# Thread: GCSE Question

1. ## GCSE Question

[IMG]file:///C:/DOCUME%7E1/SHAKEE%7E1/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG]

Can anybody help me with this question please?

2. Originally Posted by GAdams
[IMG]file:///C:/DOCUME%7E1/SHAKEE%7E1/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG]

Can anybody help me with this question please?
First let's find the point M

The formula for the midpoint is given by:

$\left( \frac {x_1 + x_2}{2}, \frac {y_1 + y_2}{2} \right)$

Using $(x_1,y_1) = A(2,9)$ and $(x_2, y_2) = B(8,7)$, the point M is:

$M = \left( \frac {2 + 8}{2}, \frac {9 + 7}{2} \right) = (5,8)$

Now we can find the slope of the lines connecting M and C, and the slope of the line connecting A and B, if there slopes are the negative inverses of each other (that is, the product of their slopes is -1), then MC is perpendicular to AB.

Let $m_1$ be the slope of the line connecting AB
Let $m_2$ be the slope of the line connecting MC

Now, using $(x_1,y_1) = (2,9) \mbox { and } (x_2,y_2) = (8,7)$

$\Rightarrow m_1 = \frac {y_2 - y_1}{x_2 - x_1} = \frac {7-9}{8-2} = - \frac {1}{3}$

Now, using $(x_1,y_1) = (5,8) \mbox { and } (x_2 , y_2) = (8,18)$

$\Rightarrow m_2 = \frac {18 - 8}{8 - 5} = \frac {10}{3}$

Thus, MC and AB are not perpendicular