# How to find the equation of the line which passes through the point of intersection of the lines 7x − 3y − 19 = 0 and 3x + 2y + 5 = 0, give that the line is parallel to the line with the equation y = 2x + 1?

Mar 27, 2017

Contd.

#### Explanation:

We will use the following well-known

Result : The eqn. of a line $l$ passing through the pt. of

intersection of the lines ${l}_{1} : {a}_{1} + {b}_{1} + {c}_{1} = 0 \mathmr{and} {l}_{2} : {a}_{2} x + {b}_{2} y + {c}_{2} = 0$ is of

the form

Result :

Mar 27, 2017

Let equation of any straight line passing through the point of intersection of two given straight line be

$k \left(7 x - 3 y - 19\right) + \left(3 x + 2 y + 5\right) = 0$

=>(7k+3)x+(2-3k)y+(5-19k)=0.. [1]

If the straight be parallel to the straight line $2 x - y + 1 = 0$

then

$\frac{7 k + 3}{2} = \frac{2 - 3 k}{- 1}$

$\implies 7 k + 3 = - 4 + 6 k$

$\implies k = - 7$

Inserting the value of k in [1]

$\left(7 \cdot \left(- 7\right) + 3\right) x + \left(2 - 3 \cdot \left(- 7\right)\right) y + \left(5 - 19 \cdot \left(- 7\right)\right) = 0$

$\implies - 46 x + 23 y + 138 = 0$

$\implies 2 x - y - 6 = 0$

This is the equation of the required straight line.

Mar 27, 2017

$y = 2 x - 6$

#### Explanation:

The first step is to find the point of intersection of the 2 lines.

$\text{Using the "color(blue)"elimination method}$

That is we attempt to eliminate the x or y term from the equations leaving us with an equation in 1 variable which we can solve.

Labelling the equations.

$\textcolor{red}{7 x} \textcolor{m a \ge n t a}{- 3 y} - 19 = 0 \to \left(1\right)$

$\textcolor{red}{3 x} \textcolor{m a \ge n t a}{+ 2 y} + 5 = 0 \to \left(2\right)$

$\text{Note:" color(magenta)(-3y)xx2=color(magenta)(-6y)" and } \textcolor{m a \ge n t a}{2 y} \times 3 = \textcolor{m a \ge n t a}{6 y}$

That is the y terms have the same coefficient but with opposing signs. Hence summing them will result in their elimination.

$\left(1\right) \times 2 : 14 x - 6 y - 38 = 0 \to \left(3\right)$

$\left(2\right) \times 3 : 9 x + 6 y + 15 = 0 \to \left(4\right)$

$\left(3\right) + \left(4\right) \text{ term by term}$

$\Rightarrow 23 x + 0 y - 23 = 0 \leftarrow \textcolor{b l u e}{\text{ equation in one variable}}$

$\Rightarrow 23 x = 23 \Rightarrow x = 1 \leftarrow \textcolor{b l u e}{\text{value for x}}$

Substitute this value into either of ( 1 ) or ( 2 ) and solve for y

$\text{Substitute " x=1" in } \left(2\right)$

$\Rightarrow \left(3 \times 1\right) + 2 y + 5 = 0$

$\Rightarrow 8 + 2 y = 0 \Rightarrow 2 y = - 8 \Rightarrow y = - 4 \leftarrow \textcolor{b l u e}{\text{value for y}}$

$\textcolor{b l u e}{\text{As a check}}$

Substitute these values into ( 1 )

$\left(7 \times 1\right) - \left(3 \times - 4\right) - 19 = 7 + 12 - 19 = 0 \to \text{ true}$

$\Rightarrow \left(1 , - 4\right) \textcolor{red}{\text{ is the point of intersection}}$

color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(" parallel lines have equal slopes")color(white)(2/2)|)))

$y = 2 x + 1 \text{ is in " color(blue)"slope-intercept form}$

$\Rightarrow \text{slope } = m = 2$

Expressing the required equation in $\textcolor{b l u e}{\text{ point-slope form}}$

$y - {y}_{1} = m \left(x - {x}_{1}\right) \text{ with " m=2" and } \left({x}_{1} , {y}_{1}\right) = \left(1 , - 4\right)$

$\Rightarrow y + 4 = 2 \left(x - 1\right) \leftarrow \textcolor{red}{\text{ in point-slope form}}$

distribute and simplify.

$y + 4 = 2 x - 2$

$\Rightarrow y = 2 x - 6 \leftarrow \textcolor{red}{\text{ in slope-intercept form}}$