Question

The vertices of a triangle OBC are O(0, 0), B(-3, -1), C(-1, -3). Find the equation of the line parallel to BC & intersecting the sides OB & OC, whose perpendicular distance from the point (0, 0) is half.

Answer 1

sqrt(2)x + sqrt(2)y + 1 = 0

Answer 2

1. Find the equation of the line BC:

The vertices of the triangle are $O(0, 0)$, $B(-3, -1)$, and $C(-1, -3)$. First, calculate the slope of the line segment BC: $m_{BC} = \frac{y_C - y_B}{x_C - x_B} = \frac{-3 - (-1)}{-1 - (-3)} = \frac{-3 + 1}{-1 + 3} = \frac{-2}{2} = -1$. Now, use the point-slope form of a linear equation with point $B(-3, -1)$ and slope $m_{BC} = -1$: $y - (-1) = -1(x - (-3))$ $y + 1 = -x - 3$ $x + y + 4 = 0$.

2. Determine the general equation of a line parallel to BC:

A line parallel to BC will have the same slope, $m = -1$. So, its equation will be of the form $x + y + k = 0$ for some constant $k$.

3. Use the given perpendicular distance from the origin (0, 0):

The perpendicular distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by the formula $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. Here, the point is the origin $(0, 0)$, and the line is $x + y + k = 0$ (so $A=1, B=1, C=k$). The given perpendicular distance is "half", which means $\frac{1}{2}$. $\frac{|1(0) + 1(0) + k|}{\sqrt{1^2 + 1^2}} = \frac{1}{2}$ $\frac{|k|}{\sqrt{2}} = \frac{1}{2}$ $|k| = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. This gives two possible values for $k$: $k = \frac{1}{\sqrt{2}}$ or $k = -\frac{1}{\sqrt{2}}$.

4. Apply the condition that the line intersects the sides OB and OC:

The line must intersect the line segments OB and OC. This implies that the line must lie "between" the origin $O(0,0)$ and the line $BC$ ($x+y+4=0$). For a line $x+y+k=0$ to separate the origin $(0,0)$ from points $B(-3,-1)$ and $C(-1,-3)$, the expression $x+y+k$ evaluated at the origin and at points B (or C) must have opposite signs. For the origin $(0,0)$, $x+y+k = 0+0+k = k$. For point $B(-3,-1)$, $x+y+k = -3-1+k = -4+k$. For point $C(-1,-3)$, $x+y+k = -1-3+k = -4+k$. So, $k$ and $(-4+k)$ must have opposite signs. If $k > 0$, then $-4+k < 0$, which implies $k < 4$. So, $0 < k < 4$. If $k < 0$, then $-4+k > 0$, which implies $k > 4$. This is a contradiction, as $k$ cannot be both less than 0 and greater than 4. Therefore, the correct value of $k$ must satisfy $0 < k < 4$. Let's check the two possible values of $k$:

• $k = \frac{1}{\sqrt{2}} \approx 0.707$. This value satisfies $0 < k < 4$.
• $k = -\frac{1}{\sqrt{2}} \approx -0.707$. This value does not satisfy $0 < k < 4$.

Thus, the correct value for $k$ is $\frac{1}{\sqrt{2}}$.

5. Write the equation of the line:

Substitute $k = \frac{1}{\sqrt{2}}$ into the general equation $x + y + k = 0$: $x + y + \frac{1}{\sqrt{2}} = 0$. To eliminate the fraction and the radical in the denominator, we can multiply the entire equation by $\sqrt{2}$: $\sqrt{2}(x + y + \frac{1}{\sqrt{2}}) = \sqrt{2}(0)$ $\sqrt{2}x + \sqrt{2}y + 1 = 0$.

This is the equation of the line.