Show that $CM$ and $RQ$ are perpendicular ?

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Show that $CM$ and $RQ$ are perpendicular ?

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Answer 1 · 2017-03-26T18:51:02

See below.

Explanation:

Given $R=(3a,a)$ and $Q=(a,-3a)$ the mid point of $RQ$ is given by

$M=1/2(R+Q)=(2a,-a)$

The circle's center is $C=(0,0)$

so $CM=(2a,-a)-(0,0)=(2a,-a)$ and

$RQ = R-Q =(2a,4a)$

Finally

$RQ cdot CM = (2a,4a) cdot (2a,-a) = 4a^2-4a^2=0$

Attached a figure

enter image source here

Answer 2 · 2017-03-26T18:54:49

The x coordinate of the chord goes from $3a$ to $a$ , therefore, the x coordinate of the midpoint is halfway between them, 2a.

The y coordinate of the chord goes from $a$ to $-3a$ , therefore, the y coordinate of the midpoint is halfway between them, -a.

The coordinates of the midpoint are $M(2a,-a)$

The slope, $m_1$ , of the chord RQ is:

$m_1 = (-3a - a)/(a - 3a)$

$m_1 = (-4a)/( -2a)$

$m_1 = 2$

The equation of the circle $x^2 + y^2 = 10a^2$ is a special case of the more general equation $(x - h)^2+(y-k)^2 = r^2$ where the center is the origin $(0,0)$ .

The slope, $m_2$ , of the line from the center to the midpoint is:

$m_2 = (-a - 0)/(2a - 0)$

$m_2 = (-a)/(2a)$

$m_2 = -1/2$

Please observe that:

$m_1m_2 = -1$

This shows that the two lines are perpendicular.

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Show that $CM$ and $RQ$ are perpendicular ?

2 Answers

Explanation:

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Science

Math

Humanities

... and beyond

Show that CMCM and RQRQ are perpendicular ?

2 Answers

Explanation:

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Impact of this question

Show that $CM$ and $RQ$ are perpendicular ?