Question

Determine the coordinates of the point on the straight line y=3x+1 that is equidistant from the origin and (−3, 4).

Answer 1

Answer:

$x = \frac{17}{18}$

$y = \frac{23}{6}$

Step-by-step explanation:

Given

$y = 3x + 1$

$Points = (0,0)\ to\ (-3,4)$

Required

Determine (x,y) equidistant from the equation and the given point

This question will be answered using distance formula;

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

Considering The first Point $(0,0)\ to\ (x,y)$

$(x_1 ,y_1) = (0,0)$

$(x_2,y_2) = (x,y)$

The distance is:

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$d = \sqrt{0 - x)^2 + (0 - y)^2}$

$d = \sqrt{(- x)^2 + (- y)^2}$

$d = \sqrt{x^2 + y^2}$

Considering the second point: $(x,y)\ to\ (-3,4)$

$(x_1,y_1) = (x,y)$

$(x_2,y_2) = (-3,4)$

The distance is:

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$d = \sqrt{(x - (-3))^2 + (y - 4)^2}$

$d = \sqrt{(x + 3)^2 + (y- 4)^2}$

$d = \sqrt{x^2 + 6x + 9 + y^2- 8y + 16}$

$d = \sqrt{x^2 + 6x + y^2- 8y + 16 + 9}$

$d = \sqrt{x^2 + 6x + y^2- 8y + 25}$

Equate both values of d

$\sqrt{x^2 + 6x + y^2- 8y + 25} = \sqrt{x^2 + y^2}$

Square both sides

$x^2 + 6x + y^2- 8y + 25 = x^2 + y^2$

Collect Like Terms

$6x - 8y + 25 = x^2 -x^2+ y^2 -y^2$

$6x - 8y + 25 =0$

Recall that $y = 3x + 1$

$6x - 8(3x + 1) + 25 = 0$

$6x - 24x - 8 + 25 = 0$

$-18x+ 17 = 0$

$-18x=- 17$

Solve for x

$x = \frac{-17}{-18}$

$x = \frac{17}{18}$

Substitute $\frac{17}{18}$ for x in $y = 3x + 1$

$y = 3 * \frac{17}{18} + 1$

$y = \frac{17}{6} + 1$

$y = \frac{17 + 6}{6}$

$y = \frac{23}{6}$