Question

Please HURRY!! The equation for QR is 5y=-4x+41. Is QR tangent to circle O at R

Answer 1

The equation of the given tangent is: $5y=-4x+41$. This can be rewritten as: $y=-0.8x+8.2$ (by dividing both sides by 5).

Therefore, the slope of the above equation is -0.8. Let us denote it by $m_1$. Thus, $m_1=-0.8$

Now, the equation of the line from the origin, (0,0) and the point (4,5) can be found as:

$\frac{y-0}{x-0}=\frac{5-0}{4-0} =\frac{5}{4}$

$\therefore \frac{y}{x}=1.25$

or, $y=1.25x$

The above is the equation of the radius of the given circle what passes through the same point (4,5) from where the tangent passes.

The slope of the radius line is 1.25. Let us depict it by $m_2$. Therefore, $m_2=1.25$.

Now, we know that "the tangent lines are always perpendicular to a circle's radius at the point of intersection". And we also know that for perpendicular lines the product of their slopes is always -1.

Thus, in our case the product of $m_1$ and $m_2$ should be -1.

When we multiply $m_1$ and $m_2$ we indeed get -1

($m_1\times m_2=-0.8\times 1.25=-1$)

Therefore, we have concluded that $\overline {QR}$ is tangent to the circle at R because the radius is perpendicular to the tangent at R.

Thus, out of the given options, the last option is the correct option.

Answer 2

Answer:

the correct answer is d

Step-by-step explanation: