The linear equation of the high ch on the edge ab of the triangle ABC is x-2y-5 = 0, and the linear equation of the middle line BM on the edge AC is 2x-y-5 = 0, and a (5,1) Find: 1) the equation of line BC 2) Area of triangle ABC

The linear equation of the high ch on the edge ab of the triangle ABC is x-2y-5 = 0, and the linear equation of the middle line BM on the edge AC is 2x-y-5 = 0, and a (5,1) Find: 1) the equation of line BC 2) Area of triangle ABC

(1) Ch x-2y-5 = 0 & # 8658; y = 1 / 2 & # 8226; X-5 / 2bm 2x-y-5 = 0 & # 8658; y = 2x-5, because ab ⊥ ch, according to the slope of each other is negative reciprocal, let AB equation be y = - 2x + B point a substitute: 1 = - 10 + B B = 11. AB equation is y = - 2x + 11 and y = 2x-5, x = 4, y = 3, y = - 2x + 11, y = 1 / 2

Given a vertex (2,3) of the triangle ABC, the linear equation of the height on the side AB is x-2y + 3 = 0, and the linear equation of the bisector of angle B is x + y-4 = 0, the linear equation of the three sides of the triangle is obtained

A is (2,3), AB is perpendicular to x-2y + 3 = 0, so AB can be set as y = - 2x + B, substituting a to get b = 7, that is, the AB equation is y = - 2x + 7; and the simultaneous solution of X + y-4 = 0 can get b (3,1)
Then find the symmetric point d (U, V) of a with respect to x + y-4 = 0, then ad is perpendicular to x + y-4 = 0, and ab = BD. that is, (v-3) / (U-2) = 1, (U-3) ^ 2 + (V-1) ^ 2 = (2-3) ^ 2 + (3-1) ^ 2, and D (1,2), C is on BD, that is, BC equation is y = - X / 2 + 5 / 2
And C is on x-2y + 3 = 0, which is combined with BC equation to get C (1,2), so AC equation is y = x + 1

In the known triangle ABC, the coordinates of point a are (1,3), and the linear equations of the middle line on the sides of AB and AC are x-2y + 1 = 0 and Y-1 = 0 respectively

Let the coordinates of point B be (m, n), then the midpoint of AB ((M + 1) / 2, (n + 3) / 2) on the straight line x-2y + 1 = 0 ｜ (M + 1) / 2-2 × (n + 3) / 2 = 0, that is, M = 2n + 5 ∵ B on the straight line Y-1 = 0 ｜ n-1 = 0 ｜ M = 7, n = 1 ｜ B coordinate is (7,1). Let the equation of line AB be y = KX + B ｜ K + B

In ABC, the high line equation on a (0,1), AB is x + 2y-4 = 0, and the middle line equation on AC is 2x + Y-3 = 0. Find the linear equation on AB, BC, AC

If the slope of the line AB is - 12, then the slope of the line AB is 2. If the line AB passes through a (0,1) ‖ AB, then the equation Y-1 = 2 (x-0) is reduced to 2x-y + 1 = 0. The equation 2x + y − 3 = 02x − y + 1 = 0 of the line AB and AC is solved to x = 12Y = 2

It is known that the three vertices of triangle ABC are a (1.2) B (3.4) C (- 2.4) to find the linear equation of ab
The equation of a circle with a point as its center and tangent to ab line

The solution a (1.2) B (3.4) i.e. k = (4-2) / (3-1) = 1 i.e. Y-2 = 1 (x-1) i.e. X-Y + 1 = 0 should take point C as the center of the circle. The distance from point C (- 2,4) to line AB is d = / - 2-4 + 1 / / / √ (1 & # 178; + 1 & # 178;) = 5 √ 2 / 2 i.e. the radius of circle C is 5 √ 2 / 2 i.e

It is known that the lengths of the three sides of the triangle ABC are ab = 6, BC = 8 and AC = 10 respectively. Take the three vertices a, B and C of the triangle as the center of the circle to make a circle a, B and C so that the three circles are tangent to each other, then what are the radii RA, Rb and R of the three centers
It is tangent in the title, there should be two kinds, one is circumscribed, the other is inscribed, circumscribed I will, how to find the inscribed

Suppose that there are three kinds of inscribes: circle a and the other two circles
The same is true for circle B and circle C
Let me take the circle a as an example. You can calculate the other two inscribes slowly
First, circles B and C must be circumscribed. (the other two are also circumscribed), so the radius is equal to 4 (because BC = 8)
When circle a and circle B are inscribed, the radius of circle a is equal to = AB + the radius of circle B
When circle a and circle C are inscribed, the radius of circle a = AC + the radius of circle C
In this case, it is obvious that the radii of circle a are not equal, so circle a cannot be tangent to circle B and circle C at the same time
You can discuss the other two slowly. You don't need to draw a picture. Space imagination is enough
To remind you, three circles cannot be inscribed
Because on the same line segment, the radius of the two circles tangent inside must be the same
The center of the circle to the other two points are not the same, so the radius is not uniform, so it cannot be tangent

Given the vertex a (1 2) B (3 4) C (1 0) (1) of triangle ABC, we can find the equation of the line where the height ad is located on the edge BC, and we can find the equation of the vertical bisector on the edge BC

K（bc）=（4-0）/（3-1）=2
（1）K（ad）=-1/2
Let the linear equation of height ad be y = (- 1 / 2) x + B
∵ straight line passes through point a (1,2)
∴2=（-1/2）*1+b
b=5/2
The linear equation y = (- 1 / 2) + 5 / 2 of high ad on BC edge
(2) K (vertical bisector) = - 1 / 2
The midpoint coordinates e (2,2) of BC edge
Let g = (- 1 / 2) x + C be the vertical bisector equation on the edge of BC
∴2=（-1/2）*2+c
c=3
The vertical bisector equation g = (- 1 / 2) x + 3 on the BC edge

Given the three vertices a (1,3) B (3,1) C (- 1,0) 1 of △ ABC, find the vertical bisector equation 2 of AB and the area of △ ABC

1）y=x
2) Area = 12-3-2-2 = 5

It is known that the square of the equation 4x - 8m + n = 0, where m and N are the waist length and bottom edge of an isosceles triangle, respectively
(1) (2) if the absolute value of the difference between the two real roots of the equation is 8, the area of the isosceles triangle is 12

(1) It is known that the square of the equation 4x - 8mx + n = 0,
Where m and N are the waist length and base of an isosceles triangle
2m>n
Discriminant = (- 8m) ^ 2-4 * 4 * n ^ 2 = 64M ^ 2-16n ^ 2 = 16 (4m ^ 2-N ^ 2) = 16 (2m-n) (2m + n) > 0
So the equation has two unequal real roots
（2）
From the root formula
The two real roots of the equation are (2m + √ (4m ^ 2-N ^ 2)) / 2 and (2m - √ (4m ^ 2-N ^ 2)) / 2
The absolute value of the difference between the two real roots of the equation is 8
|(2m+√(4m^2-n^2))/2 -(2m-√(4m^2-n^2))/2}=8
|√(4m^2-n^2)|=8
4m^2-n^2=64
The area of an isosceles triangle is 12
That is, 1 / 2n times √ (m ^ 2-N ^ 2 / 4) = 12 (the height can be obtained from Pythagorean theorem)
We get 4m ^ 2-N ^ 2 = (48 / N) ^ 2
That is 64 = = (48 / N) ^ 2
We get n = 6
Then M = 5
The perimeter of the triangle is 2m + n
So the perimeter is 16

Is the root 27 a rational number

Radical 27 = 3, radical 3, is not a rational number, but an irrational number