Show that the lines joining the origin to the point of intersection of the line

Question

Show that the lines joining the origin to the point of intersection of the line $x+y=1$ with the curve $4x^2+4y^2+4x-2y-5=0$ are at right angles to each other.

How do I approach this? Please help.

Answer 1

By the process of homogenisation the pair of lines have the equation

$4(x^2+y^2)+(4x-2y)(x+y)-5(x+y)^2=0$ which simplifies to $3x^2-8xy-3y^2=0$

We know that for the pair of lines $ax^2+2hxy+by^2=0, a+b = 0 \Rightarrow$ they are orthogonal.

Here the same holds, and hence the pair of lines are perpendicular.

Answer 2

Hint:

The equation of any quadratic curve passing through the intersection will be $$4x^2+4y^2-2y+4x+5+k(x+y-1)^2=0$$

Now this has to pass through the origin $\implies5+k=0$

Now rearrange to form a quadratic equation in $x$ or in $y$

and solve.

Answer 3

Solving of the system gives $$4x^2+4(1-x)^2+4x-2(1-x)-5=0$$ or $$8x^2-2x-3=0,$$ which gives the following points: $\left(\frac{3}{4},\frac{1}{4}\right)$ and $\left(-\frac{1}{2},\frac{3}{2}\right)$.

Now, $$m_1=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$$ and $$m_2=\frac{\frac{3}{2}}{-\frac{1}{2}}=-3$$ and since $m_1m_2=-1$, we are done!

Answer 4

Hint:

Replace $x$ with $1-y$ in the equation of the circle to find the two abcissa of the intersection.

Now we know how to find the gradient of a straight line joining two points