Introduction for geometric proofs practice:
The geometric proofs practices are generally involved in solving the proof for some problems that involving the statements from different theorems that are earlierly solved. we have to analyze the question first and then we have to choose what kind of statements are used for the problems given. After those steps are taken, we have to match it for the statements with the problems. The proof represents the way that how to we prove the given problem or some postulates.
(Source from Wikipedia)
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Examples to explain "geometric proofs practice"
We have to prove the straight line's equation is 3x + 2y = 7 and which is passing through the point (1,2) and which making the intercepts on the axes of the co-ordinate which are in the ratio 2 : 3.
Geometry proof:
The intercept form is `x/a` + `y/b` = 1 --------- (1)
The intercepts are in the ratio 2 : 3 ? a = 2k, b = 3k.
(1) becomes `x/(2k)` + `y/(3k)` = 1
i.e. 3x + 2y = 6k
We know that the point (1, 2) lies on the straight line given above, 3 + 4 = 6k
i.e. 6k = 7
Hence the required straight line's equation is 3x + 2y = 7
Thus we have proved the proof for finding the straight line's equation.
Practice:
In this problem as we are seen that it have specific points to prove the problems such that it having ratio on the concern and the point which the line passing.
Problems to explain "geometric proofs practice"
We have to prove that the point's co-ordinates are (8, 9) and (- 42, - 41) and given that it is reside on the straight line y = x + 1 which are at a distance of 5 units from the straight line 4x - 3y + 20 = 0
Geometry proof:
Let (x1, y1) be a point on y = x + 1
? y1 = x1 + 1 … (1)
The length of the perpendicular from (x1, y1) to the straight line
4x - 3y + 20 = 0 is `|(4x1 - 3y1 + 20)/sqrt(4^2 + (-3)^2)|` = `+-` `((4x1 - 3y1 + 20)/5)`
But the length of the perpendicular is given as 5.
? `+-` `((4x1 - 3y1 + 20)/5)` = 5
4x1- 3y1 + 20 = `+-` 25
Considering the positive sign, 4x1- 3y1 + 20 = `+` 25
? 4x1 - 3y1 = 5 … (2)
Considering the positive sign, 4x1- 3y1 + 20 = `-` 25
? 4x1 - 3y1 =`-` 45 … (3)
Solving (1) and (2), we get x1 = 8, y1 = 9
Solving (1) and (3), we get x1 = - 42, y1 = - 41.
? The required points co-ordinates are (8, 9) and (- 42, - 41).
Thus we have proved the proof with the given points.
Practice:
In this problem as we are seen that it have specific points to prove the problems such that it having the perpendicular distance from the two straight lines.
Practice problems on geometric proof:
We have to prove the straight line's equation is x - `sqrt(3)` y + 12 = 0, if the perpendicular comes from the origin, which makes an 120° angle with x-axis and the distance from perpendicular comes from the origin is 6 units.
We have to show that the straight lines 132x +13 y - 9 = 0 and 132x + 13y - 10 = 0 are parallel.
The geometric proofs practices are generally involved in solving the proof for some problems that involving the statements from different theorems that are earlierly solved. we have to analyze the question first and then we have to choose what kind of statements are used for the problems given. After those steps are taken, we have to match it for the statements with the problems. The proof represents the way that how to we prove the given problem or some postulates.
(Source from Wikipedia)
I like to share this Calculate Area of Polygon with you all through my article.
Examples to explain "geometric proofs practice"
We have to prove the straight line's equation is 3x + 2y = 7 and which is passing through the point (1,2) and which making the intercepts on the axes of the co-ordinate which are in the ratio 2 : 3.
Geometry proof:
The intercept form is `x/a` + `y/b` = 1 --------- (1)
The intercepts are in the ratio 2 : 3 ? a = 2k, b = 3k.
(1) becomes `x/(2k)` + `y/(3k)` = 1
i.e. 3x + 2y = 6k
We know that the point (1, 2) lies on the straight line given above, 3 + 4 = 6k
i.e. 6k = 7
Hence the required straight line's equation is 3x + 2y = 7
Thus we have proved the proof for finding the straight line's equation.
Practice:
In this problem as we are seen that it have specific points to prove the problems such that it having ratio on the concern and the point which the line passing.
Problems to explain "geometric proofs practice"
We have to prove that the point's co-ordinates are (8, 9) and (- 42, - 41) and given that it is reside on the straight line y = x + 1 which are at a distance of 5 units from the straight line 4x - 3y + 20 = 0
Geometry proof:
Let (x1, y1) be a point on y = x + 1
? y1 = x1 + 1 … (1)
The length of the perpendicular from (x1, y1) to the straight line
4x - 3y + 20 = 0 is `|(4x1 - 3y1 + 20)/sqrt(4^2 + (-3)^2)|` = `+-` `((4x1 - 3y1 + 20)/5)`
But the length of the perpendicular is given as 5.
? `+-` `((4x1 - 3y1 + 20)/5)` = 5
4x1- 3y1 + 20 = `+-` 25
Considering the positive sign, 4x1- 3y1 + 20 = `+` 25
? 4x1 - 3y1 = 5 … (2)
Considering the positive sign, 4x1- 3y1 + 20 = `-` 25
? 4x1 - 3y1 =`-` 45 … (3)
Solving (1) and (2), we get x1 = 8, y1 = 9
Solving (1) and (3), we get x1 = - 42, y1 = - 41.
? The required points co-ordinates are (8, 9) and (- 42, - 41).
Thus we have proved the proof with the given points.
Practice:
In this problem as we are seen that it have specific points to prove the problems such that it having the perpendicular distance from the two straight lines.
Practice problems on geometric proof:
We have to prove the straight line's equation is x - `sqrt(3)` y + 12 = 0, if the perpendicular comes from the origin, which makes an 120° angle with x-axis and the distance from perpendicular comes from the origin is 6 units.
We have to show that the straight lines 132x +13 y - 9 = 0 and 132x + 13y - 10 = 0 are parallel.
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