m:n, then the coordinates of P are given by x = (mx2 + nx1) / (m+n) and y = (my2 + ny1) / (m+n).
Properties of x-axis: Any point lying on the x-axis has a y-coordinate of 0.
Explanation:
Let the point where the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis be P(x, y).
Since the point P lies on the x-axis, its y-coordinate is 0. So, P(x, 0).
Let the ratio in which the line segment is divided be m:n.
Using the section formula for the y-coordinate:
0 = (m * 5 + n * (-5)) / (m + n)
0 = 5m – 5n
5m = 5n
m/n = 5/5
m/n = 1/1
So, the ratio is 1:1.
Now, using the section formula for the x-coordinate with the ratio 1:1:
x = (1 * (-4) + 1 * 1) / (1 + 1)
x = (-4 + 1) / 2
x = -3 / 2
Therefore, the ratio in which the line segment is divided by the x-axis is 1:1.
The coordinates of the point of division are (-3/2, 0).
Question:
Find the values of y for which the distance between the points P (2, -3) and Q (10, y) is 10 units.
Concept in a Minute:
Distance Formula: The distance between two points (x1, y1) and (x2, y2) in a Cartesian plane is given by the formula: d = sqrt((x2 – x1)^2 + (y2 – y1)^2).
Explanation:
Let the two points be P(x1, y1) = (2, -3) and Q(x2, y2) = (10, y).
The distance between P and Q is given as d = 10 units.
Using the distance formula:
d = sqrt((x2 – x1)^2 + (y2 – y1)^2)
10 = sqrt((10 – 2)^2 + (y – (-3))^2)
10 = sqrt((8)^2 + (y + 3)^2)
10 = sqrt(64 + (y + 3)^2)
To remove the square root, square both sides of the equation:
10^2 = 64 + (y + 3)^2
100 = 64 + (y + 3)^2
Now, isolate the term (y + 3)^2:
100 – 64 = (y + 3)^2
36 = (y + 3)^2
Take the square root of both sides:
sqrt(36) = y + 3
This gives two possible cases:
Case 1:
6 = y + 3
y = 6 – 3
y = 3
Case 2:
-6 = y + 3
y = -6 – 3
y = -9
Therefore, the values of y for which the distance between the points P (2, -3) and Q (10, y) is 10 units are 3 and -9.
Question:
Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, -3) and B is (1, 4).
Concept in a Minute:
Midpoint formula: The coordinates of the midpoint (x, y) of a line segment with endpoints (x1, y1) and (x2, y2) are given by x = (x1 + x2)/2 and y = (y1 + y2)/2.
Explanation:
Let the coordinates of point A be (x, y).
The center of the circle is the midpoint of its diameter AB.
Given:
Center of the circle (midpoint of AB) = (2, -3)
Coordinates of point B = (1, 4)
Coordinates of point A = (x, y)
Using the midpoint formula:
Midpoint x-coordinate = (x-coordinate of A + x-coordinate of B) / 2
2 = (x + 1) / 2
Multiply both sides by 2:
4 = x + 1
Subtract 1 from both sides:
x = 4 – 1
x = 3
Midpoint y-coordinate = (y-coordinate of A + y-coordinate of B) / 2
-3 = (y + 4) / 2
Multiply both sides by 2:
-6 = y + 4
Subtract 4 from both sides:
y = -6 – 4
y = -10
Therefore, the coordinates of point A are (3, -10).
Question:
Find the distance between the following pairs of points:
(a, b), (−a, −b)
Concept in a Minute:
The distance formula is used to find the distance between two points in a Cartesian coordinate system. The formula is derived from the Pythagorean theorem.
Explanation:
To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$, we use the distance formula:
$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$
In this question, our two points are $(a, b)$ and $(-a, -b)$.
Let $(x_1, y_1) = (a, b)$ and $(x_2, y_2) = (-a, -b)$.
Now, substitute these values into the distance formula:
$d = \sqrt{(-a – a)^2 + (-b – b)^2}$
$d = \sqrt{(-2a)^2 + (-2b)^2}$
$d = \sqrt{4a^2 + 4b^2}$
$d = \sqrt{4(a^2 + b^2)}$
$d = 2\sqrt{a^2 + b^2}$
Thus, the distance between the points $(a, b)$ and $(-a, -b)$ is $2\sqrt{a^2 + b^2}$.
Question:
Find the distance between the following pairs of points:
(2, 3), (4, 1)
Concept in a Minute:
The distance formula is used to find the distance between two points in a Cartesian coordinate system. If the coordinates of two points are (x1, y1) and (x2, y2), then the distance between them is given by the formula:
d = sqrt((x2 – x1)^2 + (y2 – y1)^2)
Explanation:
We are given two points: P1 = (2, 3) and P2 = (4, 1).
We can assign the coordinates as:
x1 = 2, y1 = 3
x2 = 4, y2 = 1
Now, we apply the distance formula:
d = sqrt((x2 – x1)^2 + (y2 – y1)^2)
d = sqrt((4 – 2)^2 + (1 – 3)^2)
d = sqrt((2)^2 + (-2)^2)
d = sqrt(4 + 4)
d = sqrt(8)
To simplify sqrt(8), we can write it as sqrt(4 * 2) = sqrt(4) * sqrt(2) = 2*sqrt(2).
Therefore, the distance between the points (2, 3) and (4, 1) is 2*sqrt(2) units.
Next Chapter: Introduction to Trigonometry
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