\((2, 0)\)
\((0,2)\)
\((\dfrac{√2+1}{√2},\dfrac{√2+1}{√2})\)
\((\dfrac{√5+1}{√5},\dfrac{√5+2}{√5})\)
\((\dfrac{2+√3}{2},\dfrac{3}{2})\)
We need to find the new position \( Q \) of the point \( P(1, 1) \) after translating it a distance of 1 unit in the direction of the line \( y = 2x \).
Step 1: Determine the direction vector.
The line \( y = 2x \) has a slope of 2, so its direction vector is \( \vec{v} = (1, 2) \). To get a unit vector in this direction, we normalize \( \vec{v} \):
\[ \|\vec{v}\| = \sqrt{1^2 + 2^2} = \sqrt{5} \] \[ \hat{v} = \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \]
Step 2: Compute the translation.
Translating \( P(1, 1) \) by 1 unit in the direction of \( \hat{v} \) gives:
\[ Q = P + 1 \cdot \hat{v} = \left( 1 + \frac{1}{\sqrt{5}}, 1 + \frac{2}{\sqrt{5}} \right) = \left( \frac{\sqrt{5} + 1}{\sqrt{5}}, \frac{\sqrt{5} + 2}{\sqrt{5}} \right) \]
Step 3: Verify the distance.
The distance between \( P \) and \( Q \) is:
\[ PQ = \sqrt{ \left( \frac{1}{\sqrt{5}} \right)^2 + \left( \frac{2}{\sqrt{5}} \right)^2 } = \sqrt{ \frac{1}{5} + \frac{4}{5} } = \sqrt{1} = 1 \]
This confirms that the translation is correct.
Step 4: Match with the options.
The coordinates of \( Q \) match option (D):
\[ \left( \frac{\sqrt{5} + 1}{\sqrt{5}}, \frac{\sqrt{5} + 2}{\sqrt{5}} \right) \]
Step 1: Find the direction vector of the line \( y = 2x \).
The line \( y = 2x \) has a slope of 2. A direction vector for this line is:
\[ \vec{v} = (1, 2) \]
To normalize this vector, compute its magnitude:
\[ |\vec{v}| = \sqrt{1^2 + 2^2} = \sqrt{5} \]
The unit vector in the direction of \( \vec{v} \) is:
\[ \hat{v} = \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \]
Step 2: Translate \( P(1,1) \) by 1 unit in the direction of \( \hat{v} \).
The coordinates of \( Q \) are obtained by adding the scaled unit vector \( \hat{v} \) to \( P(1,1) \):
\[ Q_x = 1 + \frac{1}{\sqrt{5}}, \quad Q_y = 1 + \frac{2}{\sqrt{5}} \]
Thus, the coordinates of \( Q \) are:
\[ Q = \left( 1 + \frac{1}{\sqrt{5}}, 1 + \frac{2}{\sqrt{5}} \right) \]
Step 3: Simplify the coordinates of \( Q \).
Rewriting \( Q \) in a compact form:
\[ Q = \left( \frac{\sqrt{5} + 1}{\sqrt{5}}, \frac{\sqrt{5} + 2}{\sqrt{5}} \right) \]
Step 4: Match with the given options.
The coordinates of \( Q \) match option \( D \):
\[ \left( \frac{\sqrt{5} + 1}{\sqrt{5}}, \frac{\sqrt{5} + 2}{\sqrt{5}} \right) \]
Final Answer:
\( \left( \frac{\sqrt{5} + 1}{\sqrt{5}}, \frac{\sqrt{5} + 2}{\sqrt{5}} \right) \)
A straight line is a figure created when two points A (x1, y1) and B (x2, y2) are connected with a minimum distance between them, and both the ends are extended to immensity (infinity). With variables x and y, the standard form of a linear equation is: ax + by = c, where a, b, and c are constants and x, and y are variables.
The equation of a straight line whose slope is m and passes through a point (x1, y1) is formed or created using the point-slope form. The equation of the point-slope form is:
y - y1 = m (x - x1),
where (x, y) = an arbitrary point on the line.
A slope of a line is the conversion in y coordinate w.r.t. the conversion in x coordinate.
The net change in the y-coordinate is demonstrated by Δy and the net change in the x-coordinate is demonstrated by Δx.
Hence, the change in y-coordinate w.r.t. the change in x-coordinate is given by,
\(m = \frac{\text{change in y}}{\text{change in x}} = \frac{Δy}{Δx}\)
Where, “m” is the slope of a line.
The slope of the line can also be shown by
\(tan θ = \frac{Δy}{Δx}\)
Read More: Slope Formula
The equation for the slope of a line and the points are known to be a point-slope form of the equation of a straight line is given by:
\(y-y_1=m(x-x_1)\)
As long as the slope-intercept form the equation of the line is given by:
\(y = mx + b\)
Where, b is the y-intercept.