Distance Point Triangle
A triangle is a closed shape
A triangle is a closed shape. A triangle having 3 angles and 3 sides.Three sides joining and covered by three points. Those three points are also called as vertex of the triangle. First we can find the distance between these three points.we use distance formula to find the distance between these points. The distance formula is mainly used to find the distance between points. With the help of the distance formula, we can identify the type of the given triangle that is a right triangle, isosceles triangle or equilateral triangle.
Brief Description of Distance Point Triangle
Let P1 (x1, y1) and P2 (x2, y2) be two distinct points in the Cartesian plane and denote the distance between P1 and P2 by dP1, P2) or by P1P2. Draw the line segment P1P2 the line segment P1P2 is neither parallel to the x-axis nor parallel to the y-axis (see Figure ).
Distance Point Triangle
Draw a line through P1 parallel to x-axis and a line through P2 parallel to y-axis. Let these lines intersect at the point P3. Then P3 (x2, y1). The length of the line segment P1P3 is 21xx- and the length of the segment P3P2 is y1y2-. We observe that the triangle ?P1P3P2 is a right triangle.
Therefore
d(P1,P2)2 =d(P1,P3)2 + d(P3,P2)2=|x1 - x2|2 + |y 1 - y2|2
= (x1 - x2)2 + (y1 - y2)2
= (x2 - x1)2 + (y2 - y1)2.
? d (P1, P2) = sqrt((x2-x1)^2 + (y2 - y1)^2)
Example Problems Distance Point Triangle
Example 1:
Find the distance between these points and find whether this points make a right triangle are not. The points are P (7, 1), Q (-4,-1) and R (4,5) are the vertices of a right triangle.
Solution:
The points P, Q, R form a triangle. To show that ?PQR is a right triangle, we have to show that one vertex angle is 90. This is done by showing that the lengths of the sides of the triangle satisfy Pythagoras theorem. Here
PQ = sqrt((-4-7)^2 + (-1 - 1)^2)=v125 = 5v5
QR = sqrt((4+4)^2 + (5+1)^2)=v100 = 10
PR = sqrt((4-7)^2 + (5 - 1)^2)=v25 = 5
We observe that QR2 + PR2 = PQ2.
? The Pythagoras formula is satisfied.
? So ?PQR is the right triangle.
Example 2:
Show that the points (0, 3), (0,1) and (v3,2) are the vertices of an equilateral triangle.
Solution:
Let the points be A, B and C respectively. One way of showing that ?ABC is an equilateral triangle is to show that all its sides are of equal length. Here we find that
d(A,B)= sqrt((0-0)^2 + (1 - 3)^2)=v4 =2
d(B,C)= sqrt((sqrt 3-0)^2 + (2 - 1)^2)=v4 = 2
d(C,A)=sqrt((0-sqrt 3)^2 + (3 - 2)^2)=v4 = 2
? d (A, B) = d (B, C) = d (C, A).
?So ?ABC - equilateral triangle.
A Point is represented as dot in a diagram which shows the main idea for location. Point has a no length, no measurements. Point denoted by letters A,B,C. A point denotes a position in space. In practice, we put a small dot on a paper or on a black board to indicate a point. But theoretically, a point has no size or shape. To learning a point can also be understood as the position where two lines intersect each other.
Point Learning-classification of Points:
Three or more points are called collinear points if they all lie on the same line.
Three or more points are called non-collinear points if at least one of them does not lie on the line passing through two of the points
If three or more lines pass through the same point P, the lines are said to be concurrent and the point P is called the point of concurrence.
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Points Learning - Example Problems:
Points learning - Example 1:
Find the distance between the points A(-15, -3) and B (7, 1).
Solution:
Let take d be the distance between A and B. (x1,y1)= (-15, -3), (x2-y2)= (7, 1).
Then d (A, B) =sqrt((x2-x1)^2+(y2-y1)^2)
=sqrt((7+15)^2 +(1+3)^2)
=sqrt(22^2+4^2)
=sqrt(484+16)
=500
=10 sqrt5
Points learning - Example 2:
Show that the points (-4, -9), (2, 0) and (4, 3) are collinear.
Solution:
Let A, B and C be the given points respectively. Then
A= (-4,-9)
B =(2, 0)
AB =sqrt((2+4)^2+(0+9)^2)
=sqrt(6^2+9^2)
=sqrt(36+81)
=sqrt117
=sqrt(9 xx13)
=3sqrt13
B (2, 0)
C (4, 3)
BC= sqrt((4-2)^2+(3-0)^2)
=sqrt(2^2+3^2)
=4+9
=13
A= (-4,-9)
C (4, 3)
AC=sqrt((4+4)^2+(3+9)^2)
=sqrt(8^2+12^2)
=sqrt(64+144)
=208 =16xx13 =4 sqrt13
AB+BC=AC
3sqrt13 +sqrt13 =4sqrt13
A,B and C are collinear.
Points learning - Example 3:
Find co-ordinate of the mid point of the line segment joining given points A(3,2) and B(7,8)
Solution:
The required mid point is
Formula ((x1+x2)/(2)) , ((y1+ y2)/(2)) here, (x1, y1) = (3,2),(x2, y2) = (7,8)
= ((( -3)+ 7)/(2))((2 +8)/(2))
=(4/2) (10/2)
= (2,5)
Points learning - Example 4:
Find the centroid of the triangle whose vertices's are the points (8, 4), (1,3) and (3,1).
Solution:
(x1 y1) = (8,4), (x2 y2) = (1,3), (x3 y3) =(3, 1)
Formula ((x1+ x2+ x3)/(3)) ,(( y1+ y2+ y3)/(3))
The centroid of the triangle is ((8+1+3)/3) ,((4+3+(-1))/3))
=(12/3),(6/3)
=(4,2)
by: mathqa
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