# Thread: Point of intersection and line equations...

1. ## Point of intersection and line equations...

I have two questions that I have trouble solving ...

1) P and Q are the points of intersection of the line,

x/a + y/b = 1 (a > 0, b > 0)

with the x- and y-axes respectively. The distance PQ is 20 and the gradient of the PQ is -3. Find the values of a and b.

2) The sides of parallelogram lie along the lines y = 2x - 4, y = 2x - 13, x + y = 5 and x + y = -4. Find the length of one side, and the perpendicular distance between this and the parallel side. Hence find the area of the parallelogram.

Have spent hours but nothing's working

2. For part 2

I would begin by finding the point of intersection between the lines

y= -x+5
y=2x-4

and

y=-x+5
y+2x-13

(solve the simaltaneous equations.)

It will give you two points to help you calculate the length of one of the sides.

Is that enough to get you going?

3. Originally Posted by struck
1) P and Q are the points of intersection of the line,

x/a + y/b = 1 (a > 0, b > 0)

with the x- and y-axes respectively. The distance PQ is 20 and the gradient of the PQ is -3. Find the values of a and b.
What have your personal efforts produced?

If the gradient of PQ is -3, which is the y-intercept, P or Q?

x/a + y/b = 1 -- Do you recognize the Intercept Form? These are P (0,b) and Q (a,0)

The Pythagorean Theorem gives the distance and it is known to be 20.

Gradient (slope) is what? Put the equation in Slope-Intercept form or calculate it from the two known points, P and Q.

4. Hello, struck!

1) P and Q are the points of intersection of the line

. . $\displaystyle \frac{x}{a} + \frac{y}{b} \:=\: 1,\quad a,b > 0$ . with the x- and y-axes respectively.

The distance $\displaystyle PQ$ is 20, and the gradient of $\displaystyle PQ$ is -3.

Find the values of $\displaystyle a\text{ and }b.$
Did you make a sketch ?
Code:
      |
(0,b) * Q
|\
| \
|  \ 20
|   \
|    \ P
- - + - - * - -
|   (a,0)

Distance $\displaystyle PQ = 20.$

. . $\displaystyle \sqrt{(0-a)^2 + (b-0)^2} \:=\:20 \quad\Rightarrow\quad a^2 + b^2 \:=\:400\;\;{\color{blue}[1]}$

Slope of $\displaystyle PQ$ is -3.

. . $\displaystyle m \:=\:\frac{b-0}{0-a} \:=\:-3\quad\Rightarrow\quad b \:=\:3a\;\;{\color{blue}[2]}$

Substitute [2] into [1]: .$\displaystyle a^2 + (3a)^2 \:=\:400 \quad\Rightarrow\quad 10a^2 \:=\:400$

. . $\displaystyle a^2 \:=\:40 \quad\Rightarrow\quad\boxed{ a \:=\:2\sqrt{10}}$

Substitute into [2]: .$\displaystyle b \:=\:3(2\sqrt{10}) \quad\Rightarrow\quad\boxed{ b \:=\:6\sqrt{10}}$