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The calculation of a distance between any two points on a plane surface can be done.
Let P(x_{1},y_{1}) and Q(x_{2},x_{2}) be two given points in the coordinate plane. Draw perpendicular PL from P to the x-axis. Then , OL = x_{1} and PL = y_{1}. Draw perpendicular QM from Q to the x-axis.
Then, OM = x_{2}and QM =y_{2}
Again, draw perpendicular PR from P to the line segment QM.
Then, PR = LM = OM-OL =x_{2}-x_{1}
QR = QM - NM = QM - PL = y_{2}-y_{1}
From right-angled ΔPRQ, by Pythagoras theorem,
PQ^{2} = PR^{2}+ QR^{2} = (x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}
PQ^{ = }\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
(Only square root is to be taken because PQ being the distance beyween two points is positive)
The distance of the point (x,y) from the origin (0,0) =\(\sqrt{(x-0)^2 + (y-0)^2}\) = \(\sqrt{x^2+y^2}\)
Remarks
To find the coordinates of the point which divides internally the line joining two points (x_{1},y_{1}) and (x_{2},y_{2}) in the given ratio m_{1}:m_{2}.
Let A(x, y) and B(x_{1}, y_{1}) be two given points. Let the point P (x_{2}, y_{2}) divide the line joining AB internally in the ratiom_{1}: m_{2}
Then AP : PB = m_{1}: m_{2}
Draw perpendiculars AL, PN and BM from A, P and B respectively to the x-axis. Then, OL = x_{1}, ON = x, OM = x_{2}, AL = y_{1}, PN = y and BM = y_{2}. Again draw respectively AQ and PR from A and P to the line segments PN and BM respectively.
The AQ =LN = ON -OL = x - x_{1} PR = NM = OM - ON = x_{2}-x
PQ = PN - QN = PN -AL = y - y_{1 }BR = BM - RM = BM - PN = y_{2} - y
In Δ^{s} PQA and BRA, ∠PQA = ∠ BRP, ∠QAP = ∠RPB and ∠APQ = ∠PBR
So, Δ^{s }PQA and BRP are similar.
Then,
\(\frac{AP}{PB}\)
= \(\frac{AQ}{PR}\)
= \(\frac{QP}{RB}\)
= \(\frac{m_1}{m_2}\)
= \(\frac{x-x_1}{x_2-x}\)
= \(\frac{y-y_1}{y_2y}\)
Taking first two relation, we get,
or, \(\frac{m_1}{m_2}\) = \(\frac{x-x_1}{x_2-x}\)
or, m_{2}x - m_{2}x_{1} = m_{1}x_{2}-m_{1}x
or, m_{2}x + m_{1}x = m_{1}x_{2} + m_{2}x_{1}
or, x(m_{1}+m_{2}) = m_{1}x_{2}+m_{2}x_{1}
or, x = \(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}\)
Similarly, from the relations
\(\frac{m_1}{m_2}\) = \(\frac{y-y_1}{y_2-y}\)
we get
y = \(\frac{m_1y_2 + m_2y_1}{m_1 + m_2}\)
∴ The coordinates of P are \(\frac{m_1x_2 + m_2x_1}{m_1 +m_2}\),\(\frac{m_1y_2 + m_2y_1}{m_1+m_2}\)
If the point P(x, y) divides AB externally in the ratio of m_{1}:m_{2} then the divided segment BP is measured in opposite direction and hence m_{2} is taken as negative.
\(\therefore\) The section formulae for external division is,
(x, y) = (\(\frac{m_1x_2 - m_2x_1}{m_1 - m_2}\)),(\(\frac{m_1y_2 - m_2y_1}{m_1 - m_2}\))
In special case, the midpoint formulae is also used'.
m_{1}:m_{2} = 1:1 i.e. m_{1} = m_{2}\(\therefore\) x = \(\frac{x_1 + x_2}{2}\) and y = \(\frac{y_1 + y_2}{2}\)
Thus, co-ordinates P(x, y) are P(\(\frac{x_1 + x_2}{2}\),\(\frac{y_1 + y_2}{2}\)) which is called mid-point formulae.
If the point P(x, y) divides the joining two points A(x_{1}, y_{1}) and B(x_{2}, y_{2}) in the ratio k : 1, then
x = \(\frac{k . x_2 + 1 . x_1}{k + 1}\) = \(\frac{kx_2 + x_1}{k + 1}\)
and, y = \(\frac{k . y_2 + 1 . y_1}{k + 1}\) = \(\frac{ky_2 + y_1}{k + 1}\)
∴ Coordinate of P are (\(\frac{kx_2 + x_1}{k + 1}, \frac{ky_2 + y_1}{k + 1}\))
For the problems in which it is required to find the ratio when a given point divides the join of two given points, it is convenient to take the ratio k : 1.
Let P(x_{1}, y_{1}), Q(x_{2}, y_{2}) and R(x_{3}, y_{3}) are the vertices of a triangle PQR. Let S, T and U are the mid - points of sides QR, RP and PQ respectively. Then, PS, QT and RU are called medians of the triangle PQR. If these medians intersect each other at a point N, then N is called the centroid of the triangle PQR.
Since S is the middle point of the side QR, then its coordinates are (\(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\))
From plane geometry, we know that the centroid of a triangle divides the median in the ratio 2 : 1. Now, if coordinates of N are (x, y), then by section formula.
x = \(\frac{2 . \frac{x_2 + x_3}{2} + 1 . x_1}{2 + 1}\) = \(\frac{x_1 + x_2 + x_3}{3}\) and y = \(\frac{2 . \frac{y_2 + y_3}{2} + 1 . y_1}{2 + 1}\) = \(\frac{y_1 + y_2 + y_3}{3}\)
Hence, coordinates of N are (\(\frac{x_1 + x_2 + x_3}{3},\frac{y_1 + y_2 + y_3}{3}\))
→ The formula remains the same if the points P (x_{1},y_{1}) and Q (x_{2},y_{2}) are taken in different quadrants.
→ If a point lies on x-axis, its ordinate is zero. Therefore, any point on x-axis can be taken as (x,0).
→ If a point lies on y-axis, its abscissa is zero. Therefore, any point on y-axis can be taken as (0,y).
→ To prove that a quadrilateral is a
1) rhombus, show that all sides are equal.
2) square, show that all sides are equal and diagonals are equal.
3) parallelogram, show that opposite sides are equal.
4) rectangle, show that opposite sides are equal and diagonals are equal.
→ To prove that a triangle is a
1) scalene, show that none of the sides are equal.
2) isosceles, show that two sides are equal.
3) equilateral, show that all sides are equal.
4) right-angled triangle, show that square on one side equals
.
Here , given P(x1 , y1) = (2 , 3) and Q(x2 , y2) = (4 , 3)
Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)
Distance , PQ = \(\sqrt{(4-2)^{2} + (3-3)^{2}}\) = \(\sqrt{2^{2} + 0^{2}}\) = \(\sqrt{2^{2}}\) = 2.
PQ = 2 units. Ans.
Here given , P(x1, ,y1) = (-1 , 3) and Q(x2 , y2) = (5 , 1)
Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)
Distance PQ = \(\sqrt{(5 -(-1)^{2} + (1 -3)^{2}}\) = \(\sqrt{6^{2} + 2^{2}}\) = \(\sqrt{36 + 4}\)
\(\therefore\) PQ = \(\sqrt{40}\) = \(\sqrt{4 \times 10}\) = 2\(\sqrt{10}\) Units. Ans.
Here given , P(x1 , y1) = (1 , -2) and Q(x2, , y2) = (-2 , 2)
Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)
Distance PQ = \(\sqrt{(-1 -3)^{2} + [-1 - (-1)]^{2}}\)
= \(\sqrt{(-4)^{2} + (-1 = 1)^{2}}\)
= \(\sqrt{4^{2} + 0^{2}}\) = \(\therefore\) PQ = \(\sqrt{4^{2}}\) = 4 units. Ans.
Here given , P(x1, y1) = (-6 , 7) and Q(x2 , y2) = (-1 , -5)
Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)
Distance PQ = \(\sqrt{[-1 -(-6)]^{2} + (-5 -7)^{2}}\)
= \(\sqrt{-1 ^{2} + 6^{2} + (-12)^{2}}\) = \(\sqrt{5 ^{2} + 144}\)
= \(\sqrt{25 + 144}\) = \(\sqrt{169}\)
\(\therefore\) PQ = 13 Units Ans.
Here given , P(x1 , y1) = (4 , 3) and Q(x2 , y2) = (3 , -6)
Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1) ^{2}}\)
Distance PQ = \(\sqrt{(-2 , -4)^{2} + (2 - 3 )^{2}}\) = \(\sqrt{(-6)^{2} + (-1)^{2}}\)
= \(\sqrt{36 + 1}\) = \(\sqrt{37}\) Units. Ans.
Here , given P(x1 , y1) = (-2 , 6) and Q(x2 , y2) = (3 , -6)
Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)
Distance PQ = \(\sqrt{[ 3 - (-2)^{2} + (-6 -6)^{2}}\)
= \(\sqrt{(3 + 2)^{2} + (-12)^{2}}\) = \(\sqrt{5 ^{2} + 144}\)
= \(\sqrt{25 + 144}\) = \(\sqrt{169}\) = 13 Units. Ans.
Here, given A(0 , 0) and B(3 , -4)
\(\therefore\) AB = \(\sqrt{(3 - 0)^{2} + (4 - 0)^{2}}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5 units.
Here given , A(0 , 0) amd C(-3 , 4)
\(\therefore\) AC = \(\sqrt{(-3 -0)^{2} + (4 - 0)^{2}}\)
= \(\sqrt{(-3)^{2} + (4)^{2}}\)
= \(\sqrt{9 + 16}\)
= \(\sqrt{25}\) = 5 units. Ans.
Here given , A(0 , 0) and C (-3 , 4)
\(\therefore\) AC = \(\sqrt{(-3 - 0)^{2}+(4 - 0)^{2} }\)
= \(\sqrt{(-3)^{2} + (4)^{2}}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5 units. Ans.
Here,A(-2,1) and B(4,3) be any two points. Using distance formula,we get (AB)2 =(x2-x1)2+(y2-y1)2 =(-2-4)2 + (1-3)2 =(-6)2 +(-2)2 =36+4 =40
Find the distance between the following points:
(2,1) and (6,4)
4 units
5 units
9 units
2 units
Find the distance between the following points:
(2,3) and (6,6)
5 units
9 units
4 units
2 units
Find the distance between the following points:
(-4,1) and (0,4)
5 units
9 units
3 units
7 units
Find the distance between the following point and the origin:
(3,4)
10 units
5 units
3 units
2 units
Find the distance between the following point and the origin:
(-3,4)
10 units
6 units
2 units
5 units
The X-co-ordinate of a point P on X-axis is 4 and Y-Co-ordinates of a point Q and Y-axis is -3.Find the distance of PQ.
8 units
1 units
10 units
5 units
The X-co-ordinate of a point P on X-axis is 6 and Y-Co-ordinates of a point Q and Y-axis is 8.Find the distance of PQ.
5 units
10 units
8 units
2 units
The two continuous vertices of an equilateral of an equilateral triangle are (2,4) and (2,6) respectively.Find the semiperimeter of the triangle.
2 units
3 units
8 units
4 units
The two continuous vertices of a rhombus are (3,-3) and (-3,3) respectively.Find the perimeter of the rhombus.
24√1
29√1
24√2
44√3
Find the point (a,b) which is equidistant from the point (-4,7),)(4,7) and (4,1).
(6,3)
(1,5)
5,3)
(0,4)
Find the co-ordinates of midpoint of the line segment joining the points A(4,5) and B(6,5).
The co-ordinates of two ends of a line segment AB are (6,2) and (8,6) respectively.Find the co-ordinates of midpoint of the line segment AB.
(5,5)
(3,4)
(2,2)
(7,4)
(2,3) is the midpoint between the points (a,4) and (2,b).Find the values of a and b.
(1,1)
(2,2)
(3,3)
(4,4)
In what ratio does the point (5,3)divide the line segment joining (2,3) &(7,3)?
2:2
8:9
3:2
4:8
In what ratio does the point (2,b)divide the line segment joining points (-4,3) &(6,3)?
3:4
4:2
4:1
3:2
Three continuous corners of a parallelogram are (8,5),(-7,-5) and (-5,-5).Find the co-ordinates of the remaining corner.
(9,6)
(10,5)
(7,7)
(8,8)
Three continuous corners of a parallelogram are (8,5),(-7,-5) and (-5,-5).Find the co-ordinates of the remaining corner.
(8,8)
(10,5)
(7,7)
(9,6)
Three continuous corners of a parallelogram are p(3,4), Q(2,2) and R(5,2).Find the co-ordinates of the remaining corner.
(3,2)
(6,4)
(9,3)
(7 ,4)
The points p(3,5) divides the line segment joining the points A(2,4) and B in the ratio 1:3.Find the co-ordinates of midnight of the line segment AB.
(3,3)
(4,6)
(2,2)
(5,5)
The points p(2,-1) divides the line segment joining the points A(1,-3) and B in the ratio 1:2.Find the midpoint of co-ordinate the line segment AB.
(8,9)
(4,7)
(1,1)
(2.5,0)
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Could you provide some examples for locus's problems pleasant.
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