Given the slope of the straight line k = 1 / 2, P1 (- 2,3), P2 (X2, - 2), P3 (1 / 2, Y3) three points on a straight line, find X2, Y3

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• 1. It is known that the straight line passes through P1 (x1,5), P2 (4, Y2), P3 (- 1, - 3), and the slope of the straight line is - 3
• 2. If two points P1 (4,9) and P2 (6,3) are known, then the equation for a circle with diameter p1p2 is______ ．
• 3. Given that the straight line with slope 2 intersects with hyperbola x ^ 2-y ^ 2 = 12 at P1 and P2, find the trajectory equation of the midpoint of line p1p2
• 4. Given the hyperbola x ^ 2-y ^ 2 / 2 = 1, the line L passing through point a (2,1) intersects with the given hyperbola at two points P1 and P2, and the trajectory equation of point P in line p1p2 is obtained
• 5. A straight line L passing through the point (- 5, - 4) intersects two coordinate axes and is tangent to the triangle formed by the two axes. The area of the triangle is 5. The equation of L is obtained . urgent
• 6. If the line L passes through the point P (3,2), and the midpoint of the line cut by the coordinate axis is exactly P, the equation of the line L is obtained
• 7. A straight line passes through the point (1.3), which is exactly the midpoint of the line cut by the two coordinate axes
• 8. If the point P (- 1.3) is the midpoint of the line L cut by two axes, the equation of the line is obtained
• 9. Point a (- 4. - 2) is the midpoint of the line segment of line L cut by two coordinate axes, then the equation of line L is Such as the title
• 10. A straight line passes through the point m (- 2,2) and the area of the triangle surrounded by the two coordinate axes is 1. The equation of the straight line is obtained First floor: why x = 0 and y = 0? I get the following formula: s △ = 1 / 2 (2k + 2) * (- 2 / K - 2) = 1
• 11. If we know that a line passes through P (2a, 3b) and Q (4b, 6a) and a is not zero, we can find the slope of the line
• 12. Given that a straight line passes through points P (2a, 3b) and Q (4b, 6a), and a is not equal to 0, the slope of the straight line is calculated
• 13. It is known that the slope k of the straight line L is obtained through two points P1 (2,1) P2 (m, 2) (m ∈ R), and the inclination angle a of L and its value range are obtained?
• 14. Given the slope of the straight line k = 2, P1 (3,5), P2 (x2,7), P3 (- 1, Y3) are the three points on the straight line, find X2, Y3 Does anyone know
• 15. P1 (x1, Y1) P2 (X2, Y2) are two points on a straight line with a slope of K Prove that ip1p2i = 1 + k * 2 times ix1-x2i = 1 + k * 2 times (x1 + x2) - 4x1x2 I is the vertical line, ip1p2i is the absolute value of p1p2, and ix1-x2i is the absolute value of x1-x2
• 16. Given P1 (- 1, a), P2 (3, 6) and the slope of p1p2 k = 2, | p1p2 | =? To find the detailed explanation
• 17. If a straight line passes through the point P (- 5, - 4) and the area of the triangle enclosed by the two coordinate axes is 5, the equation of the straight line is obtained
• 18. Solve the linear equation of point P (- 5,4) which is surrounded by two coordinate axes and has a triangle area of 5
• 19. When crossing point a (- 5, - 4) on a straight line L, it intersects two coordinate axes and the area of the triangle enclosed by the two axes is 5, the equation for solving the straight line L is obtained 1. Let the slope of the line be K y+4=k(x+5) x=0,y=5k-4 y=0,x=4/k-5=(4-5k)/k So area = | 5k-4 | * | (4-5k) / K} / 2 = 5 |(5k-4)^2/k|=10 (5k-4)^2=±10k 25k^2-40k+16=±10k -No solution at 10K 25k^2-50k+16=0 (5k-8)(5k-2)=0 k=8/5,k=2/5 8x-5y+20=0 2x-5y-10=0 Or? 2. Let y = KX + 5k-4. (k is not equal to 0) Let y = 0, x = (4-5k) / K Let x = 0.y = 5k-4 S = 1 / 2 * {5k-4} * {(4-5k) / K} ({} is absolute value.) Let k > 4 / 5, then (5k-4) ^ 2 / k = 10, k = 8 / 5, k = 2 / 5 (rounding) Then y = 8 / 5x + 4 When 0
• 20. Make a straight line L through point a (- 5, - 4) so that it intersects with the two coordinate axes and the area of the triangle enclosed by the two axes is 5