If the equation of the base of an equilateral triangle is \(x + y = 2\) and the vertex is \((2, -1)\), find the length of the side of the triangle.
The distance from the vertex \((2, -1)\) to the line \(x + y = 2\) is the altitude of the equilateral triangle.
Using the perpendicular distance formula:
\[ d = \dfrac{|2 + (-1) - 2|}{\sqrt{1^2 + 1^2}} = \dfrac{1}{\sqrt{2}} \]
In an equilateral triangle, altitude \(h = \dfrac{\sqrt{3}}{2} a\), where \(a\) is the side.
So, \( \dfrac{\sqrt{3}}{2} a = \dfrac{1}{\sqrt{2}} \Rightarrow a = \sqrt{\dfrac{2}{3}} \).
Therefore, the side length is \( \sqrt{\dfrac{2}{3}} \).
A variable line passes through a fixed point \(P\). The algebraic sum of the perpendiculars drawn from the points \((2,0)\), \((0,2)\), and \((1,1)\) on the line is zero. Find the coordinates of point \(P\).
Let the line through \(P(x_1,y_1)\) be \(y - y_1 = m(x - x_1)\).
Convert to normal form and compute perpendicular distances from \((2,0)\), \((0,2)\), and \((1,1)\).
Using the condition that the algebraic sum of perpendiculars is zero gives:
\[ (2 - x_1)m - (y_1) + (0 - x_1)m + (2 - y_1) + (1 - x_1)m + (1 - y_1) = 0 \]
Simplifying yields \(x_1 = 1, y_1 = 1\).
Thus, the fixed point is \((1,1)\).
In what direction should a line be drawn through the point \((1, 2)\) so that its point of intersection with the line \(x + y = 4\) is at a distance \(\dfrac{\sqrt{6}}{3}\) from the given point?
Let the required line through \((1,2)\) make angle \(\theta\) with the positive x-axis.
Its parametric form is:
\[ (x, y) = (1, 2) + t(\cos\theta, \sin\theta). \]
Substitute into \(x + y = 4\):
\[ 1 + 2 + t(\cos\theta + \sin\theta) = 4 \Rightarrow t = \dfrac{1}{\cos\theta + \sin\theta}. \]
The distance from \((1,2)\) to the intersection point is:
\[ |t| = \dfrac{1}{\cos\theta + \sin\theta} = \dfrac{\sqrt{6}}{3}. \]
Solving this gives \(\theta = 15^\circ\) or \(75^\circ\).
A straight line moves so that the sum of the reciprocals of its intercepts on the coordinate axes is constant. Show that the line passes through a fixed point.
The intercept form of the line is:
\[ \dfrac{x}{a} + \dfrac{y}{b} = 1. \]
Given that
\[ \dfrac{1}{a} + \dfrac{1}{b} = k, \]
a constant.
Rewrite the line equation:
\[ bx + ay = ab. \]
Using \(ab(\dfrac{1}{a} + \dfrac{1}{b}) = abk\):
\[ b + a = abk. \]
Hence the equation becomes:
\[ bx + ay = a + b. \]
Dividing both sides by \(k\):
\[ b(x - k) + a(y - k) = 0. \]
This holds for every valid choice of \(a\) and \(b\), implying all such lines pass through the fixed point:
\[ (k, k). \]
Find the equation of the line which passes through the point \((-4, 3)\) and the portion of the line intercepted between the axes is divided internally in the ratio \(5:3\) by this point.
Let the line intersect the axes at \((a,0)\) and \((0,b)\).
The point \((-4,3)\) divides the intercepts in the ratio \(5:3\):
\[ \dfrac{-4}{a} = \dfrac{5}{5+3} = \dfrac{5}{8}, \quad \dfrac{3}{b} = \dfrac{3}{8}. \]
Thus, \(a = -\dfrac{32}{5}\) and \(b = 8\).
The intercept form of the line becomes:
\[ \dfrac{x}{-32/5} + \dfrac{y}{8} = 1. \]
Simplifying yields the equation:
\[ 9x - 20y + 96 = 0. \]
Find the equations of the lines through the point of intersection of the lines \(x - y + 1 = 0\) and \(2x - 3y + 5 = 0\) and whose distance from the point \((3,2)\) is \(\dfrac{7}{5}\).
Let the required line be:
\[ (x - y + 1) + \lambda (2x - 3y + 5) = 0. \]
Simplify to obtain:
\[ (1 + 2\lambda)x + (-1 - 3\lambda)y + (1 + 5\lambda) = 0. \]
Apply perpendicular distance formula to point \((3,2)\) and equate to \(\dfrac{7}{5}\).
Solving for \(\lambda\) produces two values, giving two lines:
\[ 3x - 4y + 6 = 0, \]
\[ 4x - 3y + 1 = 0. \]
If the sum of the distances of a moving point in a plane from the axes is 1, find the locus of the point.
The distance of \((x,y)\) from the x-axis is \(|y|\), and from the y-axis is \(|x|\).
Given:
\[ |x| + |y| = 1. \]
This is the equation of a square of side 1, oriented at 45° with respect to the coordinate axes.
\(P_1, P_2\) are points on either of the two lines \(y - \sqrt{3}|x| = 2\) at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from \(P_1, P_2\) on the bisector of the angle between the given lines.
The lines are:
\[ y = \sqrt{3}x + 2 \quad (x \ge 0), \]
\[ y = -\sqrt{3}x + 2 \quad (x < 0). \]
Their intersection is at \((0,2)\).
Points \(P_1, P_2\) lie 5 units away from \((0,2)\) on these lines.
The angle bisector is the y-axis. The foot of perpendicular from \((x_1,y_1)\) on the y-axis has x-coordinate 0 and y-coordinate equal to the point’s y-value.
The y-coordinates are:
\[ y = 2 \pm 5\cos 30^\circ = 2 \pm \dfrac{5\sqrt{3}}{2}. \]
Thus the feet of perpendiculars are:
\[ \left(0, 2 + \dfrac{5\sqrt{3}}{2}\right), \, \left(0, 2 - \dfrac{5\sqrt{3}}{2}\right). \]
If \(p\) is the length of perpendicular from the origin on the line \(\dfrac{x}{a} + \dfrac{y}{b} = 1\) and \(a^2, p^2, b^2\) are in A.P., show that \(a^4 + b^4 = 0\).
The perpendicular distance from origin to the line \(\dfrac{x}{a} + \dfrac{y}{b} = 1\) is:
\[ p = \dfrac{ab}{\sqrt{a^2 + b^2}}. \]
Given \(a^2, p^2, b^2\) in A.P., so:
\[ 2p^2 = a^2 + b^2. \]
Substitute the expression for \(p^2\):
\[ 2 \left( \dfrac{a^2 b^2}{a^2 + b^2} \right) = a^2 + b^2. \]
Cross-multiplying gives:
\[ 2a^2 b^2 = (a^2 + b^2)^2. \]
Expanding the right-hand side:
\[ 2a^2 b^2 = a^4 + 2a^2 b^2 + b^4. \]
Simplifying:
\[ a^4 + b^4 = 0. \]