# How to answer these question ?

May 25, 2018

See explanation below

#### Explanation:

Given $x + p y = q$ we know that passes trough (1,2), then

$1 + 2 p = q$

By other hand, we know that our line is perpendicular to $2 x - y + 7 = 0$

Lets check the slopes of both lines

Slope of first line: $y = - \frac{x}{p} + \frac{q}{p}$, so slope is $- \frac{1}{p}$

Slope of second line $y = 2 x + 7$, so slope is $2$

We know that two perpendicular lines verifies m·m´=-1

With $m$ and m´ both slopes, Then -1/p·2=-1, then $p = 2$

Finally $1 + 2 p = q$; and $1 + 4 = 5 = q$

Our line is $x + 2 y = 5$

May 25, 2018

$p = 2 , q = 5$.

#### Explanation:

Let, the given lines be ${l}_{1} : x + p y = q \mathmr{and} {l}_{2} : 2 x - y + 7 = 0$.

Let the given point be $P = P \left(1 , 2\right)$.

Rewriting ${l}_{1} : y = - \frac{1}{p} \cdot x + \frac{1}{p} \cdot q \mathmr{and} {l}_{2} : y = 2 x + 7$, we find

that their respective slopes are ${m}_{1} = - \frac{1}{p} \mathmr{and} {m}_{2} = 2$.

Knowing that, ${l}_{1} \bot {l}_{2} , \text{ we must have, } {m}_{1} \cdot {m}_{2} = - 1$.

$\therefore - \frac{1}{p} \cdot 2 = - 1 \Rightarrow p = 2$.

Further, $P \left(1 , 2\right) \in {l}_{1} : x + p y = q \mathmr{and} p = 2$.

$\Rightarrow q = 1 + 2 \left(2\right) = 1 + 4 = 5$.

Enjoy Maths.!

May 25, 2018

$p = 2 , q = 5$

#### Explanation:

In my old age I prefer avoiding explicitly writing the slope. The perpendicular family comes from swapping the coefficients on $x$ and $y$, negating one. We set the constant to choose the family member through $\left(1 , 2\right) .$

So the perpendicular to $2 x - y = - 7$ through $\left(1 , 2\right)$ is

$x + 2 y = 1 \left(1\right) + 2 \left(2\right) = 5$

$p = 2 , q = 5$