# Thread: coordinate geometry of straight lines

1. ## coordinate geometry of straight lines

The point P has coordinates (1,10) and the point Q has coordinates (4,4).

(a) Show the length of PQ is 3 squared 5.

(b) (i) Find the equation of the perpendicular bisector of PQ.
(ii) This perpendicular bisector intersects the x-axis at the point A. Find the coordinates of A.

2. Originally Posted by Sam16
The point P has coordinates (1,10) and the point Q has coordinates (4,4).

(a) Show the length of PQ is 3 squared 5.
$\displaystyle PQ =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\displaystyle PQ =\sqrt{(4-1)^2+(4-10)^2}$

$\displaystyle PQ =\sqrt{(3)^2+(-6)^2}$

$\displaystyle PQ =\sqrt{9+36}$

$\displaystyle PQ =\sqrt{45}=\sqrt{9 \cdot 5}=\sqrt{9} \cdot \sqrt{5}=3\sqrt{5}$

3. Originally Posted by Sam16
The point P has coordinates (1,10) and the point Q has coordinates (4,4).

(a) Show the length of PQ is 3 squared 5.

(b) (i) Find the equation of the perpendicular bisector of PQ.
(ii) This perpendicular bisector intersects the x-axis at the point A. Find the coordinates of A.
to (a): Use the distance formula

$\displaystyle d(PQ)=\sqrt{(x_P-x_Q)^2+(y_P -y_Q)^2}$

to (b)(i)
1. Calculate the slope $\displaystyle m_{PQ}$
2. Calculate the perpendicular direction to this slope
3. Calculate the coordinates of the mid point of PQ: $\displaystyle M_{PQ}\left(\frac{x_P+x_Q}2~,~\frac{y_P+y_Q}2\righ t)$
4. Use the point-slope-formula to get the equation of the perpendicular bisector.

to (b)(ii)
Solve the equation y = 0 using the equation of (b)(i)4

4. Originally Posted by Sam16
The point P has coordinates (1,10) and the point Q has coordinates (4,4).

(b) (i) Find the equation of the perpendicular bisector of PQ.
(ii) This perpendicular bisector intersects the x-axis at the point A. Find the coordinates of A.
Find the midpoint of PQ.

$\displaystyle Midpoint = \left(\frac{1+4}{2} \ \ , \ \ \frac{10+4}{2}\right) = \left(\frac{5}{2} \ \ , \ \ 7\right)$

Find the slop of PQ.

$\displaystyle m=\frac{4-10}{4-1}=-\frac{6}{3}=-2$

The slope of the perpendicular will be the negative reciprocal of that slope which is $\displaystyle \frac{1}{2}$

The equation of the perpendicular will have slope = 1/2 and pass through the midpoint of PQ which is (5/2, 7).

y = mx + b

$\displaystyle 7=\frac{1}{2}\left(\frac{5}{2}\right)+b$

$\displaystyle 7=\frac{5}{4}+b$

$\displaystyle \frac{23}{4}=b$

$\displaystyle y=\frac{1}{2}x+\frac{23}{4}$

The above is the equation of the perpendicular in slope-intercept form.

General form would be $\displaystyle 2x-4y+23=0$

To find the x-intercept, simply assign 0 to y and solve for x. The coordinates will be in the form (x, 0).