Question: "Find the point on the line y=4x+5 that..."

villane0

Answered question

2021-12-05

Find the point on the line

y = 4 x + 5

that is closest to the origin.

(x, y) =

Answer & Explanation

Alfonso Miller

Beginner2021-12-06Added 20 answers

Step 1
Find the point on the line $y = 4 x + 5$ that is closest to the origin.
Solution:
let's say $(h, k)$ is the closest to the origin $(0, 0)$ and lies on the line $y = 4 x + 5$ . So $(h, k)$ will satisfy the line $y = 4 x + 5$ equation
Put $x = h$ , $y = k$ ,
$k = 4 h + 5$ (1)
distance of point (h,k) from (0,0).
$d = \sqrt{{(h - 0)}^{2} + {(k - 0)}^{2}}$
$d = \sqrt{h^{2} + k^{2}}$ (2) NST Shortest distance of the given line from origin
$\frac{[4 X (0) (0) + 5]}{\sqrt{4^{2} + {(- 1)}^{2}}}$
Put $4 x + 5 - y = 0$
Put $x = 0, y = 0$
$d = \frac{5}{\sqrt{17}}$ (3)
from equation $(2) & (3)$ ,
$\sqrt{h^{2} + k^{2}} = \frac{5}{\sqrt{17}}$
$h^{2} + k^{2} = \frac{25}{17}$ (4)
Step 2
Put $k = 4 h + 5$ into equation (4)
$h^{2} + {(4 h + 5)}^{2} = \frac{25}{17}$
$h^{2} + 16^{2} + 25 + 40 h = \frac{25}{17}$
$17 h^{2} + 40 h + 25 - \frac{25}{17} = 0$
$\frac{289 h^{2} + 680 h + 425 - 25}{17} = 0$
$289 h^{2} + 680 h + 400 = 0$
Use sridharacharya method to find root,
$A = 289, B = 680, C = 400$
$h = \frac{- B \pm \sqrt{B^{2} - 4 A C}}{2 A} = \frac{- 680 \pm \sqrt{{(680)}^{2} - 4 \times 289 \times 400}}{2 \times 289} = \frac{- 680 \pm 0}{2 \times 289} = \frac{- 40}{2 \times 17} = \frac{- 20}{17}$
$h = \frac{- 20}{17}$
so, $k = 4 h + 5 =$

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