Question

The vertices of a triangle $OBC$ are $0(0, 0), B(-3, -1), C(-1, -3)$. Equation of line parallel to $BC$ & intersecting the sides $OB$ & $OC$ whose perpendicular distance from the point $(0,0)$ is $\frac{1}{\sqrt{2}}$, is $ax + by + 2 = 0$, then the value of $\frac{a^4+b^4}{4}$ is

Answer 1

8

Answer 2

The vertices of the triangle $OBC$ are $O(0, 0)$, $B(-3, -1)$, and $C(-1, -3)$.

First, find the equation of the line $BC$. The slope of $BC$ is $m_{BC} = \frac{-3 - (-1)}{-1 - (-3)} = \frac{-2}{2} = -1$.

Using the point-slope form with point $B(-3, -1)$: $y - (-1) = -1(x - (-3))$ $y + 1 = -x - 3$ $x + y + 4 = 0$.

A line parallel to $BC$ has the same slope, $-1$. Its equation is of the form $x + y + k = 0$ for some constant $k$.

The perpendicular distance from the origin $(0,0)$ to the line $x + y + k = 0$ is given by the formula $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. Here $(x_0, y_0) = (0,0)$, $A=1$, $B=1$, $C=k$. The distance is $\frac{|1(0) + 1(0) + k|}{\sqrt{1^2 + 1^2}} = \frac{|k|}{\sqrt{2}}$.

We are given that this distance is $\frac{1}{\sqrt{2}}$. So, $\frac{|k|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$, which implies $|k| = 1$. Thus, $k = 1$ or $k = -1$. The possible equations for the line are $x + y + 1 = 0$ and $x + y - 1 = 0$.

The problem states that the line intersects the sides $OB$ and $OC$. This means the line must pass "between" the origin and the points $B$ and $C$. The line segment $OB$ connects $O(0,0)$ and $B(-3,-1)$. Points on this segment have x-coordinates between $-3$ and $0$ and y-coordinates between $-1$ and $0$. The line segment $OC$ connects $O(0,0)$ and $C(-1,-3)$. Points on this segment have x-coordinates between $-1$ and $0$ and y-coordinates between $-3$ and $0$.

Let's test the line $x + y + 1 = 0$. Intersection with line $OB$: The equation of line $OB$ is $y = \frac{-1-0}{-3-0}x = \frac{1}{3}x$. Substitute $y = x/3$ into $x + y + 1 = 0$: $x + \frac{x}{3} + 1 = 0$ $\frac{4x}{3} = -1 \implies x = -\frac{3}{4}$. Then $y = \frac{1}{3}x = \frac{1}{3}(-\frac{3}{4}) = -\frac{1}{4}$. The intersection point is $(-\frac{3}{4}, -\frac{1}{4})$. Check if this point lies on the segment $OB$: $-3 < -\frac{3}{4} < 0$ and $-1 < -\frac{1}{4} < 0$. Yes, it does.

Intersection with line $OC$: The equation of line $OC$ is $y = \frac{-3-0}{-1-0}x = 3x$. Substitute $y = 3x$ into $x + y + 1 = 0$: $x + 3x + 1 = 0$ $4x = -1 \implies x = -\frac{1}{4}$. Then $y = 3x = 3(-\frac{1}{4}) = -\frac{3}{4}$. The intersection point is $(-\frac{1}{4}, -\frac{3}{4})$. Check if this point lies on the segment $OC$: $-1 < -\frac{1}{4} < 0$ and $-3 < -\frac{3}{4} < 0$. Yes, it does. Since the line $x + y + 1 = 0$ intersects both segments $OB$ and $OC$, this is the required line.

Now let's test the line $x + y - 1 = 0$. Intersection with line $OB$ ($y = x/3$): $x + \frac{x}{3} - 1 = 0$ $\frac{4x}{3} = 1 \implies x = \frac{3}{4}$. Then $y = \frac{1}{3}x = \frac{1}{3}(\frac{3}{4}) = \frac{1}{4}$. The intersection point is $(\frac{3}{4}, \frac{1}{4})$. Check if this point lies on the segment $OB$: Is $\frac{3}{4}$ between $-3$ and $0$? No. Is $\frac{1}{4}$ between $-1$ and $0$? No. This point is outside the segment $OB$. Thus, $x + y - 1 = 0$ is not the required line.

The equation of the line is $x + y + 1 = 0$. The problem states the equation is $ax + by + 2 = 0$. To match the constant term, multiply the equation $x + y + 1 = 0$ by 2: $2(x + y + 1) = 2(0)$ $2x + 2y + 2 = 0$. Comparing this with $ax + by + 2 = 0$, we get $a = 2$ and $b = 2$.

We need to find the value of $\frac{a^4+b^4}{4}$. Substitute $a=2$ and $b=2$: $\frac{a^4+b^4}{4} = \frac{2^4 + 2^4}{4} = \frac{16 + 16}{4} = \frac{32}{4} = 8$.

The final answer is $\boxed{8}$.

Explanation of the solution:

1. Find the equation of line $BC$.
2. A line parallel to $BC$ has the form $x+y+k=0$.
3. Use the given perpendicular distance from the origin to find the possible values of $k$. The distance is $\frac{|k|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$, so $|k|=1$, meaning $k=1$ or $k=-1$.
4. Determine which value of $k$ corresponds to the line intersecting the sides $OB$ and $OC$. This requires checking if the intersection points with lines $OB$ and $OC$ lie on the segments $OB$ and $OC$.
5. The line $x+y+1=0$ intersects segments $OB$ and $OC$. The line $x+y-1=0$ does not.
6. The equation of the line is $x+y+1=0$.
7. Compare this equation with the given form $ax+by+2=0$ by scaling the equation. Multiplying by 2 gives $2x+2y+2=0$.
8. This gives $a=2$ and $b=2$.
9. Calculate the required expression $\frac{a^4+b^4}{4} = \frac{2^4+2^4}{4} = \frac{16+16}{4} = \frac{32}{4} = 8$.

The final answer is $\boxed{8}$.