Proof:
Draw AL, BM and CN perpendicular to x-axis. LM = x 2 -x 1 MN = x 3 -x 2 LN = x 3 -x 1 Area of D ABC = area of trap {ALMB + BMNC - ALNC} This can be expressed in the form of a determinant Another form which is convenient to use for the area of triangles but which is very much useful wh..
Draw AL, BM and CN perpendicular to x-axis. LM = x 2 -x 1 MN = x 3 -x 2 LN = x 3 -x 1 Area of D ABC = area of trap {ALMB + BMNC - ALNC} This can be expressed in the form of a determinant Another form which is convenient to use for the area of triangles but which is very much useful wh..Proof
Let A (x 1 , y 1 ) and B (x 2 , y 2 ) be two points in the plane. Let d = distance between the points A and B. Draw AL and BM perpendicular to x-axis (parallel to y-axis). Draw AC perpendicular to BM to cut BM at C. In the figure, OL = x 1 , OM = x 2 [AC = LM = OM - OL = x 2 - x 1 ] MB = y 2 , MC =..
Let A (x 1 , y 1 ) and B (x 2 , y 2 ) be two points in the plane. Let d = distance between the points A and B. Draw AL and BM perpendicular to x-axis (parallel to y-axis). Draw AC perpendicular to BM to cut BM at C. In the figure, OL = x 1 , OM = x 2 [AC = LM = OM - OL = x 2 - x 1 ] MB = y 2 , MC =..Proof
The course of reasoning, which establishes the truth or falsity of a statement is called proof..
Proof:
..
..Proof:
(i) ( Theorem) From (i) and (ii), ..
(i) ( Theorem) From (i) and (ii), ..Proof:
(Vertically opposite angles) From (i) and (ii), ..
(Vertically opposite angles) From (i) and (ii), ..Proof:
( Linear pair) ( corresponding angles axiom) Similarly, we can prove that..
( Linear pair) ( corresponding angles axiom) Similarly, we can prove that..Proof:
MN is a transversal. (Alternate angles) . . (2) (Alternate angles) From the figure, ( is a straight line and sum of the angles at M = 180 o ) From (1) and (2)..
MN is a transversal. (Alternate angles) . . (2) (Alternate angles) From the figure, ( is a straight line and sum of the angles at M = 180 o ) From (1) and (2)..Proof:
( Given) (Vertically opposite angles) (corresponding angles) AB||CD ( corresponding angles axio..
( Given) (Vertically opposite angles) (corresponding angles) AB||CD ( corresponding angles axio..Proof:
Now in triangles, ABD and ABC, AB=AB (common side) (each angle is a right angle) AD=BC (opposite sides of parallelogram) BD=AC (corresponding parts of corresponding triangles) Hence the theorem is prov..
Now in triangles, ABD and ABC, AB=AB (common side) (each angle is a right angle) AD=BC (opposite sides of parallelogram) BD=AC (corresponding parts of corresponding triangles) Hence the theorem is prov..See what our Users say :
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