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Line RS passes through points R (5, 3) and S (-1, 0).

Line PQ is parallel to line RS and passes through points P (3, -1) and Q (-2, y).

What are the coordinates of point Q?

Guest Apr 14, 2015

#2
+20025
+8

Line RS passes through points R (5, 3) and S (-1, 0).

Line PQ is parallel to line RS and passes through points P (3, -1) and Q (-2, y).

What are the coordinates of point Q ?

$$\small{\text{Line PQ is parallel to line RS - cross product }} \boxed{|(\vec{S}-\vec{R}) \times (\vec{Q}-\vec{P})| = 0}\\ \\ \left| \left[ \binom{-1}{0}-\binom{5}{3} \right] \times \left[ \binom{-2}{y}-\binom{3}{-1} \right] \right| = 0\\\\ \left| \binom{-6}{-3} \times \binom{-5}{y+1} \right| = 0\\\\ (-6)\cdot(y+1)-(-3)\cdot(-5) = 0\\ (-6)\cdot(y+1)-15 = 0\\ (-6)\cdot(y+1)=15\\\\ y+1=-\frac{15}{6}\\\\ y=-\frac{15}{6}-1\\\\ y=-\frac{21}{6}\\\\ y= -\frac{7}{2}\\\\ y=-3.5\\\\ \boxed{\vec{Q}=\binom{-2}{-3.5}}$$

heureka  Apr 14, 2015
#1
+981
+5

The lines are parralel. This means they have the same gradient.

First we find the gradient of RS.

The formula for gradient is: m = (y{2}-y{1})/(x{2}-x{1}).

y{2}=0, y{1}=3, x{2}=-1, x{1}=5

$${\mathtt{m}} = {\frac{\left({\mathtt{0}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{5}}\right)}}$$

$${\mathtt{m}} = {\frac{\left(-{\mathtt{3}}\right)}{\left(-{\mathtt{6}}\right)}}$$

$${\mathtt{m}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$

Now we can sub our gradient into the equation for line PQ's gradient.

m = (y{2}-y{1})/(x{2}-x{1})

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\frac{\left({\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}}$$ (keep in mind the y corresponds to point Q)

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\frac{\left({\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\left(-{\mathtt{5}}\right)}}$$ Now we solve for y.

$$-{\mathtt{5}} = {\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)$$

$$-{\mathtt{5}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}$$

$$-{\mathtt{7}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{y}}$$

$${\mathtt{y}} = -{\mathtt{3.5}}$$

We can check this by substituting y into the equation. Does (-3.5 --1)/(-5) = 1/2?

This is our y coordinate for point Q.

So point Q = (-2, -3.5)

Hope this helps :)

zacismyname  Apr 14, 2015
#2
+20025
+8

Line RS passes through points R (5, 3) and S (-1, 0).

Line PQ is parallel to line RS and passes through points P (3, -1) and Q (-2, y).

What are the coordinates of point Q ?

$$\small{\text{Line PQ is parallel to line RS - cross product }} \boxed{|(\vec{S}-\vec{R}) \times (\vec{Q}-\vec{P})| = 0}\\ \\ \left| \left[ \binom{-1}{0}-\binom{5}{3} \right] \times \left[ \binom{-2}{y}-\binom{3}{-1} \right] \right| = 0\\\\ \left| \binom{-6}{-3} \times \binom{-5}{y+1} \right| = 0\\\\ (-6)\cdot(y+1)-(-3)\cdot(-5) = 0\\ (-6)\cdot(y+1)-15 = 0\\ (-6)\cdot(y+1)=15\\\\ y+1=-\frac{15}{6}\\\\ y=-\frac{15}{6}-1\\\\ y=-\frac{21}{6}\\\\ y= -\frac{7}{2}\\\\ y=-3.5\\\\ \boxed{\vec{Q}=\binom{-2}{-3.5}}$$

heureka  Apr 14, 2015