The area of the triangle formed by the line passing through point a (2, - 3) and the coordinate axis is 4

The area of the triangle formed by the line passing through point a (2, - 3) and the coordinate axis is 4

Let y = KX + B replace a with B = 2K + 3
That is, the line y = KX + 2K + 3
The intercept of the line on the x-axis is | - (2k + 3) / K|
The intercept on the Y axis is | 2K + 3|
Then we discuss it by category
If the intercept of the line on X and Y axes is positive and negative, then ((2k + 3) / k) * (2k + 3) = 8 has no solution
If the intercept of the line on X and Y axes are both positive or negative, then
（-（2k+3）/k）*（2k+3）=8
The solution is k = - 1 / 2 or - 9 / 2
So the line is y = (- 1 / 2) x + 2 or y = (- 9 / 2) X-6

Find the linear equation that the area of triangle passing through point a (- 2,2) and surrounded by two coordinate axes is 1

Let the straight line be Y-2 = K (x + 2), intersect the X axis at the point (− 2K − 2, 0), intersect the Y axis at the point (0, 2K + 2), s = 12 ×| 2K + 2 ×| 2K + 2 | = 1, | 4 + 2K + 2K | = 1, then 2k2 + 3K + 2 = 0, or 2k2 + 5K + 2 = 0. The solution is k = - 12, or K = - 2, | x + 2y-2 = 0, or 2x + y + 2 = 0

Find the linear equation that the area of triangle passing through point a (- 2,2) and surrounded by two coordinate axes is 1

Let the straight line be Y-2 = K (x + 2), intersect the X axis at the point (− 2K − 2, 0), intersect the Y axis at the point (0, 2K + 2), s = 12 ×| 2K + 2 ×| 2K + 2 | = 1, | 4 + 2K + 2K | = 1, then 2k2 + 3K + 2 = 0, or 2k2 + 5K + 2 = 0. The solution is k = - 12, or K = - 2, | x + 2y-2 = 0, or 2x + y + 2 = 0

Find the linear equation of area 4 which is parallel to the line x-2y + 1 = 0 and surrounded by two coordinate axes

The slope of the line parallel to the line x-2y + 1 = 0, that is, y = 1 / 2 * x + 1 / 2, is also 1 / 2. Therefore, in the intersection of the line and the two axes, the distance between the intersection on the Y axis and the origin Y0 is 1 / 2 of the distance between the intersection on the X axis and the origin x0. Therefore, s = 1 / 2 * x0y0 = Y0 ^ 2 = 4 = > Y0 = 2 or - 2, because the slope is 1 / 2, the corresponding x0 is: x0

What is the area of the triangle enclosed by the line x + 2y-4 = 0 and the coordinate axis

When y = 0
x=4
The abscissa is (4,0)
When x = 0
y=2
The ordinate is (0,2)
∴S△=4*2/2=4

Point a (3,4) is known. Find the linear equation with the smallest area of triangle formed by the line and the positive half axis of two coordinate axes,

Let y = KX + B4 = 3K + B discuss: when k = 0, y = 4 has no intersection with X axis, intersects with y axis at point (0,4), cannot encircle triangle, does not conform to the situation; when k ≠ 0, y and X axis intersect with y axis at point (- B / K, 0), (0, b); triangle area s = (1 / 2

The equation of line L is ()
A. 3x+y-6=0B. 3x-y=0C. x+3y-10=0D. x-3y+8=0

Let the line be Y-3 = K (x-1), so the line L intersects the x-axis at the point (1-3k, 0), and intersects the y-axis at the point (0, 3-K). Because the line L intersects the positive half axis of the two coordinate axes, so s = 12 × (1 − 3K) × (3 − K) = 6, the solution is k = - 3, so the linear equation is 3x + y-6 = 0

A straight line passes through point a (1,6), and the area of the triangle enclosed by the positive half axis of the two coordinate axes is 16

Let y-6 = K (x-1)
x=0 y=6-k
y=0 x=-6/k +1=(k-6)/k
So | 6-k | * | (K-6) / K | = 2 * 16
(k-6)²=32|k|
From the known K

A straight line passes through point a (1,6), and the area of the triangle surrounded by the positive half axis of the two coordinate axes is 16. The equation of the straight line is obtained

Let y = KX + 6-k, and K

Find the linear equation that passes through the point P (2.1) and the area of the triangle enclosed by the positive half axis of the two coordinate axes is 9 / 2
My math is not good. Please help me```

Let the linear equation be y = ax + B
Through point P (2.1), we get 1 = 2A + B
Let the coordinates of the intersection of the line and the x-axis be a (m, 0), and the coordinates of the intersection of the line and the y-axis be B (0, n)
1/2mn=9/2 mn=9
Yes, because a and B are in a straight line
am+b=0,b=n
The solution is a = - 1, B = 3
So the linear equation is y = - x + 3