Cho Tam giác ABC có A(-1;4) B(2;5) C(3;0) Tìm D thuộc (d) : x-y+1=0 sao cho AD=5 05/12/2021 Bởi Valerie Cho Tam giác ABC có A(-1;4) B(2;5) C(3;0) Tìm D thuộc (d) : x-y+1=0 sao cho AD=5
Ta có: $(d): x – y + 1 =0$ $\to (d): y = x + 1$ $\to D(a;a+1)\in (d)$ $\to AD = \sqrt{(a+1)^2 + (a + 1 – 4)^2}$ $\to 5 = \sqrt{(a+1)^2 + (a-3)^2}$ $\to 25 = 2a^2 – 4a + 10$ $\to 2a^2 – 4a -15 = 0$ $\to \left[\begin{array}{l}a = 1 + \sqrt{\dfrac{17}{2}}\\a = 1 – \sqrt{\dfrac{17}{2}}\end{array}\right.$ Vậy $D\left(1 + \sqrt{\dfrac{17}{2}};2+ \sqrt{\dfrac{17}{2}}\right)$ hoặc $D\left(1 – \sqrt{\dfrac{17}{2}};2- \sqrt{\dfrac{17}{2}}\right)$ Bình luận
Ta có: $(d): x – y + 1 =0$
$\to (d): y = x + 1$
$\to D(a;a+1)\in (d)$
$\to AD = \sqrt{(a+1)^2 + (a + 1 – 4)^2}$
$\to 5 = \sqrt{(a+1)^2 + (a-3)^2}$
$\to 25 = 2a^2 – 4a + 10$
$\to 2a^2 – 4a -15 = 0$
$\to \left[\begin{array}{l}a = 1 + \sqrt{\dfrac{17}{2}}\\a = 1 – \sqrt{\dfrac{17}{2}}\end{array}\right.$
Vậy $D\left(1 + \sqrt{\dfrac{17}{2}};2+ \sqrt{\dfrac{17}{2}}\right)$ hoặc $D\left(1 – \sqrt{\dfrac{17}{2}};2- \sqrt{\dfrac{17}{2}}\right)$