Co-ordinate Geometry is a branch of geometry which is used to identify a point on a plane. It was invented by RENE DESCARTES.
The two mutually perpendicular number lines which are used to find the position of a point on a plane is called rectangular axis. In the graph, XOX' is called x-axis and the point YOY' is called y-axis. The two lines XOX' and YOY' are also called rectangular co-ordinate axis which divide the plane into four equal parts which are called quadrant.
Plotting points is the process of locating a point whose co-ordinates are given. The plotting of points can be done on graph paper.
If the elements or co-ordinates of any two points are given, the distance between them can be found with the help of distance formulae.
Suppose that,
P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) are any two points in the co-ordinates plane and 'd' is the distance between them.
Now,
Draw PT and QK perpendicular on x-axis and PM perpendicular to QK.
Then,
OK = x_{2}, KQ = y_{2}, OT = x, PT = y.
PM = TK= OK - OT = (x_{2}- x_{1})
Also,
OM = OK - MK = QK - PT
= (y_{2}- y_{1}) [∴ MK = PT]
Since, ∠PMQ is a right angle, so \(\triangle\)PMQ is a right angle triangle.
Now,
\(\triangle\)PMQ using Pythagoras Theorem,
PQ^{2}= PM^{2}= QM^{2}
= (x_{2}- x_{1})^{2}+ (y_{2}- y_{1})^{2}
or, PQ = (x_{2}− x_{1})^{2} + (y_{2}− y_{1})^{2} ................... \(\sqrt{(x_2 − x_1)^2 + (y_2 − y_1)^2}\)
∴ Distance (d) = (x_{2}− x_{1})^{2}+ (y_{2}− y_{1})^{2} ...................\(\sqrt{(x_2 − x_1)^2 + (y_2 − y_1)^2}\)
Again,
The distance of a point A(x, y) from the origin O(0, 0) is,
or, OA = (x−0)^{2}+ (y−0)^{2} .................. \(\sqrt{(x−0)^2 + (y−0)^2}\)
∴ OA = x^{2}+ y^{2} .................... \(\sqrt{x^2 + y^2}\)
Again,
Slope of the line = PQ = Tanθ
\(\frac{OM}{PQ}\) = \(\frac{y_2 − y_1}{x_2 − x_1}\)
a) |
Scalene |
No sides are equal i.e. in ΔABC, AB≠BC, BC≠CA and CA≠AB. |
b) |
Isosceles |
Two sides are equal i.e. in ΔABC, AB=BC or BC=CA or AB=AC. |
c) |
Equilateral |
All sides are equal i.e. in ΔABC, AB = BC = CA. |
d) |
Right-Angled Triangle |
Sum of squares of two shorter sides is equal to the square of the longest side. |
e) |
Right Angled Isosceles Triangle |
Two shorter sides are equal and the sum of the squares of two shortest sides is equal to the square of the longest side. |
a) |
Parrallelogram |
Opposite sides are equal. In quadrilateral ABCD, AB=CD and BC=AD |
b) |
Rectangle |
Opposites sides are equal and diagonals are equal i.e. in quadrilateral, ABCD, AB=CD, AD=BC and AC=BD. |
c) |
Rhombus |
All sides are equal but diagonals are not equal i.e. in quadrilateral, ABCD, AB=BC=CD=DA and AC≠BD. |
d) |
Square |
All sides are equal and diagonals are also equal i.e. in quadrilateral ABCD, AB=BC=CD=DA and AC=BD. |
Solution:
A(−4, 0) = (x1, y1)
B(0,−3) = (x2, y2)
AB = ?
We know that,
or, AB = \(\sqrt{(x_2− x_1)^2 + (y_2−y_1)^2}\)
or, AB = \(\sqrt{(0 + 4)^2 + (− 3 − 0)^2}\)
or, AB = \(\sqrt{4)^2 + (3)^2}\)
or, AB = \(\sqrt{16 + 9}\)
or, AB = \(\sqrt{25}\)
∴ AB = 5 units.
Solution:
Here,
P(−8, 0) = (x_{1}, y_{1})
B(0, 6) =(x_{2}, y_{2})
PB = ?
We know that,
or, PB = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, PB = \(\sqrt{(0 + 8)^2 + (6 − 0)^2}\)
or, PB = \(\sqrt{(8)^2 + (6)^2}\)
or, PB = \(\sqrt{64 + 36}\)
or, PB = \(\sqrt{100}\)
∴ PB = 10 units.
Solution:
Here,
A(0,−5) = (x_{1}, y_{1})
B(4 , 8).= (x_{2}, y_{2})
Slope (m) = ?
We know that,
Slope (m) = \(\frac{y_2− y_1}{x_2− x_1}\)
=\(\frac{8 + 5}{4 − 0}\)
= \(\frac{13}{4}\)
Solution:
Here,
E(4,−7) = (x_{2}, y_{1})
F(−3, 4) = (x_{2}, y_{2})
Slope (m) = ?
we know that,
Slope (m) = \(\frac{y_2− y_1}{x_2− x_1}\)
=\(\frac{4 + 7}{−3 − 4}\)
= \(\frac{11}{−7}\)
Solution:
Let,
A (−1, 7) = (x_{1}, y_{1})
B (3, 10) = (x_{2}, y_{2})
Distance of AB = ?
We know that,
or, AB = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, AB = \(\sqrt{(3 + 1)^2 + (10 − 7)^2}\)
or, AB = \(\sqrt{(4)^2 + (3)^2}\)
or, AB = \(\sqrt{16+ 9}\)
or, AB = \(\sqrt{25}\)
∴ AB = 5 units
Solution:
Let, Q(0, 0) and R(−6,−4)
Distance of QR = ?
We know that,
or, QR = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, QR = \(\sqrt{(− 6 − 0)^2 + (− 4 − 0)^2}\)
or, QR = \(\sqrt{(6)^2 + (4)^2}\)
or, QR = \(\sqrt{36 + 16}\)
or, QR = \(\sqrt{52}\)
∴ QR = 2\(\sqrt{3}\)
Solution:
Here,
A(4, 2) = (x_{1}, y_{1})
B(6, 8) = (x_{2}, y_{2})
Slope (m) = ?
We know that,
Slope (m) = \(\frac{y_2− y_1}{x_2− x_1}\)
=\(\frac{8 − 2}{6 − 4}\)
=\(\frac{6}{2}\)
= 3
Find the distance.
(2, 1) and (6, 4)
Find the distance.
(2, 1) and (6, 4)
Find the distance.
(5, 6) and (8, 10)
Find the distance.
(3, -9) and (7, -6)
Find the distance.
(2, -7) and (6,- 4)
Find the distance from origin (3, 4)
Find the distance from origin (-3, 4)
Find the distance from origin (1, 2(sqrt{6}))
Find the distance from origin (2, (sqrt{21}))
Find the distance from origin (4,3)
Find the distance.
(-7, -7) and (-3, -4)
Find distance: (3, 4) and (5, 7)
Find distance: (5, 6 ) and (8, 9)
Find distance: (2, 0) and (0, 2)
Find distance: (0, 0) and (-6, -4)
Find the distance between point (0, 0) and (−6, −4)
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Sajeena
If the points (-1,b) and (b,a) are at equidistant from the point (m,n), then show that m(a b) = n(b-a)
Jan 24, 2017
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