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The slope of the straight line L is known to be 1 6, and forms a triangle with an area of 3 with the two coordinate axes, then the equation of straight line L is __
From the meaning of the question, we can assume that the equation of the straight line L is y = 16x + B. obviously, the intersection points of the straight line and the two coordinate axes are (0, b), (- 6B, 0) respectively. Then, by enclosing the straight line and the two coordinate axes into a triangle with an area of 3, we can get 12|b | •| - 6B | = 3 B = ± 1, so the equation of the straight line is y = 16x ± 1, i.e. X -
A straight line passes through point P (- 5,4) and the area of the triangle surrounded by the two coordinate axes is 5. Find the equation of this straight line?
Y-4 = K (x + 5). It is transformed into intercept formula: Y / (5K + 4) + X / [- (5K + 4) / k] = 1. So: (1 / 2) (5K + 4) [- (5K + 4) / k] = ± 5. That is: (5K + 4) ²＝ ± 5K. K ≥ o: 25K ²+ 30K + 16 = 0.900-1600 ＜ 0, no real solution. When k ＜ 0: 25K ²+ 50k+16＝0.k＝-...
Make a straight line L through point P (4,3), which intersects the two coordinate axes and the area of the triangle surrounded by the two coordinate axes is 3 square units. Find the equation of straight line L
Let the linear equation be the intersection of Y-3 = K (x-4) and the two coordinate axes: y = 0, x = 4-3 / KX = 0, y = 3-4k. The area of the triangle surrounded by the two coordinate axes is 3 square units. 1 / 2 * | 4-3 / k * | 3-4k| = 31 / 2 * | 3-4k| ²/ k =3|3-4k| ²= 6K, then k = 3 / 2, or 3 / 8 ‡ the equation is 3x-8y + 12 = 0 or 3x-2y-6 = 0
The slope of the straight line L is known to be 1 6, and forms a triangle with an area of 3 with the two coordinate axes, then the equation of straight line L is __
From the meaning of the question, we can set the equation of straight line L as y = 1
6X + B, obviously the intersection of this line and the two coordinate axes is (0, b), (- 6B, 0) respectively
Then a triangle with an area of 3 is formed by a straight line and two coordinate axes, and 1 can be obtained
2 | B | • - 6B | = 3, the solution is obtained b=±1，
Therefore, the equation of the straight line is y = 1
6X ± 1, i.e. x-6y + 6 = 0, or x-6y-6 = 0,
So the answer is X-6y + 6 = 0, or x-6y-6 = 0
It is known that the area of the triangle surrounded by the straight line L and the two coordinate axes is 3, and the equations of the straight line L satisfying the following conditions are solved respectively It is known that the area of the triangle formed by the straight line L and the two coordinate axes is 3, and the equations of the straight line L satisfying the following conditions are solved respectively (1) Over point a (- 3,4) (2) The slope is 1 / 6 (1) 2X + 3y-6 = 0 or 8x + 3Y + 12 = 0 (2) x-6y + 6 = 0 or x-6y-6 = 0 is the answer
(1) Let the linear equation: y = ax + B passes through the intersection of X and Y axes as (- B / A, 0), (0, b) respectively
Area s= ½× bottom × High so s= ½ （-b/a） × b=3 →b ²= 6a, substitute point a into the equation to obtain: - 3A + B = 4 → 2x + 3y-6 = 0 or 8x + 3Y + 12 = 0
(2) Let the linear equation: y = 1 / 6x + B (the slope is the value of a obtained from the standard linear equation), intersect on the x-axis (6b, 0) or (- 6B, 0) intersect on the y-axis (0, b), and get 6B from the area equal to three ²= 6, get b = ± 1, so as to get the equation: x-6y + 6 = 0 or x-6y-6 = 0
If the three sides of a triangle are 6, 8 and 10, the area of the triangle surrounded by the three median lines of the triangle is
This triangle is a right triangle. The three sides of the triangle surrounded by the median line are 3,4,5, which is also a right triangle, with an area of 1 / 2 * 3 * 4 = 6
RELATED INFORMATIONS
- 1. The area of a triangle is 48cm2, and the area of the triangle surrounded by its three median lines is
- 2. In △ ABC, it is known that BD and CE are the midlines on both sides respectively, and BD ⊥ CE, BD = 4, CE = 6, then the area of △ ABC is equal to () A. 12 B. 14 C. 16 D. 18
- 3. Draw four small triangles in the square (as shown in the figure). The area ratio of triangles I and II is 2:1; The areas of triangles III and IV are equal; The sum of the areas of triangles I, II and III is 1 4 square meters; The sum of the areas of triangles II, III and IV is 1 6 square meters; So what is the sum of the four small triangles______ Square meters
- 4. The three sides of a right triangle are 3cm, 4cm and 5cm respectively. The area of this triangle is () square centimeters; The height on the hypotenuse is () cm Such as the title! Urgent~
- 5. The three sides of a right triangle are 3cm, 4cm and 5cm respectively. How many square centimeters is the area of the right triangle?
- 6. Given that the lengths of both sides of the triangle are 5 and 7 respectively, the value range of the center line length x of the third side is () A. 2＜x＜12 B. 5＜x＜7 C. 1＜x＜6 D. Unable to determine
- 7. Given that the lengths of both sides of the triangle are 5 and 7 respectively, the value range of the center line length x of the third side is () A. 2＜x＜12 B. 5＜x＜7 C. 1＜x＜6 D. Unable to determine
- 8. In triangle ABC, the sides a, B and C opposite to a, B and C, side C = 7 / 2, and Tana + tanb = root 3 × Tanatan B - radical 3, and triangle ABC In triangle ABC, the sides a, B and C opposite to a, B and C, side C = 7 / 2, and Tana + tanb = root 3 × Tanatanb - root sign 3, and the area of triangle ABC is (3 times root sign 3) / 2. Find the value of a + B
- 9. In triangle ABC, ab = 17, AC = 15, the center line ad on the side of BC = 4, calculate the area of triangle ABC
- 10. The perimeter of a right triangle is 2+ 6. If the length of the center line on the hypotenuse is 1, the area of the triangle is equal to () A. 1 B. 1 two C. 1 four D. 3 four
- 11. The lengths of both sides of the triangle are 8 and 6 respectively. The length of the third side is a real root of the univariate quadratic equation x2-12x + 20 = 0. Find the area of the triangle
- 12. If the area of the triangle enclosed by the straight line y = 2x + m on the two coordinate axes is equal to m, the value of M is
- 13. Given that the perimeter of a right triangle is 30 and the area is 30, find the oblique side length of the triangle
- 14. If the perimeter of a right triangle is 5cm and the center line on the hypotenuse is 1cm, the area of this right triangle is?
- 15. In △ ABC, a = 60 °, C: B = 8:5, the area of the inscribed circle is 12 π, and the radius of the circumscribed circle of △ ABC is calculated
- 16. It is known that in △ ABC, the side opposite the acute angle B = 7, the radius of the circumscribed circle r = 7 times the root number 3 / 3, and the triangle area s = 10 times the root number 3. Find the length of the other two sides of the triangle
- 17. In triangular row ABC, a equals 1, C equals 2, and B equals 60 degrees. What is the area of triangular ABC
- 18. In the acute triangle ABC, the opposite sides of angles a, B and C are a, B and C respectively, and the root number 3B = 2asinb is known to calculate the size of the angle; If a = 6, find the range of B + C
- 19. Cut out a square as large as possible, the largest square, from an isosceles right triangle cardboard with an inclined edge of 18cm long What is the side length of the?
- 20. As shown in the figure, the side length of the regular triangle ABC is 6. Cut off three corners to get a regular hexagon. Find the side length and area of the regular hexagon